1. (i). If H and K are two sub groups of G such that G is an internal direct product of H and K . Then find $H\cap K$$H\cap K$H nn KH \cap K$H\cap K$.
   (ii). Write class equation for the finite group.
   (iii). Define normal extension of a field.
   (iv). Write Schwartz inequality.

2. Prove that every Euclidean ring is a principal ideal domain.
3. If W is any subspace of a finite dimensional inner product space V , the prove that ${\left(W^{\perp }\right)}^{\perp }=W$${\left(W^{\perp }\right)}^{\perp }=W$(W^(_|_))^(_|_)=W\left(W^{\perp}\right)^{\perp}=W${\left(W^{\perp }\right)}^{\perp }=W$.
4. Prove that every additive abelian group is a module over the ring Z of integers.
5. If $K$$K$KK$K$ is a field and $d_1,d_2,\dots .d_n$$d_1,d_2,\dots .d_n$d_(1),d_(2),dots.d_(n)d_1, d_2, \ldots . d_n$d_1,d_2,\dots .d_n$ are distinct automorphisms of $K$$K$KK$K$. Then prove that it is impossible to find elements $b_1,b_2,\dots .b_n$$b_1,b_2,\dots .b_n$b_(1),b_(2),dots.b_(n)b_1, b_2, \ldots . b_n$b_1,b_2,\dots .b_n$ not all zero in $K$$K$KK$K$ such that $b_1{\varphi }_1(a)+b_2{\varphi }_2(a)+\dots \dots \dots +b_n{\varphi }_n(a)=0\mathrm{\forall }a\in K$$b_1{\varphi }_1(a)+b_2{\varphi }_2(a)+\dots \dots \dots +b_n{\varphi }_n(a)=0\mathrm{\forall }a\in K$b_(1)phi_(1)(a)+b_(2)phi_(2)(a)+dots dots dots+b_(n)phi _(n)(a)=0AA a in Kb_1 \phi_1(a)+b_2 \phi_2(a)+\ldots \ldots \ldots+b_n \phi_n(a)=0 \forall a \in K$b_1{\varphi }_1(a)+b_2{\varphi }_2(a)+\dots \dots \dots +b_n{\varphi }_n(a)=0\mathrm{\forall }a\in K$.

6. Let $M_1$$M_1$M_(1)M_1$M_1$ and $M_2$$M_2$M_(2)M_2$M_2$ are submodule of an R-module M . Then prove that:

7. Let V and $V^{\prime }$$V^{\prime }$V^(‘)V^{\prime}$V^{\prime }$ be any two finite dimensional vector space over the same field F . Then prove that the vector space $\mathrm{Hom}(\mathrm{V},V^{\prime })$$\mathrm{Hom}\left(\mathrm{V},V^{\prime }\right)$Hom(V,V^(‘))\operatorname{Hom}\left(\mathrm{V}, V^{\prime}\right)$\mathrm{Hom}(\mathrm{V},V^{\prime })$ of all linear transformation of V to $V^{\prime }$$V^{\prime }$V^(‘)V^{\prime}$V^{\prime }$ is also finite dimensional and dim $\mathrm{Hom}(\mathrm{V},V^{\prime })=\dim \mathrm{V}\times \dim V^{\prime }$$\mathrm{Hom}\left(\mathrm{V},V^{\prime }\right)=\dim \mathrm{V}\times \dim V^{\prime }$Hom(V,V^(‘))=dimVxx dimV^(‘)\operatorname{Hom}\left(\mathrm{V}, V^{\prime}\right)=\operatorname{dim} \mathrm{V} \times \operatorname{dim} V^{\prime}$\mathrm{Hom}(\mathrm{V},V^{\prime })=\dim \mathrm{V}\times \dim V^{\prime }$.

Answer:

Question:-01(a)

If H and K are two subgroups of G such that G is an internal direct product of H and K. Then find $H\cap K$$H\cap K$H nn KH \cap K$H\cap K$.

Answer:

1. Every element of $G$$G$GG$G$ can be uniquely written as a product $hk$$hk$hkh k$hk$ where $h\in H$$h\in H$h in Hh \in H$h\in H$ and $k\in K$$k\in K$k in Kk \in K$k\in K$.
2. $H\cap K=\{e\}$$H\cap K=\{e\}$H nn K={e}H \cap K = \{ e \}$H\cap K=\{e\}$, where $e$$e$ee$e$ is the identity element of $G$$G$GG$G$.

Question:-01(b)

Write class equation for the finite group.

Answer:

• $|G|$$|G|$|G||G|$|G|$: The order (size) of the group $G$$G$GG$G$,
• $Z(G)$$Z(G)$Z(G)Z(G)$Z(G)$: The center of $G$$G$GG$G$, i.e., the set of elements that commute with every element of $G$$G$GG$G$,
• $|Z(G)|$$|Z(G)|$|Z(G)||Z(G)|$|Z(G)|$: The size (order) of the center $Z(G)$$Z(G)$Z(G)Z(G)$Z(G)$,
• $g_i$$g_i$g_(i)g_i$g_i$: A representative of each conjugacy class outside the center,
• $C_G(g_i)$$C_G(g_i)$C_(G)(g_(i))C_G(g_i)$C_G(g_i)$: The centralizer of $g_i$$g_i$g_(i)g_i$g_i$ in $G$$G$GG$G$, i.e., the set of elements in $G$$G$GG$G$ that commute with $g_i$$g_i$g_(i)g_i$g_i$,
• $[G:C_G(g_i)]$$[G:C_G(g_i)]$[G:C_(G)(g_(i))][G : C_G(g_i)]$[G:C_G(g_i)]$: The index of the centralizer $C_G(g_i)$$C_G(g_i)$C_(G)(g_(i))C_G(g_i)$C_G(g_i)$ in $G$$G$GG$G$, which equals the size of the conjugacy class containing $g_i$$g_i$g_(i)g_i$g_i$.

Interpretation:

1. The elements in $Z(G)$$Z(G)$Z(G)Z(G)$Z(G)$ form singleton conjugacy classes (since they commute with all elements).
2. The remaining conjugacy classes contribute terms of the form $[G:C_G(g_i)]$$[G:C_G(g_i)]$[G:C_(G)(g_(i))][G : C_G(g_i)]$[G:C_G(g_i)]$, representing their sizes.

Question:-01(c)

Define normal extension of a field.

Answer:

1. Every irreducible polynomial $f(x)\in F[x]$$f(x)\in F[x]$f(x)in F[x]f(x) \in F[x]$f(x)\in F[x]$ that has at least one root in $E$$E$EE$E$ splits completely into linear factors over $E$$E$EE$E$, i.e., all the roots of $f(x)$$f(x)$f(x)f(x)$f(x)$ are contained in $E$$E$EE$E$.

