Which of the following statements are true and which are false? Justify your answer with a short proof or a counterexample.
i) The function f:RrarrRf: \mathbf{R} \rightarrow \mathbf{R} defined by f(x)=cos xf(x)=\cos x is 1-11-1.
Answer:
To determine whether the statement “The function f:RrarrRf: \mathbf{R} \rightarrow \mathbf{R} defined by f(x)=cos xf(x) = \cos x is 1-1″ is true or false, we need to understand what a 1-1 (one-to-one) function is and then apply this definition to the cosine function.
Definition of a 1-1 Function:
A function f:A rarr Bf: A \rightarrow B is called one-to-one (or injective) if for every x_(1),x_(2)in Ax_1, x_2 \in A, whenever f(x_(1))=f(x_(2))f(x_1) = f(x_2), it must be the case that x_(1)=x_(2)x_1 = x_2. In simpler terms, no two different inputs in the domain of the function should map to the same output in the codomain.
Applying the Definition to f(x)=cos xf(x) = \cos x:
To test if f(x)=cos xf(x) = \cos x is 1-1, we need to see if there are any distinct values x_(1)x_1 and x_(2)x_2 in the domain R\mathbf{R} (the set of all real numbers) such that cos x_(1)=cos x_(2)\cos x_1 = \cos x_2 but x_(1)!=x_(2)x_1 \neq x_2.
Counterexample:
Consider the values x_(1)=0x_1 = 0 and x_(2)=2pix_2 = 2\pi. These are distinct values in R\mathbf{R}, but:
cos 0=1\cos 0 = 1
cos 2pi=1\cos 2\pi = 1
Here, cos x_(1)=cos x_(2)\cos x_1 = \cos x_2 even though x_(1)!=x_(2)x_1 \neq x_2. This shows that the function f(x)=cos xf(x) = \cos x is not 1-1, as it violates the definition of a one-to-one function.
Conclusion:
The statement “The function f:RrarrRf: \mathbf{R} \rightarrow \mathbf{R} defined by f(x)=cos xf(x) = \cos x is 1-1″ is false. The counterexample of x_(1)=0x_1 = 0 and x_(2)=2pix_2 = 2\pi demonstrates that two different inputs can yield the same output, which contradicts the definition of a one-to-one function.
ii) The operation * defined by x**y=log(xy)x * y=\log (x y) is a binary operation on SS, where SS is the set {x inR∣x > 0}\{x \in \mathbf{R} \mid x>0\}.
Answer:
To determine whether the statement “The operation *** defined by x**y=log(xy)x * y = \log(xy) is a binary operation on SS, where SS is the set {x inR∣x > 0}\{x \in \mathbf{R} \mid x > 0\}” is true or false, we need to understand what a binary operation is and then apply this definition to the given operation.
Definition of a Binary Operation:
A binary operation on a set SS is a rule that assigns to each ordered pair of elements in SS a unique element in SS. In other words, if *** is a binary operation on SS, then for every x,y in Sx, y \in S, it must be the case that x**y in Sx * y \in S.
Applying the Definition to x**y=log(xy)x * y = \log(xy):
We need to check if for every x,y in Sx, y \in S, where S={x inR∣x > 0}S = \{x \in \mathbf{R} \mid x > 0\}, the result of x**y=log(xy)x * y = \log(xy) is also in SS.
Closure Property: For x,y in Sx, y \in S, we have x > 0x > 0 and y > 0y > 0. The product xyxy is also greater than 0 since the product of two positive numbers is positive. The logarithm function, log\log, is defined for positive real numbers. Therefore, log(xy)\log(xy) is a real number. Since log(xy)\log(xy) is defined and results in a real number for all x,y > 0x, y > 0, the operation *** is closed in SS.
Result in SS: We need to ensure that log(xy)\log(xy) is also greater than 0 to be in SS. However, this is not necessarily the case. For example, if x=y=(1)/(2)x = y = \frac{1}{2}, then xy=(1)/(4)xy = \frac{1}{4}, and log((1)/(4))\log\left(\frac{1}{4}\right) is negative, which is not in SS.
Conclusion:
The statement “The operation *** defined by x**y=log(xy)x * y = \log(xy) is a binary operation on SS, where SS is the set {x inR∣x > 0}\{x \in \mathbf{R} \mid x > 0\}” is false. The operation *** is not closed in SS because there exist x,y in Sx, y \in S such that x**y=log(xy)x * y = \log(xy) is not in SS. The example x=y=(1)/(2)x = y = \frac{1}{2} demonstrates that the result of the operation can be outside of SS, violating the definition of a binary operation.
iii) The set {(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}\left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1=2 x_2+3\right\} is a subspace of R^(n)\mathbf{R}^n.
Answer:
To determine whether the statement “The set {(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}\left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1 = 2x_2 + 3\right\} is a subspace of R^(n)\mathbf{R}^n” is true or false, we need to understand what a subspace is and then apply this definition to the given set.
Definition of a Subspace:
A subset WW of a vector space VV is a subspace of VV if and only if it satisfies three conditions:
Zero Vector: The zero vector of VV is in WW.
Closed under Addition: For every u,v in Wu, v \in W, the sum u+vu + v is in WW.
Closed under Scalar Multiplication: For every u in Wu \in W and every scalar cc, the product cucu is in WW.
Applying the Definition to the Given Set:
Let’s denote the set as S={(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}S = \left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1 = 2x_2 + 3\right\} and check if it satisfies the subspace criteria in R^(n)\mathbf{R}^n.
