Sample Solution

MST-022 Solved Assignment

1. (a) Using Gram Schmidt orthogonalisation process find orthogonal basis for the subspace of ${\mathbb{R}}^4$${\mathbb{R}}^4$R^(4)\mathbb{R}^4${\mathbb{R}}^4$ having basis ${\mathrm{w}}_1=\left[\begin{array}{l}1\\ 0\\ 1\\ 0\end{array}\right],{\mathrm{w}}_2=\left[\begin{array}{l}2\\ 0\\ 0\\ 2\end{array}\right]$${\mathrm{w}}_1=\left[\begin{array}{l}1\\ 0\\ 1\\ 0\end{array}\right],{\mathrm{w}}_2=\left[\begin{array}{l}2\\ 0\\ 0\\ 2\end{array}\right]$w_(1)=[[1],[0],[1],[0]],w_(2)=[[2],[0],[0],[2]]\mathrm{w}_1=\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 0\end{array}\right], \mathrm{w}_2=\left[\begin{array}{l}2 \\ 0 \\ 0 \\ 2\end{array}\right]${\mathrm{w}}_1=\left[\begin{array}{l}1\\ 0\\ 1\\ 0\end{array}\right],{\mathrm{w}}_2=\left[\begin{array}{l}2\\ 0\\ 0\\ 2\end{array}\right]$, and ${\mathrm{w}}_3=\left[\begin{array}{l}0\\ 0\\ 4\\ 4\end{array}\right]$${\mathrm{w}}_3=\left[\begin{array}{l}0\\ 0\\ 4\\ 4\end{array}\right]$w_(3)=[[0],[0],[4],[4]]\mathrm{w}_3=\left[\begin{array}{l}0 \\ 0 \\ 4 \\ 4\end{array}\right]${\mathrm{w}}_3=\left[\begin{array}{l}0\\ 0\\ 4\\ 4\end{array}\right]$.
   (b) Gradient of the function $f(x,y)=2x^2-\frac{1}{3}{y}^2$$f(x,y)=2x^2-\frac{1}{3}{y}^2$f(x,y)=2x^(2)-(1)/(3)y^(2)f(x, y)=2 x^2-\frac{1}{3} y^2$f(x,y)=2x^2-\frac{1}{3}{y}^2$ at the point $P(1,2)$$P(1,2)$P(1,2)P(1,2)$P(1,2)$ is:
   (A) $4\mathrm{i}-\frac{4}{3}\hat{\mathrm{j}}$$4\mathrm{i}-\frac{4}{3}\widehat{\mathrm{j}}$4i-(4)/(3) hat(j)4 \mathrm{i}-\frac{4}{3} \hat{\mathrm{j}}$4\mathrm{i}-\frac{4}{3}\hat{\mathrm{j}}$
   (B) $4\mathrm{i}+\frac{4}{3}\hat{\mathrm{j}}$$4\mathrm{i}+\frac{4}{3}\widehat{\mathrm{j}}$4i+(4)/(3) hat(j)4 \mathrm{i}+\frac{4}{3} \hat{\mathrm{j}}$4\mathrm{i}+\frac{4}{3}\hat{\mathrm{j}}$
   (C) $4\mathrm{i}-\frac{3}{4}\hat{\mathrm{j}}$$4\mathrm{i}-\frac{3}{4}\widehat{\mathrm{j}}$4i-(3)/(4) hat(j)4 \mathrm{i}-\frac{3}{4} \hat{\mathrm{j}}$4\mathrm{i}-\frac{3}{4}\hat{\mathrm{j}}$
   (D) $4\mathrm{i}+\frac{3}{4}\hat{\mathrm{j}}$$4\mathrm{i}+\frac{3}{4}\widehat{\mathrm{j}}$4i+(3)/(4) hat(j)4 \mathrm{i}+\frac{3}{4} \hat{\mathrm{j}}$4\mathrm{i}+\frac{3}{4}\hat{\mathrm{j}}$
   (c) Using Taylor’s formula for $\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{x}{\mathrm{e}}^{\mathrm{y}}$$\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{x}{\mathrm{e}}^{\mathrm{y}}$f(x,y)=xe^(y)\mathrm{f}(\mathrm{x}, \mathrm{y})=\mathrm{x} \mathrm{e}^{\mathrm{y}}$\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{x}{\mathrm{e}}^{\mathrm{y}}$ the quadratic approximation of $\mathrm{f}(\mathrm{x},\mathrm{y})$$\mathrm{f}(\mathrm{x},\mathrm{y})$f(x,y)\mathrm{f}(\mathrm{x}, \mathrm{y})$\mathrm{f}(\mathrm{x},\mathrm{y})$ near origin is:
   (A) $x+xy$$x+xy$x+xyx+x y$x+xy$
   (B) $x-xy$$x-xy$x-xyx-x y$x-xy$
   (C) $2\mathrm{x}+\mathrm{y}$$2\mathrm{x}+\mathrm{y}$2x+y2 \mathrm{x}+\mathrm{y}$2\mathrm{x}+\mathrm{y}$
   (D) $xy-x$$xy-x$xy-xx y-x$xy-x$
   (d) If $w=\tan (xy+\pi ),x=\ln (1+2t),y=e^{t/2}$$w=\tan (xy+\pi ),x=\ln (1+2t),y=e^{t/2}$w=tan(xy+pi),x=ln(1+2t),y=e^(t//2)w=\tan (x y+\pi), x=\ln (1+2 t), y=e^{t / 2}$w=\tan (xy+\pi ),x=\ln (1+2t),y=e^{t/2}$, then $\frac{dw}{dt}$$\frac{dw}{dt}$(dw)/(dt)\frac{d w}{d t}$\frac{dw}{dt}$ at the point $t=0$$t=0$t=0t=0$t=0$ is equal to:
   (A) 2
   (B) 0
   (C) $-\frac{3}{4}$$-\frac{3}{4}$-(3)/(4)-\frac{3}{4}$-\frac{3}{4}$
   (D) -1
2. Obtain SVD of the matrix $A=\left[\begin{array}{ll}2& 1\\ 1& 0\\ 0& 1\end{array}\right]$$A=\left[\begin{array}{l}2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\\ 1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\\ 0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1\end{array}\right]$A=[[2,1],[1,0],[0,1]]A=\left[\begin{array}{ll}2 & 1 \\ 1 & 0 \\ 0 & 1\end{array}\right]$A=\left[\begin{array}{ll}2& 1\\ 1& 0\\ 0& 1\end{array}\right]$.
3. Explain how the principle of gradient descent works.
4. Show that ${\mathbb{R}}^n$${\mathbb{R}}^n$R^(n)\mathbb{R}^n${\mathbb{R}}^n$ is a vector space over $\mathbb{R}$$\mathbb{R}$R\mathbb{R}$\mathbb{R}$, under the vector addition and scalar multiplication defined as follows.
   For all $\mathbf{x}=({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,\dots ,{\mathrm{x}}_{\mathrm{n}}),\mathrm{y}=({\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3,\dots ,{\mathrm{y}}_{\mathrm{n}})\in {\mathbb{R}}^{\mathrm{n}}$$\mathbf{x}=\left({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,\dots ,{\mathrm{x}}_{\mathrm{n}}\right),\mathrm{y}=\left({\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3,\dots ,{\mathrm{y}}_{\mathrm{n}}\right)\in {\mathbb{R}}^{\mathrm{n}}$x=(x_(1),x_(2),x_(3),dots,x_(n)),y=(y_(1),y_(2),y_(3),dots,y_(n))inR^(n)\mathbf{x}=\left(\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \ldots, \mathrm{x}_{\mathrm{n}}\right), \mathrm{y}=\left(\mathrm{y}_1, \mathrm{y}_2, \mathrm{y}_3, \ldots, \mathrm{y}_{\mathrm{n}}\right) \in \mathbb{R}^{\mathrm{n}}$\mathbf{x}=({\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,\dots ,{\mathrm{x}}_{\mathrm{n}}),\mathrm{y}=({\mathrm{y}}_1,{\mathrm{y}}_2,{\mathrm{y}}_3,\dots ,{\mathrm{y}}_{\mathrm{n}})\in {\mathbb{R}}^{\mathrm{n}}$ and $\alpha \in \mathbb{R}$$\alpha \in \mathbb{R}$alpha inR\alpha \in \mathbb{R}$\alpha \in \mathbb{R}$