Equivalent Definitions:

1. $E/F$$E/F$E//FE/F$E/F$ is normal if $E$$E$EE$E$ is the splitting field of some family of polynomials in $F[x]$$F[x]$F[x]F[x]$F[x]$.
2. $E/F$$E/F$E//FE/F$E/F$ is normal if it is closed under conjugation by automorphisms, i.e., for any field automorphism $\sigma$$\sigma$sigma\sigma$\sigma$ of a larger field containing $E$$E$EE$E$, if $\sigma (e)\in E$$\sigma (e)\in E$sigma(e)in E\sigma(e) \in E$\sigma (e)\in E$ for some $e\in E$$e\in E$e in Ee \in E$e\in E$, then $\sigma (e)$$\sigma (e)$sigma(e)\sigma(e)$\sigma (e)$ is also in $E$$E$EE$E$.

Examples:

1. The extension $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$$\mathbb{Q}(\sqrt{2})/\mathbb{Q}$Q(sqrt2)//Q\mathbb{Q}(\sqrt{2})/\mathbb{Q}$\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ is not normal, because the irreducible polynomial $x^2-2$$x^2-2$x^(2)-2x^2 – 2$x^2-2$ over $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$ has a root $\sqrt{2}$$\sqrt{2}$sqrt2\sqrt{2}$\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})$$\mathbb{Q}(\sqrt{2})$Q(sqrt2)\mathbb{Q}(\sqrt{2})$\mathbb{Q}(\sqrt{2})$, but its other root, $-\sqrt{2}$$-\sqrt{2}$-sqrt2-\sqrt{2}$-\sqrt{2}$, is not in $\mathbb{Q}(\sqrt{2})$$\mathbb{Q}(\sqrt{2})$Q(sqrt2)\mathbb{Q}(\sqrt{2})$\mathbb{Q}(\sqrt{2})$.
2. The extension $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})/\mathbb{Q}$$\mathbb{Q}(\sqrt{2},\sqrt[3]{3})/\mathbb{Q}$Q(sqrt2,root(3)(3))//Q\mathbb{Q}(\sqrt{2}, \sqrt[3]{3})/\mathbb{Q}$\mathbb{Q}(\sqrt{2},\sqrt[3]{3})/\mathbb{Q}$ is normal, because it is the splitting field of $(x^2-2)(x^3-3)$$(x^2-2)(x^3-3)$(x^(2)-2)(x^(3)-3)(x^2 – 2)(x^3 – 3)$(x^2-2)(x^3-3)$ over $\mathbb{Q}$$\mathbb{Q}$Q\mathbb{Q}$\mathbb{Q}$.

Question:-01(d)

Write Schwartz inequality.

Answer:

• $⟨\mathbf{u},\mathbf{v}⟩$$⟨\mathbf{u},\mathbf{v}⟩$(:u,v:)\langle \mathbf{u}, \mathbf{v} \rangle$⟨\mathbf{u},\mathbf{v}⟩$: The inner product of $\mathbf{u}$$\mathbf{u}$u\mathbf{u}$\mathbf{u}$ and $\mathbf{v}$$\mathbf{v}$v\mathbf{v}$\mathbf{v}$,
• $\Vert \mathbf{u}\Vert =\sqrt{⟨\mathbf{u},\mathbf{u}⟩}$$\Vert \mathbf{u}\Vert =\sqrt{⟨\mathbf{u},\mathbf{u}⟩}$||u||=sqrt((:u,u:))\|\mathbf{u}\| = \sqrt{\langle \mathbf{u}, \mathbf{u} \rangle}$\Vert \mathbf{u}\Vert =\sqrt{⟨\mathbf{u},\mathbf{u}⟩}$: The norm (magnitude) of $\mathbf{u}$$\mathbf{u}$u\mathbf{u}$\mathbf{u}$,
• $\Vert \mathbf{v}\Vert =\sqrt{⟨\mathbf{v},\mathbf{v}⟩}$$\Vert \mathbf{v}\Vert =\sqrt{⟨\mathbf{v},\mathbf{v}⟩}$||v||=sqrt((:v,v:))\|\mathbf{v}\| = \sqrt{\langle \mathbf{v}, \mathbf{v} \rangle}$\Vert \mathbf{v}\Vert =\sqrt{⟨\mathbf{v},\mathbf{v}⟩}$: The norm (magnitude) of $\mathbf{v}$$\mathbf{v}$v\mathbf{v}$\mathbf{v}$.

Equality Condition:

Special Cases:

1. In Euclidean space ${\mathbb{R}}^n$${\mathbb{R}}^n$R^(n)\mathbb{R}^n${\mathbb{R}}^n$, with the standard dot product:$$|\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert ,$$$$|\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert ,$$|u*v| <= ||u||||v||,|\mathbf{u} \cdot \mathbf{v}| \leq \|\mathbf{u}\| \|\mathbf{v}\|,$$|\mathbf{u}\cdot \mathbf{v}|\le \Vert \mathbf{u}\Vert \Vert \mathbf{v}\Vert ,$$where $\Vert \mathbf{u}\Vert =\sqrt{\sum _{i=1}^n{u}_i^2}$$\Vert \mathbf{u}\Vert =\sqrt{\sum _{i=1}^n u_i^2}$||u||=sqrt(sum_(i=1)^(n)u_(i)^(2))\|\mathbf{u}\| = \sqrt{\sum_{i=1}^n u_i^2}$\Vert \mathbf{u}\Vert =\sqrt{\sum _{i=1}^n{u}_i^2}$.
2. For functions $f,g$$f,g$f,gf, g$f,g$ in an inner product space $L^2[a,b]$$L^2[a,b]$L^(2)[a,b]L^2[a, b]$L^2[a,b]$:$$|{\int }_a^{b}f(x)g(x)dx|\le \sqrt{{\int }_a^{b}f(x{)}^{2}dx}\cdot \sqrt{{\int }_a^{b}g(x{)}^{2}dx}.$$$$\left|{\int }_a^b f(x)g(x)dx\right|\le \sqrt{{\int }_a^b f(x{)}^{2}dx}\cdot \sqrt{{\int }_a^b g(x{)}^{2}dx}.$$|int_(a)^(b)f(x)g(x)dx| <= sqrt(int_(a)^(b)f(x)^(2)dx)*sqrt(int_(a)^(b)g(x)^(2)dx).\left| \int_a^b f(x)g(x) \, dx \right| \leq \sqrt{\int_a^b f(x)^2 \, dx} \cdot \sqrt{\int_a^b g(x)^2 \, dx}.$$|{\int }_a^{b}f(x)g(x)dx|\le \sqrt{{\int }_a^{b}f(x{)}^{2}dx}\cdot \sqrt{{\int }_a^{b}g(x{)}^{2}dx}.$$

Question:-02

Prove that every Euclidean ring is a principal ideal domain.