Zero Vector: The zero vector in R^(n)\mathbf{R}^n is (0,0,dots,0)(0, 0, \ldots, 0). For this vector to be in SS, we need x_(1)=2x_(2)+3=0x_1 = 2x_2 + 3 = 0, which is not possible since 2x_(2)+32x_2 + 3 is never zero for any real value of x_(2)x_2. Therefore, the zero vector is not in SS.
Since the zero vector is not in SS, SS cannot be a subspace of R^(n)\mathbf{R}^n. This alone is sufficient to conclude that the statement is false. However, for completeness, let’s briefly consider the other two properties:
Closed under Addition: Even if we were to consider this property, the set SS is not necessarily closed under addition. For example, if we take two different vectors from SS that satisfy x_(1)=2x_(2)+3x_1 = 2x_2 + 3, their sum may not satisfy this condition.
Closed under Scalar Multiplication: Similarly, scalar multiplication of a vector in SS may not result in a vector that still satisfies x_(1)=2x_(2)+3x_1 = 2x_2 + 3.
Conclusion:
The statement “The set {(x_(1)*x_(2),dots,x_(n))∣x_(1),x_(2),dots,x_(n)inR,x_(1)=2x_(2)+3}\left\{\left(x_1 \cdot x_2, \ldots, x_n\right) \mid x_1, x_2, \ldots, x_n \in \mathbf{R}, x_1 = 2x_2 + 3\right\} is a subspace of R^(n)\mathbf{R}^n” is false. The primary reason is that the zero vector of R^(n)\mathbf{R}^n is not in the set, violating the first and most fundamental condition for being a subspace.
iv) There is no 7xx57 \times 5 matrix of rank 6 .
Answer:
To evaluate the statement “There is no 7xx57 \times 5 matrix of rank 6,” we need to understand the concept of the rank of a matrix and the constraints imposed by the dimensions of the matrix.
Definition of Matrix Rank:
The rank of a matrix is defined as the maximum number of linearly independent column vectors in the matrix. It can also be equivalently defined as the maximum number of linearly independent row vectors in the matrix.
Analyzing the 7xx57 \times 5 Matrix:
A 7xx57 \times 5 matrix has 7 rows and 5 columns.
The rank of a matrix cannot exceed the number of rows or the number of columns. In other words, the rank of a matrix is limited by the smaller of these two dimensions.
Applying the Definition to a 7xx57 \times 5 Matrix:
Since the matrix in question is a 7xx57 \times 5 matrix, the maximum number of linearly independent columns it can have is 5 (the number of columns).
Similarly, the maximum number of linearly independent rows it can have is also 5 (since there are only 5 columns to form linearly independent combinations).
Conclusion:
The statement “There is no 7xx57 \times 5 matrix of rank 6″ is true. The rank of a 7xx57 \times 5 matrix cannot exceed the number of its columns, which is 5. Therefore, it is impossible for such a matrix to have a rank of 6.
v) If VV and V^(‘)V^{\prime} are vector spaces and T:V rarrV^(‘)T: V \rightarrow V^{\prime} is a linear transformation, then whenever u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_k are linearly independent, Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_k are also linearly independent.
Answer:
The statement “If VV and V^(‘)V’ are vector spaces and T:V rarrV^(‘)T: V \rightarrow V’ is a linear transformation, then whenever u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_k are linearly independent, Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_k are also linearly independent” needs to be evaluated for its truthfulness.
Definition of Linear Independence:
A set of vectors u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_k in a vector space is said to be linearly independent if the only solution to the linear equation c_(1)u_(1)+c_(2)u_(2)+dots+c_(k)u_(k)=0c_1 u_1 + c_2 u_2 + \ldots + c_k u_k = 0 (where 00 is the zero vector) is c_(1)=c_(2)=dots=c_(k)=0c_1 = c_2 = \ldots = c_k = 0.
Definition of a Linear Transformation:
A function T:V rarrV^(‘)T: V \rightarrow V’ is a linear transformation if for all u,v in Vu, v \in V and scalars cc, the following two properties hold:
T(u+v)=T(u)+T(v)T(u + v) = T(u) + T(v)
T(cu)=cT(u)T(cu) = cT(u)
Analyzing the Statement:
The statement claims that if a set of vectors u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_k in VV is linearly independent, then their images under TT, namely Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_k in V^(‘)V’, are also linearly independent.
Counterexample:
To show that the statement is false, we can provide a counterexample. Consider a linear transformation TT that is not injective (one-to-one). For instance, let TT be the zero transformation, which maps every vector in VV to the zero vector in V^(‘)V’.
In this case, even if u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_k are linearly independent in VV, their images under TT will all be the zero vector in V^(‘)V’. The set {Tu_(1),Tu_(2),dots,Tu_(k)}\{T u_1, T u_2, \ldots, T u_k\} will then consist of only the zero vector, which is not a set of linearly independent vectors, as multiple non-zero scalar combinations of the zero vector will result in the zero vector.
Conclusion:
The statement “If VV and V^(‘)V’ are vector spaces and T:V rarrV^(‘)T: V \rightarrow V’ is a linear transformation, then whenever u_(1),u_(2),dots,u_(k)u_1, u_2, \ldots, u_k are linearly independent, Tu_(1),Tu_(2),dots,Tu_(k)T u_1, T u_2, \ldots, T u_k are also linearly independent” is false. The existence of linear transformations that are not injective, such as the zero transformation, provides a counterexample where linearly independent vectors in VV can map to vectors in V^(‘)V’ that are not linearly independent.