Answer:

Question:-1(a)

Using Gram Schmidt orthogonalisation process find orthogonal basis for the subspace of ${\mathbb{R}}^4$${\mathbb{R}}^4$R^(4)\mathbb{R}^4${\mathbb{R}}^4$ having basis ${\mathrm{w}}_1=\left[\begin{array}{l}1\\ 0\\ 1\\ 0\end{array}\right],{\mathrm{w}}_2=\left[\begin{array}{l}2\\ 0\\ 0\\ 2\end{array}\right]$${\mathrm{w}}_1=\left[\begin{array}{l}1\\ 0\\ 1\\ 0\end{array}\right],{\mathrm{w}}_2=\left[\begin{array}{l}2\\ 0\\ 0\\ 2\end{array}\right]$w_(1)=[[1],[0],[1],[0]],w_(2)=[[2],[0],[0],[2]]\mathrm{w}_1=\left[\begin{array}{l}1 \\ 0 \\ 1 \\ 0\end{array}\right], \mathrm{w}_2=\left[\begin{array}{l}2 \\ 0 \\ 0 \\ 2\end{array}\right]${\mathrm{w}}_1=\left[\begin{array}{l}1\\ 0\\ 1\\ 0\end{array}\right],{\mathrm{w}}_2=\left[\begin{array}{l}2\\ 0\\ 0\\ 2\end{array}\right]$, and ${\mathrm{w}}_3=\left[\begin{array}{l}0\\ 0\\ 4\\ 4\end{array}\right]$${\mathrm{w}}_3=\left[\begin{array}{l}0\\ 0\\ 4\\ 4\end{array}\right]$w_(3)=[[0],[0],[4],[4]]\mathrm{w}_3=\left[\begin{array}{l}0 \\ 0 \\ 4 \\ 4\end{array}\right]${\mathrm{w}}_3=\left[\begin{array}{l}0\\ 0\\ 4\\ 4\end{array}\right]$.

Answer:

Step 1: Set the first orthogonal vector

Step 2: Compute the second orthogonal vector

• ${\mathbf{w}}_2\cdot {\mathbf{v}}_1=\left[\begin{array}{c}2\\ 0\\ 0\\ 2\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=2\cdot 1+0\cdot 0+0\cdot 1+2\cdot 0=2$${\mathbf{w}}_2\cdot {\mathbf{v}}_1=\left[\begin{array}{c}2\\ 0\\ 0\\ 2\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=2\cdot 1+0\cdot 0+0\cdot 1+2\cdot 0=2$w_(2)*v_(1)=[[2],[0],[0],[2]]*[[1],[0],[1],[0]]=2*1+0*0+0*1+2*0=2\mathbf{w}_2 \cdot \mathbf{v}_1 = \begin{bmatrix} 2 \\ 0 \\ 0 \\ 2 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} = 2 \cdot 1 + 0 \cdot 0 + 0 \cdot 1 + 2 \cdot 0 = 2${\mathbf{w}}_2\cdot {\mathbf{v}}_1=\left[\begin{array}{c}2\\ 0\\ 0\\ 2\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=2\cdot 1+0\cdot 0+0\cdot 1+2\cdot 0=2$,
• ${\mathbf{v}}_1\cdot {\mathbf{v}}_1=\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=1^2+0^2+1^2+0^2=2$${\mathbf{v}}_1\cdot {\mathbf{v}}_1=\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=1^2+0^2+1^2+0^2=2$v_(1)*v_(1)=[[1],[0],[1],[0]]*[[1],[0],[1],[0]]=1^(2)+0^(2)+1^(2)+0^(2)=2\mathbf{v}_1 \cdot \mathbf{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} = 1^2 + 0^2 + 1^2 + 0^2 = 2${\mathbf{v}}_1\cdot {\mathbf{v}}_1=\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=1^2+0^2+1^2+0^2=2$.

Step 3: Compute the third orthogonal vector

• Projection onto ${\mathbf{v}}_1$${\mathbf{v}}_1$v_(1)\mathbf{v}_1${\mathbf{v}}_1$:

• ${\mathbf{w}}_3\cdot {\mathbf{v}}_1=\left[\begin{array}{c}0\\ 0\\ 4\\ 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=0\cdot 1+0\cdot 0+4\cdot 1+4\cdot 0=4$${\mathbf{w}}_3\cdot {\mathbf{v}}_1=\left[\begin{array}{c}0\\ 0\\ 4\\ 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=0\cdot 1+0\cdot 0+4\cdot 1+4\cdot 0=4$w_(3)*v_(1)=[[0],[0],[4],[4]]*[[1],[0],[1],[0]]=0*1+0*0+4*1+4*0=4\mathbf{w}_3 \cdot \mathbf{v}_1 = \begin{bmatrix} 0 \\ 0 \\ 4 \\ 4 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} = 0 \cdot 1 + 0 \cdot 0 + 4 \cdot 1 + 4 \cdot 0 = 4${\mathbf{w}}_3\cdot {\mathbf{v}}_1=\left[\begin{array}{c}0\\ 0\\ 4\\ 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]=0\cdot 1+0\cdot 0+4\cdot 1+4\cdot 0=4$,
• ${\mathbf{v}}_1\cdot {\mathbf{v}}_1=2$${\mathbf{v}}_1\cdot {\mathbf{v}}_1=2$v_(1)*v_(1)=2\mathbf{v}_1 \cdot \mathbf{v}_1 = 2${\mathbf{v}}_1\cdot {\mathbf{v}}_1=2$ (from before).