Answer:

Statement:

Proof:

Definitions:

1. Euclidean Ring: A commutative ring $R$$R$RR$R$ with unity is a Euclidean ring if there exists a function $\delta :R\setminus \{0\}\to \mathbb{N}$$\delta :R\setminus \{0\}\to \mathbb{N}$delta:R\\{0}rarrN\delta: R \setminus \{0\} \to \mathbb{N}$\delta :R\setminus \{0\}\to \mathbb{N}$ (called a Euclidean norm) such that for all $a,b\in R$$a,b\in R$a,b in Ra, b \in R$a,b\in R$ with $b\ne 0$$b\ne 0$b!=0b \neq 0$b\ne 0$:
   • There exist $q,r\in R$$q,r\in R$q,r in Rq, r \in R$q,r\in R$ such that $a=bq+r$$a=bq+r$a=bq+ra = bq + r$a=bq+r$, where either $r=0$$r=0$r=0r = 0$r=0$ or $\delta (r)<\delta (b)$$\delta (r)<\delta (b)$delta(r) < delta(b)\delta(r) < \delta(b)$\delta (r)<\delta (b)$.
2. Principal Ideal Domain (PID): A commutative ring $R$$R$RR$R$ with unity is a PID if every ideal $I\subseteq R$$I\subseteq R$I sube RI \subseteq R$I\subseteq R$ is a principal ideal, i.e., there exists some $a\in R$$a\in R$a in Ra \in R$a\in R$ such that $I=(a)=\{ra:r\in R\}$$I=(a)=\{ra:r\in R\}$I=(a)={ra:r in R}I = (a) = \{ ra : r \in R \}$I=(a)=\{ra:r\in R\}$.

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Proof:

1. Since $I$$I$II$I$ is nonzero, there exists some $b\in I$$b\in I$b in Ib \in I$b\in I$ with $b\ne 0$$b\ne 0$b!=0b \neq 0$b\ne 0$. Among all nonzero elements of $I$$I$II$I$, choose an element $d\in I$$d\in I$d in Id \in I$d\in I$ such that $\delta (d)$$\delta (d)$delta(d)\delta(d)$\delta (d)$ is minimal (this is possible because $\delta$$\delta$delta\delta$\delta$ maps elements of $R\setminus \{0\}$$R\setminus \{0\}$R\\{0}R \setminus \{0\}$R\setminus \{0\}$ to $\mathbb{N}$$\mathbb{N}$N\mathbb{N}$\mathbb{N}$, which is well-ordered).
2. We claim that $I=(d)$$I=(d)$I=(d)I = (d)$I=(d)$. To prove this, we need to show that every element $a\in I$$a\in I$a in Ia \in I$a\in I$ can be written as $a=rd$$a=rd$a=rda = rd$a=rd$ for some $r\in R$$r\in R$r in Rr \in R$r\in R$.
   • Since $d\in I$$d\in I$d in Id \in I$d\in I$, clearly $(d)\subseteq I$$(d)\subseteq I$(d)sube I(d) \subseteq I$(d)\subseteq I$.
   • Let $a\in I$$a\in I$a in Ia \in I$a\in I$. By the division algorithm in $R$$R$RR$R$, there exist $q,r\in R$$q,r\in R$q,r in Rq, r \in R$q,r\in R$ such that:
     $$a=qd+r,$$$$a=qd+r,$$a=qd+r,a = qd + r,$$a=qd+r,$$
     where either $r=0$$r=0$r=0r = 0$r=0$ or $\delta (r)<\delta (d)$$\delta (r)<\delta (d)$delta(r) < delta(d)\delta(r) < \delta(d)$\delta (r)<\delta (d)$.
   • Since $a,d\in I$$a,d\in I$a,d in Ia, d \in I$a,d\in I$ and $I$$I$II$I$ is an ideal, $qd\in I$$qd\in I$qd in Iqd \in I$qd\in I$. Thus, $r=a-qd\in I$$r=a-qd\in I$r=a-qd in Ir = a – qd \in I$r=a-qd\in I$.
   • If $r\ne 0$$r\ne 0$r!=0r \neq 0$r\ne 0$, then $\delta (r)<\delta (d)$$\delta (r)<\delta (d)$delta(r) < delta(d)\delta(r) < \delta(d)$\delta (r)<\delta (d)$, which contradicts the minimality of $\delta (d)$$\delta (d)$delta(d)\delta(d)$\delta (d)$. Hence, $r=0$$r=0$r=0r = 0$r=0$, and $a=qd$$a=qd$a=qda = qd$a=qd$.
3. Therefore, $a\in (d)$$a\in (d)$a in(d)a \in (d)$a\in (d)$, and we conclude $I\subseteq (d)$$I\subseteq (d)$I sube(d)I \subseteq (d)$I\subseteq (d)$.
4. Combining $(d)\subseteq I$$(d)\subseteq I$(d)sube I(d) \subseteq I$(d)\subseteq I$ and $I\subseteq (d)$$I\subseteq (d)$I sube(d)I \subseteq (d)$I\subseteq (d)$, we get $I=(d)$$I=(d)$I=(d)I = (d)$I=(d)$, showing that $I$$I$II$I$ is a principal ideal.

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Conclusion:

Question:-03

If W is any subspace of a finite dimensional inner product space V, prove that ${\left(W^{\perp }\right)}^{\perp }=W$${\left(W^{\perp }\right)}^{\perp }=W$(W^(_|_))^(_|_)=W\left(W^{\perp}\right)^{\perp}=W${\left(W^{\perp }\right)}^{\perp }=W$.

Answer:

Statement:

Proof:

Definitions:

1. The orthogonal complement $W^{\perp }$$W^{\perp }$W^( _|_)W^\perp$W^{\perp }$ of a subspace $W\subseteq V$$W\subseteq V$W sube VW \subseteq V$W\subseteq V$ is defined as:
   $$W^{\perp }=\{v\in V:⟨v,w⟩=0\text{for all}w\in W\}.$$$$W^{\perp }=\{v\in V:⟨v,w⟩=0\text{for all}w\in W\}.$$W^( _|_)={v in V:(:v,w:)=0″for all “w in W}.W^\perp = \{ v \in V : \langle v, w \rangle = 0 \ \text{for all } w \in W \}.$$W^{\perp }=\{v\in V:⟨v,w⟩=0\text{for all}w\in W\}.$$
2. The space ${\left(W^{\perp }\right)}^{\perp }$${\left(W^{\perp }\right)}^{\perp }$(W^( _|_))^(_|_)\left(W^\perp\right)^\perp${\left(W^{\perp }\right)}^{\perp }$ is the orthogonal complement of $W^{\perp }$$W^{\perp }$W^( _|_)W^\perp$W^{\perp }$, defined as:
   $${\left(W^{\perp }\right)}^{\perp }=\{v\in V:⟨v,u⟩=0\text{for all}u\in W^{\perp }\}.$$$${\left(W^{\perp }\right)}^{\perp }=\{v\in V:⟨v,u⟩=0\text{for all}u\in W^{\perp }\}.$$(W^( _|_))^(_|_)={v in V:(:v,u:)=0″for all “u inW^( _|_)}.\left(W^\perp\right)^\perp = \{ v \in V : \langle v, u \rangle = 0 \ \text{for all } u \in W^\perp \}.$${\left(W^{\perp }\right)}^{\perp }=\{v\in V:⟨v,u⟩=0\text{for all}u\in W^{\perp }\}.$$

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Key Observations:

1. Direct Sum Decomposition: In a finite-dimensional inner product space $V$$V$VV$V$, any subspace $W\subseteq V$$W\subseteq V$W sube VW \subseteq V$W\subseteq V$ satisfies:
   $$V=W\oplus W^{\perp },$$$$V=W\oplus W^{\perp },$$V=W o+W^( _|_),V = W \oplus W^\perp,$$V=W\oplus W^{\perp },$$
   where:
   • $W\cap W^{\perp }=\{0\}$$W\cap W^{\perp }=\{0\}$W nnW^( _|_)={0}W \cap W^\perp = \{0\}$W\cap W^{\perp }=\{0\}$,
   • Every $v\in V$$v\in V$v in Vv \in V$v\in V$ can be uniquely written as $v=w+w^{\perp }$$v=w+w^{\perp }$v=w+w^( _|_)v = w + w^\perp$v=w+w^{\perp }$ for $w\in W$$w\in W$w in Ww \in W$w\in W$ and $w^{\perp }\in W^{\perp }$$w^{\perp }\in W^{\perp }$w^( _|_)inW^( _|_)w^\perp \in W^\perp$w^{\perp }\in W^{\perp }$.
2. Dimension Relation: The dimensions of $W$$W$WW$W$, $W^{\perp }$$W^{\perp }$W^( _|_)W^\perp$W^{\perp }$, and $V$$V$VV$V$ are related as:
   $$\dim (W)+\dim (W^{\perp })=\dim (V).$$$$\dim (W)+\dim (W^{\perp })=\dim (V).$$dim(W)+dim(W^( _|_))=dim(V).\dim(W) + \dim(W^\perp) = \dim(V).$$\dim (W)+\dim (W^{\perp })=\dim (V).$$
3. Orthogonality Property: If $v\in W$$v\in W$v in Wv \in W$v\in W$, then $⟨v,u⟩=0$$⟨v,u⟩=0$(:v,u:)=0\langle v, u \rangle = 0$⟨v,u⟩=0$ for all $u\in W^{\perp }$$u\in W^{\perp }$u inW^( _|_)u \in W^\perp$u\in W^{\perp }$.

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Proof Steps:

1. Inclusion $W\subseteq (W^{\perp }{)}^{\perp }$$W\subseteq (W^{\perp }{)}^{\perp }$W sube(W^( _|_))^(_|_)W \subseteq (W^\perp)^\perp$W\subseteq (W^{\perp }{)}^{\perp }$:
   • Take any $w\in W$$w\in W$w in Ww \in W$w\in W$. For all $u\in W^{\perp }$$u\in W^{\perp }$u inW^( _|_)u \in W^\perp$u\in W^{\perp }$, by the definition of orthogonal complement:$$⟨w,u⟩=0.$$$$⟨w,u⟩=0.$$(:w,u:)=0.\langle w, u \rangle = 0.$$⟨w,u⟩=0.$$
   • Thus, $w\in {\left(W^{\perp }\right)}^{\perp }$$w\in {\left(W^{\perp }\right)}^{\perp }$w in(W^( _|_))^(_|_)w \in \left(W^\perp\right)^\perp$w\in {\left(W^{\perp }\right)}^{\perp }$. Therefore, $W\subseteq {\left(W^{\perp }\right)}^{\perp }$$W\subseteq {\left(W^{\perp }\right)}^{\perp }$W sube(W^( _|_))^(_|_)W \subseteq \left(W^\perp\right)^\perp$W\subseteq {\left(W^{\perp }\right)}^{\perp }$.
2. Inclusion $(W^{\perp }{)}^{\perp }\subseteq W$$(W^{\perp }{)}^{\perp }\subseteq W$(W^( _|_))^(_|_)sube W(W^\perp)^\perp \subseteq W$(W^{\perp }{)}^{\perp }\subseteq W$:
   • Let $v\in {\left(W^{\perp }\right)}^{\perp }$$v\in {\left(W^{\perp }\right)}^{\perp }$v in(W^( _|_))^(_|_)v \in \left(W^\perp\right)^\perp$v\in {\left(W^{\perp }\right)}^{\perp }$. Then, $v$$v$vv$v$ satisfies $⟨v,u⟩=0$$⟨v,u⟩=0$(:v,u:)=0\langle v, u \rangle = 0$⟨v,u⟩=0$ for all $u\in W^{\perp }$$u\in W^{\perp }$u inW^( _|_)u \in W^\perp$u\in W^{\perp }$.
   • By the direct sum decomposition $V=W\oplus W^{\perp }$$V=W\oplus W^{\perp }$V=W o+W^( _|_)V = W \oplus W^\perp$V=W\oplus W^{\perp }$, $v$$v$vv$v$ can be uniquely written as $v=w+w^{\perp }$$v=w+w^{\perp }$v=w+w^( _|_)v = w + w^\perp$v=w+w^{\perp }$, where $w\in W$$w\in W$w in Ww \in W$w\in W$ and $w^{\perp }\in W^{\perp }$$w^{\perp }\in W^{\perp }$w^( _|_)inW^( _|_)w^\perp \in W^\perp$w^{\perp }\in W^{\perp }$.
   • Since $v\in {\left(W^{\perp }\right)}^{\perp }$$v\in {\left(W^{\perp }\right)}^{\perp }$v in(W^( _|_))^(_|_)v \in \left(W^\perp\right)^\perp$v\in {\left(W^{\perp }\right)}^{\perp }$, we have:
     $$0=⟨v,w^{\perp }⟩=⟨w+w^{\perp },w^{\perp }⟩=⟨w^{\perp },w^{\perp }⟩,$$$$0=⟨v,w^{\perp }⟩=⟨w+w^{\perp },w^{\perp }⟩=⟨w^{\perp },w^{\perp }⟩,$$0=(:v,w^( _|_):)=(:w+w^( _|_),w^( _|_):)=(:w^( _|_),w^( _|_):),0 = \langle v, w^\perp \rangle = \langle w + w^\perp, w^\perp \rangle = \langle w^\perp, w^\perp \rangle,$$0=⟨v,w^{\perp }⟩=⟨w+w^{\perp },w^{\perp }⟩=⟨w^{\perp },w^{\perp }⟩,$$
     because $⟨w,w^{\perp }⟩=0$$⟨w,w^{\perp }⟩=0$(:w,w^( _|_):)=0\langle w, w^\perp \rangle = 0$⟨w,w^{\perp }⟩=0$ (orthogonality of $W$$W$WW$W$ and $W^{\perp }$$W^{\perp }$W^( _|_)W^\perp$W^{\perp }$).
   • This implies $\Vert w^{\perp }{\Vert }^2=0$$\Vert w^{\perp }{\Vert }^2=0$||w^( _|_)||^(2)=0\|w^\perp\|^2 = 0$\Vert w^{\perp }{\Vert }^2=0$, so $w^{\perp }=0$$w^{\perp }=0$w^( _|_)=0w^\perp = 0$w^{\perp }=0$. Hence, $v=w\in W$$v=w\in W$v=w in Wv = w \in W$v=w\in W$.
   • Therefore, $(W^{\perp }{)}^{\perp }\subseteq W$$(W^{\perp }{)}^{\perp }\subseteq W$(W^( _|_))^(_|_)sube W(W^\perp)^\perp \subseteq W$(W^{\perp }{)}^{\perp }\subseteq W$.
3. Conclusion:
   Combining $W\subseteq (W^{\perp }{)}^{\perp }$$W\subseteq (W^{\perp }{)}^{\perp }$W sube(W^( _|_))^(_|_)W \subseteq (W^\perp)^\perp$W\subseteq (W^{\perp }{)}^{\perp }$ and $(W^{\perp }{)}^{\perp }\subseteq W$$(W^{\perp }{)}^{\perp }\subseteq W$(W^( _|_))^(_|_)sube W(W^\perp)^\perp \subseteq W$(W^{\perp }{)}^{\perp }\subseteq W$, we have:
   $$(W^{\perp }{)}^{\perp }=W.$$$$(W^{\perp }{)}^{\perp }=W.$$(W^( _|_))^(_|_)=W.(W^\perp)^\perp = W.$$(W^{\perp }{)}^{\perp }=W.$$