• Projection onto ${\mathbf{v}}_2$${\mathbf{v}}_2$v_(2)\mathbf{v}_2${\mathbf{v}}_2$:

• ${\mathbf{w}}_3\cdot {\mathbf{v}}_2=\left[\begin{array}{c}0\\ 0\\ 4\\ 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]=0\cdot 1+0\cdot 0+4\cdot (-1)+4\cdot 2=-4+8=4$${\mathbf{w}}_3\cdot {\mathbf{v}}_2=\left[\begin{array}{c}0\\ 0\\ 4\\ 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]=0\cdot 1+0\cdot 0+4\cdot (-1)+4\cdot 2=-4+8=4$w_(3)*v_(2)=[[0],[0],[4],[4]]*[[1],[0],[-1],[2]]=0*1+0*0+4*(-1)+4*2=-4+8=4\mathbf{w}_3 \cdot \mathbf{v}_2 = \begin{bmatrix} 0 \\ 0 \\ 4 \\ 4 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ -1 \\ 2 \end{bmatrix} = 0 \cdot 1 + 0 \cdot 0 + 4 \cdot (-1) + 4 \cdot 2 = -4 + 8 = 4${\mathbf{w}}_3\cdot {\mathbf{v}}_2=\left[\begin{array}{c}0\\ 0\\ 4\\ 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]=0\cdot 1+0\cdot 0+4\cdot (-1)+4\cdot 2=-4+8=4$,
• ${\mathbf{v}}_2\cdot {\mathbf{v}}_2=\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]=1^2+0^2+(-1{)}^2+2^2=1+1+4=6$${\mathbf{v}}_2\cdot {\mathbf{v}}_2=\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]=1^2+0^2+(-1{)}^2+2^2=1+1+4=6$v_(2)*v_(2)=[[1],[0],[-1],[2]]*[[1],[0],[-1],[2]]=1^(2)+0^(2)+(-1)^(2)+2^(2)=1+1+4=6\mathbf{v}_2 \cdot \mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \\ -1 \\ 2 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ -1 \\ 2 \end{bmatrix} = 1^2 + 0^2 + (-1)^2 + 2^2 = 1 + 1 + 4 = 6${\mathbf{v}}_2\cdot {\mathbf{v}}_2=\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]=1^2+0^2+(-1{)}^2+2^2=1+1+4=6$.

• ${\mathbf{v}}_1\cdot {\mathbf{v}}_3=\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]=1\cdot (-1)+0\cdot 0+1\cdot 1+0\cdot 1=-1+1=0$${\mathbf{v}}_1\cdot {\mathbf{v}}_3=\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]=1\cdot (-1)+0\cdot 0+1\cdot 1+0\cdot 1=-1+1=0$v_(1)*v_(3)=[[1],[0],[1],[0]]*[[-1],[0],[1],[1]]=1*(-1)+0*0+1*1+0*1=-1+1=0\mathbf{v}_1 \cdot \mathbf{v}_3 = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} -1 \\ 0 \\ 1 \\ 1 \end{bmatrix} = 1 \cdot (-1) + 0 \cdot 0 + 1 \cdot 1 + 0 \cdot 1 = -1 + 1 = 0${\mathbf{v}}_1\cdot {\mathbf{v}}_3=\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]=1\cdot (-1)+0\cdot 0+1\cdot 1+0\cdot 1=-1+1=0$,
• ${\mathbf{v}}_2\cdot {\mathbf{v}}_3=\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]=1\cdot (-1)+0\cdot 0+(-1)\cdot 1+2\cdot 1=-1-1+2=0$${\mathbf{v}}_2\cdot {\mathbf{v}}_3=\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]=1\cdot (-1)+0\cdot 0+(-1)\cdot 1+2\cdot 1=-1-1+2=0$v_(2)*v_(3)=[[1],[0],[-1],[2]]*[[-1],[0],[1],[1]]=1*(-1)+0*0+(-1)*1+2*1=-1-1+2=0\mathbf{v}_2 \cdot \mathbf{v}_3 = \begin{bmatrix} 1 \\ 0 \\ -1 \\ 2 \end{bmatrix} \cdot \begin{bmatrix} -1 \\ 0 \\ 1 \\ 1 \end{bmatrix} = 1 \cdot (-1) + 0 \cdot 0 + (-1) \cdot 1 + 2 \cdot 1 = -1 – 1 + 2 = 0${\mathbf{v}}_2\cdot {\mathbf{v}}_3=\left[\begin{array}{c}1\\ 0\\ -1\\ 2\end{array}\right]\cdot \left[\begin{array}{c}-1\\ 0\\ 1\\ 1\end{array}\right]=1\cdot (-1)+0\cdot 0+(-1)\cdot 1+2\cdot 1=-1-1+2=0$.

Step 4: Verify linear independence

Final Answer