Question:-04

Prove that every additive abelian group is a module over the ring $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$ of integers.

Answer:

Statement:

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Proof:

Definitions:

1. Additive Abelian Group: A set $G$$G$GG$G$ with a binary operation $+$$+$++$+$ is called an additive abelian group if:
   • $G$$G$GG$G$ is closed under addition,
   • Addition is associative: $(a+b)+c=a+(b+c)$$(a+b)+c=a+(b+c)$(a+b)+c=a+(b+c)(a + b) + c = a + (b + c)$(a+b)+c=a+(b+c)$ for all $a,b,c\in G$$a,b,c\in G$a,b,c in Ga, b, c \in G$a,b,c\in G$,
   • There exists an identity element $0\in G$$0\in G$0in G0 \in G$0\in G$ such that $a+0=a$$a+0=a$a+0=aa + 0 = a$a+0=a$ for all $a\in G$$a\in G$a in Ga \in G$a\in G$,
   • Every element $a\in G$$a\in G$a in Ga \in G$a\in G$ has an additive inverse $-a\in G$$-a\in G$-a in G-a \in G$-a\in G$ such that $a+(-a)=0$$a+(-a)=0$a+(-a)=0a + (-a) = 0$a+(-a)=0$,
   • Addition is commutative: $a+b=b+a$$a+b=b+a$a+b=b+aa + b = b + a$a+b=b+a$ for all $a,b\in G$$a,b\in G$a,b in Ga, b \in G$a,b\in G$.
2. Module over $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$: A module over the ring $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$ is a set $G$$G$GG$G$ with:
   • An additive abelian group structure,
   • A scalar multiplication $\cdot :\mathbb{Z}\times G\to G$$\cdot :\mathbb{Z}\times G\to G$*:Zxx G rarr G\cdot : \mathbb{Z} \times G \to G$\cdot :\mathbb{Z}\times G\to G$ such that:
     1. $(m+n)\cdot g=m\cdot g+n\cdot g$$(m+n)\cdot g=m\cdot g+n\cdot g$(m+n)*g=m*g+n*g(m + n) \cdot g = m \cdot g + n \cdot g$(m+n)\cdot g=m\cdot g+n\cdot g$ for all $m,n\in \mathbb{Z}$$m,n\in \mathbb{Z}$m,n inZm, n \in \mathbb{Z}$m,n\in \mathbb{Z}$, $g\in G$$g\in G$g in Gg \in G$g\in G$,
     2. $m\cdot (g+h)=m\cdot g+m\cdot h$$m\cdot (g+h)=m\cdot g+m\cdot h$m*(g+h)=m*g+m*hm \cdot (g + h) = m \cdot g + m \cdot h$m\cdot (g+h)=m\cdot g+m\cdot h$ for all $m\in \mathbb{Z}$$m\in \mathbb{Z}$m inZm \in \mathbb{Z}$m\in \mathbb{Z}$, $g,h\in G$$g,h\in G$g,h in Gg, h \in G$g,h\in G$,
     3. $(m\cdot n)\cdot g=m\cdot (n\cdot g)$$(m\cdot n)\cdot g=m\cdot (n\cdot g)$(m*n)*g=m*(n*g)(m \cdot n) \cdot g = m \cdot (n \cdot g)$(m\cdot n)\cdot g=m\cdot (n\cdot g)$ for all $m,n\in \mathbb{Z}$$m,n\in \mathbb{Z}$m,n inZm, n \in \mathbb{Z}$m,n\in \mathbb{Z}$, $g\in G$$g\in G$g in Gg \in G$g\in G$,
     4. $1\cdot g=g$$1\cdot g=g$1*g=g1 \cdot g = g$1\cdot g=g$ for all $g\in G$$g\in G$g in Gg \in G$g\in G$.

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Proof Steps:

1. Define Scalar Multiplication:
   Define scalar multiplication $\cdot :\mathbb{Z}\times G\to G$$\cdot :\mathbb{Z}\times G\to G$*:Zxx G rarr G\cdot : \mathbb{Z} \times G \to G$\cdot :\mathbb{Z}\times G\to G$ as:
   $$m\cdot g=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}\text{if}m>0,$$$$m\cdot g=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}\text{if}m>0,$$m*g=ubrace(g+g+cdots+gubrace)_(m” times”)quad”if “m > 0,m \cdot g = \underbrace{g + g + \dots + g}_{m \text{ times}} \quad \text{if } m > 0,$$m\cdot g=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}\text{if}m>0,$$
   $$m\cdot g=0\text{if}m=0,$$$$m\cdot g=0\text{if}m=0,$$m*g=0quad”if “m=0,m \cdot g = 0 \quad \text{if } m = 0,$$m\cdot g=0\text{if}m=0,$$
   $$m\cdot g=-(\underset{|m|\text{times}}{\underbrace{g+g+\cdots +g}})\text{if}m<0.$$$$m\cdot g=-(\underset{|m|\text{times}}{\underbrace{g+g+\cdots +g}})\text{if}m<0.$$m*g=-(ubrace(g+g+cdots+gubrace)_(|m|” times”))quad”if “m < 0.m \cdot g = -(\underbrace{g + g + \dots + g}_{|m| \text{ times}}) \quad \text{if } m < 0.$$m\cdot g=-(\underset{|m|\text{times}}{\underbrace{g+g+\cdots +g}})\text{if}m<0.$$
   This scalar multiplication is well-defined because $G$$G$GG$G$ is an additive abelian group, and we can repeatedly add or subtract elements.
2. Check Module Axioms:
   We need to verify that the scalar multiplication satisfies the module axioms.
   • (1) Distributivity over $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$-addition:
     $$(m+n)\cdot g=\underset{(m+n)\text{times}}{\underbrace{g+g+\cdots +g}}=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}+\underset{n\text{times}}{\underbrace{g+g+\cdots +g}}=m\cdot g+n\cdot g.$$$$(m+n)\cdot g=\underset{(m+n)\text{times}}{\underbrace{g+g+\cdots +g}}=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}+\underset{n\text{times}}{\underbrace{g+g+\cdots +g}}=m\cdot g+n\cdot g.$$(m+n)*g=ubrace(g+g+cdots+gubrace)_((m+n)” times”)=ubrace(g+g+cdots+gubrace)_(m” times”)+ubrace(g+g+cdots+gubrace)_(n” times”)=m*g+n*g.(m + n) \cdot g = \underbrace{g + g + \dots + g}_{(m + n) \text{ times}} = \underbrace{g + g + \dots + g}_{m \text{ times}} + \underbrace{g + g + \dots + g}_{n \text{ times}} = m \cdot g + n \cdot g.$$(m+n)\cdot g=\underset{(m+n)\text{times}}{\underbrace{g+g+\cdots +g}}=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}+\underset{n\text{times}}{\underbrace{g+g+\cdots +g}}=m\cdot g+n\cdot g.$$
   • (2) Distributivity over group addition:
     $$m\cdot (g+h)=\underset{m\text{times}}{\underbrace{(g+h)+(g+h)+\cdots +(g+h)}}=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}+\underset{m\text{times}}{\underbrace{h+h+\cdots +h}}=m\cdot g+m\cdot h.$$$$m\cdot (g+h)=\underset{m\text{times}}{\underbrace{(g+h)+(g+h)+\cdots +(g+h)}}=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}+\underset{m\text{times}}{\underbrace{h+h+\cdots +h}}=m\cdot g+m\cdot h.$$m*(g+h)=ubrace((g+h)+(g+h)+cdots+(g+h)ubrace)_(m” times”)=ubrace(g+g+cdots+gubrace)_(m” times”)+ubrace(h+h+cdots+hubrace)_(m” times”)=m*g+m*h.m \cdot (g + h) = \underbrace{(g + h) + (g + h) + \dots + (g + h)}_{m \text{ times}} = \underbrace{g + g + \dots + g}_{m \text{ times}} + \underbrace{h + h + \dots + h}_{m \text{ times}} = m \cdot g + m \cdot h.$$m\cdot (g+h)=\underset{m\text{times}}{\underbrace{(g+h)+(g+h)+\cdots +(g+h)}}=\underset{m\text{times}}{\underbrace{g+g+\cdots +g}}+\underset{m\text{times}}{\underbrace{h+h+\cdots +h}}=m\cdot g+m\cdot h.$$
   • (3) Compatibility with multiplication in $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$:
     $$(m\cdot n)\cdot g=\underset{(m\cdot n)\text{times}}{\underbrace{g+g+\cdots +g}}=m\cdot (\underset{n\text{times}}{\underbrace{g+g+\cdots +g}})=m\cdot (n\cdot g).$$$$(m\cdot n)\cdot g=\underset{(m\cdot n)\text{times}}{\underbrace{g+g+\cdots +g}}=m\cdot (\underset{n\text{times}}{\underbrace{g+g+\cdots +g}})=m\cdot (n\cdot g).$$(m*n)*g=ubrace(g+g+cdots+gubrace)_((m*n)” times”)=m*(ubrace(g+g+cdots+gubrace)_(n” times”))=m*(n*g).(m \cdot n) \cdot g = \underbrace{g + g + \dots + g}_{(m \cdot n) \text{ times}} = m \cdot (\underbrace{g + g + \dots + g}_{n \text{ times}}) = m \cdot (n \cdot g).$$(m\cdot n)\cdot g=\underset{(m\cdot n)\text{times}}{\underbrace{g+g+\cdots +g}}=m\cdot (\underset{n\text{times}}{\underbrace{g+g+\cdots +g}})=m\cdot (n\cdot g).$$
   • (4) Action of 1:
     $$1\cdot g=\underset{1\text{time}}{\underbrace{g}}=g.$$$$1\cdot g=\underset{1\text{time}}{\underbrace{g}}=g.$$1*g=ubrace(gubrace)_(1″ time”)=g.1 \cdot g = \underbrace{g}_{1 \text{ time}} = g.$$1\cdot g=\underset{1\text{time}}{\underbrace{g}}=g.$$
3. Conclusion:
   Since $G$$G$GG$G$ satisfies the axioms of a module over $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$, every additive abelian group is indeed a module over $\mathbb{Z}$$\mathbb{Z}$Z\mathbb{Z}$\mathbb{Z}$.

Question:-05

If $K$$K$KK$K$ is a field and $d_1,d_2,\dots ,d_n$$d_1,d_2,\dots ,d_n$d_(1),d_(2),dots,d_(n)d_1, d_2, \ldots, d_n$d_1,d_2,\dots ,d_n$ are distinct automorphisms of $K$$K$KK$K$. Then prove that it is impossible to find elements $b_1,b_2,\dots ,b_n$$b_1,b_2,\dots ,b_n$b_(1),b_(2),dots,b_(n)b_1, b_2, \ldots, b_n$b_1,b_2,\dots ,b_n$ not all zero in $K$$K$KK$K$ such that

Answer:

Statement:

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Proof:

1. Assume the contrary:

2. Linearity of $L$$L$LL$L$:

3. Evaluating $L(a)=0$$L(a)=0$L(a)=0L(a) = 0$L(a)=0$ for all $a\in K$$a\in K$a in Ka \in K$a\in K$:

4. Distinct automorphisms imply independence:

5. Conclusion:

Question:-06

Let $M_1$$M_1$M_(1)M_1$M_1$ and $M_2$$M_2$M_(2)M_2$M_2$ be submodules of an $R$$R$RR$R$-module $M$$M$MM$M$. Then prove that:

Answer:

Statement:

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Detailed Proof:

Step 1: Define a Map

• $\phi (m_1)$$\phi (m_1)$varphi(m_(1))\varphi(m_1)$\phi (m_1)$ represents the coset $m_1+M_2$$m_1+M_2$m_(1)+M_(2)m_1 + M_2$m_1+M_2$, which is a member of the quotient module $\frac{M_1+M_2}{M_2}$$\frac{M_1+M_2}{M_2}$(M_(1)+M_(2))/(M_(2))\frac{M_1 + M_2}{M_2}$\frac{M_1+M_2}{M_2}$.

Step 2: Verify $\phi$$\phi$varphi\varphi$\phi$ is Well-Defined

Step 3: Check Linearity of $\phi$$\phi$varphi\varphi$\phi$

1. Additivity:$$\phi (m_1+m_1^{\prime })=(m_1+m_1^{\prime })+M_2=(m_1+M_2)+(m_1^{\prime }+M_2)=\phi (m_1)+\phi (m_1^{\prime }).$$$$\phi (m_1+m_1^{\prime })=(m_1+m_1^{\prime })+M_2=(m_1+M_2)+(m_1^{\prime }+M_2)=\phi (m_1)+\phi (m_1^{\prime }).$$varphi(m_(1)+m_(1)^(‘))=(m_(1)+m_(1)^(‘))+M_(2)=(m_(1)+M_(2))+(m_(1)^(‘)+M_(2))=varphi(m_(1))+varphi(m_(1)^(‘)).\varphi(m_1 + m_1′) = (m_1 + m_1′) + M_2 = (m_1 + M_2) + (m_1′ + M_2) = \varphi(m_1) + \varphi(m_1′).$$\phi (m_1+m_1^{\prime })=(m_1+m_1^{\prime })+M_2=(m_1+M_2)+(m_1^{\prime }+M_2)=\phi (m_1)+\phi (m_1^{\prime }).$$
2. Scalar Multiplication:$$\phi (r{m}_1)=(r{m}_1)+M_2=r(m_1+M_2)=r\phi (m_1).$$$$\phi (r{m}_1)=(r{m}_1)+M_2=r(m_1+M_2)=r\phi (m_1).$$varphi(rm_(1))=(rm_(1))+M_(2)=r(m_(1)+M_(2))=r varphi(m_(1)).\varphi(rm_1) = (rm_1) + M_2 = r(m_1 + M_2) = r\varphi(m_1).$$\phi (r{m}_1)=(r{m}_1)+M_2=r(m_1+M_2)=r\phi (m_1).$$

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Step 4: Kernel of $\phi$$\phi$varphi\varphi$\phi$

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Step 5: Image of $\phi$$\phi$varphi\varphi$\phi$

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Step 6: Apply the First Isomorphism Theorem

• $\ker (\phi )=M_1\cap M_2$$\ker (\phi )=M_1\cap M_2$ker(varphi)=M_(1)nnM_(2)\ker(\varphi) = M_1 \cap M_2$\ker (\phi )=M_1\cap M_2$,
• $\text{im}(\phi )=\frac{M_1+M_2}{M_2}$$\text{im}(\phi )=\frac{M_1+M_2}{M_2}$“im”(varphi)=(M_(1)+M_(2))/(M_(2))\text{im}(\varphi) = \frac{M_1 + M_2}{M_2}$\text{im}(\phi )=\frac{M_1+M_2}{M_2}$,

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Conclusion:

Question:-07

Let $V$$V$VV$V$ and $V^{\prime }$$V^{\prime }$V^(‘)V^{\prime}$V^{\prime }$ be any two finite dimensional vector spaces over the same field $F$$F$FF$F$. Then prove that the vector space $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V^{\prime})$\mathrm{Hom}(V,V^{\prime })$ of all linear transformations from $V$$V$VV$V$ to $V^{\prime }$$V^{\prime }$V^(‘)V^{\prime}$V^{\prime }$ is also finite dimensional and

Answer:

Statement:

1. The vector space $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V’)$\mathrm{Hom}(V,V^{\prime })$, consisting of all linear transformations from $V$$V$VV$V$ to $V^{\prime }$$V^{\prime }$V^(‘)V’$V^{\prime }$, is finite-dimensional.
2. Its dimension is given by:$$\dim \mathrm{Hom}(V,V^{\prime })=\dim V\times \dim V^{\prime }=n\times m.$$$$\dim \mathrm{Hom}(V,V^{\prime })=\dim V\times \dim V^{\prime }=n\times m.$$dim Hom(V,V^(‘))=dim V xx dim V^(‘)=n xx m.\dim \operatorname{Hom}(V, V’) = \dim V \times \dim V’ = n \times m.$$\dim \mathrm{Hom}(V,V^{\prime })=\dim V\times \dim V^{\prime }=n\times m.$$

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Detailed Proof:

Step 1: Define $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V’)$\mathrm{Hom}(V,V^{\prime })$:

• The set $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V’)$\mathrm{Hom}(V,V^{\prime })$ is the collection of all linear transformations $T:V\to V^{\prime }$$T:V\to V^{\prime }$T:V rarrV^(‘)T: V \to V’$T:V\to V^{\prime }$.
• For any $T\in \mathrm{Hom}(V,V^{\prime })$$T\in \mathrm{Hom}(V,V^{\prime })$T in Hom(V,V^(‘))T \in \operatorname{Hom}(V, V’)$T\in \mathrm{Hom}(V,V^{\prime })$, the action of $T$$T$TT$T$ is completely determined by how it maps basis elements of $V$$V$VV$V$. This is because $T$$T$TT$T$ is linear, and any vector $v\in V$$v\in V$v in Vv \in V$v\in V$ can be written as a linear combination of basis elements.

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Step 2: Basis of $V$$V$VV$V$ and $V^{\prime }$$V^{\prime }$V^(‘)V’$V^{\prime }$:

• Let $\{v_1,v_2,\dots ,v_n\}$$\{v_1,v_2,\dots ,v_n\}${v_(1),v_(2),dots,v_(n)}\{v_1, v_2, \ldots, v_n\}$\{v_1,v_2,\dots ,v_n\}$ be a basis of $V$$V$VV$V$, so every $v\in V$$v\in V$v in Vv \in V$v\in V$ can be written uniquely as:
  $$v=\sum _{i=1}^n{c}_i{v}_i,\text{where}{c}_i\in F.$$$$v=\sum _{i=1}^n c_i{v}_i,\text{where}{c}_i\in F.$$v=sum_(i=1)^(n)c_(i)v_(i),quad”where “c_(i)in F.v = \sum_{i=1}^n c_i v_i, \quad \text{where } c_i \in F.$$v=\sum _{i=1}^n{c}_i{v}_i,\text{where}{c}_i\in F.$$
• Let $\{w_1,w_2,\dots ,w_m\}$$\{w_1,w_2,\dots ,w_m\}${w_(1),w_(2),dots,w_(m)}\{w_1, w_2, \ldots, w_m\}$\{w_1,w_2,\dots ,w_m\}$ be a basis of $V^{\prime }$$V^{\prime }$V^(‘)V’$V^{\prime }$, so every $w\in V^{\prime }$$w\in V^{\prime }$w inV^(‘)w \in V’$w\in V^{\prime }$ can be written uniquely as:
  $$w=\sum _{j=1}^m{d}_j{w}_j,\text{where}{d}_j\in F.$$$$w=\sum _{j=1}^m d_j{w}_j,\text{where}{d}_j\in F.$$w=sum_(j=1)^(m)d_(j)w_(j),quad”where “d_(j)in F.w = \sum_{j=1}^m d_j w_j, \quad \text{where } d_j \in F.$$w=\sum _{j=1}^m{d}_j{w}_j,\text{where}{d}_j\in F.$$

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Step 3: Represent $T\in \mathrm{Hom}(V,V^{\prime })$$T\in \mathrm{Hom}(V,V^{\prime })$T in Hom(V,V^(‘))T \in \operatorname{Hom}(V, V’)$T\in \mathrm{Hom}(V,V^{\prime })$:

• Since $T$$T$TT$T$ is linear, for any $i=1,2,\dots ,n$$i=1,2,\dots ,n$i=1,2,dots,ni = 1, 2, \ldots, n$i=1,2,\dots ,n$, $T(v_i)$$T(v_i)$T(v_(i))T(v_i)$T(v_i)$ must lie in $V^{\prime }$$V^{\prime }$V^(‘)V’$V^{\prime }$.
• Write $T(v_i)$$T(v_i)$T(v_(i))T(v_i)$T(v_i)$ as a linear combination of the basis elements of $V^{\prime }$$V^{\prime }$V^(‘)V’$V^{\prime }$:
  $$T(v_i)=\sum _{j=1}^m{a}_{ij}{w}_j,\text{where}{a}_{ij}\in F.$$$$T(v_i)=\sum _{j=1}^m a_{ij}{w}_j,\text{where}{a}_{ij}\in F.$$T(v_(i))=sum_(j=1)^(m)a_(ij)w_(j),quad”where “a_(ij)in F.T(v_i) = \sum_{j=1}^m a_{ij} w_j, \quad \text{where } a_{ij} \in F.$$T(v_i)=\sum _{j=1}^m{a}_{ij}{w}_j,\text{where}{a}_{ij}\in F.$$
• The coefficients $\{a_{ij}{\}}_{1\le i\le n,1\le j\le m}$$\{a_{ij}{\}}_{1\le i\le n,1\le j\le m}${a_(ij)}_(1 <= i <= n,1 <= j <= m)\{a_{ij} \}_{1 \leq i \leq n, 1 \leq j \leq m}$\{a_{ij}{\}}_{1\le i\le n,1\le j\le m}$ completely determine the linear transformation $T$$T$TT$T$.

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Step 4: Matrix Representation of $T$$T$TT$T$:

• The map $T$$T$TT$T$ can be represented as a matrix $A$$A$AA$A$ of size $m\times n$$m\times n$m xx nm \times n$m\times n$, where the $(j,i)$$(j,i)$(j,i)(j, i)$(j,i)$-th entry of $A$$A$AA$A$ is $a_{ij}$$a_{ij}$a_(ij)a_{ij}$a_{ij}$.
  $$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{m1}& a_{m2}& \cdots & a_{mn}\end{array}\right].$$$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{m1}& a_{m2}& \cdots & a_{mn}\end{array}\right].$$A=[[a_(11),a_(12),cdots,a_(1n)],[a_(21),a_(22),cdots,a_(2n)],[vdots,vdots,ddots,vdots],[a_(m1),a_(m2),cdots,a_(mn)]].A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}.$$A=\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}& \cdots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{m1}& a_{m2}& \cdots & a_{mn}\end{array}\right].$$
• Each column of $A$$A$AA$A$ corresponds to the image of a basis element $v_i$$v_i$v_(i)v_i$v_i$ of $V$$V$VV$V$ under $T$$T$TT$T$.

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Step 5: Vector Space Structure of $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V’)$\mathrm{Hom}(V,V^{\prime })$:

• The set $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V’)$\mathrm{Hom}(V,V^{\prime })$ has a natural structure as a vector space:
  1. Addition of linear transformations: For $T_1,T_2\in \mathrm{Hom}(V,V^{\prime })$$T_1,T_2\in \mathrm{Hom}(V,V^{\prime })$T_(1),T_(2)in Hom(V,V^(‘))T_1, T_2 \in \operatorname{Hom}(V, V’)$T_1,T_2\in \mathrm{Hom}(V,V^{\prime })$ and $v\in V$$v\in V$v in Vv \in V$v\in V$:$$(T_1+T_2)(v)=T_1(v)+T_2(v).$$$$(T_1+T_2)(v)=T_1(v)+T_2(v).$$(T_(1)+T_(2))(v)=T_(1)(v)+T_(2)(v).(T_1 + T_2)(v) = T_1(v) + T_2(v).$$(T_1+T_2)(v)=T_1(v)+T_2(v).$$
  2. Scalar multiplication: For $c\in F$$c\in F$c in Fc \in F$c\in F$ and $T\in \mathrm{Hom}(V,V^{\prime })$$T\in \mathrm{Hom}(V,V^{\prime })$T in Hom(V,V^(‘))T \in \operatorname{Hom}(V, V’)$T\in \mathrm{Hom}(V,V^{\prime })$:$$(cT)(v)=c\cdot T(v).$$$$(cT)(v)=c\cdot T(v).$$(cT)(v)=c*T(v).(cT)(v) = c \cdot T(v).$$(cT)(v)=c\cdot T(v).$$
• The space $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V’)$\mathrm{Hom}(V,V^{\prime })$ is isomorphic to the space of $m\times n$$m\times n$m xx nm \times n$m\times n$ matrices over $F$$F$FF$F$.

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Step 6: Dimension of $\mathrm{Hom}(V,V^{\prime })$$\mathrm{Hom}(V,V^{\prime })$Hom(V,V^(‘))\operatorname{Hom}(V, V’)$\mathrm{Hom}(V,V^{\prime })$:

• The space of $m\times n$$m\times n$m xx nm \times n$m\times n$ matrices over $F$$F$FF$F$ has dimension $m\cdot n$$m\cdot n$m*nm \cdot n$m\cdot n$, as there are $m\times n$$m\times n$m xx nm \times n$m\times n$ independent entries in the matrix.
• Each entry of the matrix corresponds to a coefficient $a_{ij}\in F$$a_{ij}\in F$a_(ij)in Fa_{ij} \in F$a_{ij}\in F$, and choosing these coefficients independently determines a unique linear transformation $T$$T$TT$T$.
• Thus:
  $$\dim \mathrm{Hom}(V,V^{\prime })=m\cdot n=\dim V^{\prime }\cdot \dim V.$$$$\dim \mathrm{Hom}(V,V^{\prime })=m\cdot n=\dim V^{\prime }\cdot \dim V.$$dim Hom(V,V^(‘))=m*n=dim V^(‘)*dim V.\dim \operatorname{Hom}(V, V’) = m \cdot n = \dim V’ \cdot \dim V.$$\dim \mathrm{Hom}(V,V^{\prime })=m\cdot n=\dim V^{\prime }\cdot \dim V.$$

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Conclusion: