Practice Questions for NET JRF Linear Algebra : Vector Space and Subspaces - III

Practice Questions for NET JRF JRF Linear Algebra
Assignment: Vector Space and Subspaces

21. Let \(\left\{e_{1}, e_{2}, e_{3}\right\}\) be a basis of a vector space \(V\) over \(\mathbb{R}\). Consider the following sets:

\(A=\left\{e_{2}, e_{1}+e_{2}, e_{1}+e_{2}+e_{3}\right\}\)

\(B=\left\{e_{1}, e_{1}+e_{2}, e_{1}+e_{2}+e_{3}\right\}\)

\(C=\left\{2 e_{1}, 3 e_{1}+e_{3}, 6 e_{1}+3 e_{2}+e_{3}\right\}\)

(a.) \(A\) and \(B\) are bases of \(V\)
(b.) \(A\) and \(C\) are bases of \(V\)
(c.) \(B\) and \(C\) are bases of \(V\)
(d.) Only \(B\) is a basis of \(V\)

Explanation for Question 21 (a, b, c):
For \( A \): Let \( \alpha e_{2}+\beta\left(e_{1}+e_{2}\right)+r\left(e_{1}+e_{2}+e_{3}\right)=0 \) \[ \begin{aligned} & \Rightarrow \beta e(\beta+r) e_{1}+(\alpha+\beta+r) e_{2}+r e_{3}=0 \\ & \Rightarrow \beta=-\gamma, \alpha=-\beta-\gamma, \gamma=0 \\ & \Rightarrow \alpha=0=\beta=r \end{aligned} \] \( \Rightarrow A \) is a basis of \( V \). Similarly, \( B \) is also a basis of \( V \)
For \( C \): \[ \begin{aligned} & \text { let } \alpha\left(2 e_{1}\right)+\beta\left(3 e_{1}+e_{3}\right)+\gamma \left(6 e_{1}+3 e_{2}+e_{3}\right)=0 \\ & \Rightarrow(2 \alpha+3 \beta+6 \gamma) e_{1}+(\beta+3 \gamma) e_{2}+\gamma e_{3}=0 \\ & \Rightarrow \alpha=0=\beta=\gamma \\ \text { C is a basis of } V \end{aligned} \] Alternate : one can make matrix for each and check rank.

22. Let \(V\) be the vector space of all \(5 \times 5\) real skew-symmetric matrices. Then the dimension of \(V\) is

(a.) 20
(b.) 15
(c.) 10
(d.) 5

Explanation for Question 22 (c):
Same as Question 16.

23. \(U\) is a subset of \(\mathbb{R}^{4}\) given by \(\left.\left(x_{1}-x_{2}+x_{3}=0=x_{1}+x_{2}+x_{4}\right)\right\}\) then

(a.) \(U\) is not a subspace of \(\mathbb{R}^{4}\)
(b.) \(U\) is a subspace of \(\mathbb{R}^{4}\) of dimension 1
(c.) \(U\) is a subspace of \(\mathbb{R}^{4}\) of dimension 2
(d.) \(U\) is a subspace of \(\mathbb{R}^{4}\) of dimension 3

Explanation for Question 23 (c):
\[ \begin{aligned} U & =\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) / x_{1}-x_{2}+x_{3}=0, x_{1}+x_{2}+x_{4}=0 ) \right\} \\ & =\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}\right) / x_{3}=x_{2}-x_{1}, x_{4}=-\left(x_{1}+x_{2}\right) \right\} \\ & \Rightarrow \operatorname{dim}(U)=\left|\left\{x_{1}, x_{2}\right\}\right| \\ & =2 . \end{aligned} \] Let \( T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4} \) be defined as a linear transformation such that \( T\left(\left(x_{1}, x_{2}, x_{3}, x_{4}\right)\right)=\left(x_{1}, x_{2}, x_{2}-x_{1},-\left(x_{1}+x_{2}\right)\right) \) \[ U=T\left(\mathbb{R}^{4}\right) \] \( \Rightarrow U \) is the image of \( T \) \( \Rightarrow U \) is a vector space

24. Let \(S=\left\{x_{1}, x_{2}, \ldots, x_{m}\right\}\) and \(T=\left\{y_{1}, \ldots ; y_{n}\right\}\) be subsets of the vector space \(V\). Then

(a.) If \(S\) and \(T\) are both linearly independent then \(m=n\)
(b.) If \(S\) is a basis for \(V\) and if \(T\) spans \(V\) then \(m \geq n\)
(c.) If \(S\) is a basis for \(V\) and if \(T\) is linearly independent, then \(m \geq n\)
(d.) If \(S\) is linearly independent and if \(T\) spans \(V\), then \(m \leq n\)

Explanation for Question 24 (c, d):
For option a: \( S=\left\{x_{1}, x_{2}, \ldots, x_{m}\right\} \) and \( T=\left\{y_{1}, \ldots, y_{m}\right\} \) \( s=\left\{e_{1}\right\} \) and \( T=\left\{e_{1}, e_{2}\right\} \) are linearly independent in \(\mathbb{R}^{2}\) but \( |S| \neq |T| \) For option b: A basis is the smallest cardinality spanning set, so \( m \leq n \) For option c: A basis is the largest cardinality linearly independent set, so \( m \geqslant n \) For option d: Every linearly independent set can be extended to a basis, and a basis is the smallest cardinality spanning set, so \( m \leq d i m v \leq n \) In other words, \( m \leq n \).

25. Let \(S=\{(0,1, \alpha),(\alpha, 1,0),(1, \alpha, 1)\}\). Then \(S\) is a basis for \(\mathbb{R}^{3}\) if and only if

(a.) \(\alpha \neq 0\)
(b.) \(\alpha \neq 1\)
(c.) \(\alpha \neq 0\) and \(\alpha^{2} \neq 2\)
(d.) \(-1 \leq \alpha \leq 1\)

Explanation for Question 25 (c):
Let \( S^{\prime}=\left[\begin{array}{lll}0 & 1 & \alpha \\ \alpha & 1 & 0 \\ 1 & \alpha & 1\end{array}\right] \) \( S \) is a basis if \( \left|S^{\prime}\right| \neq 0 \) \[ \begin{aligned} &\left|S^{\prime}\right| =-1(\alpha-0)+\alpha\left(\alpha^{2}-1\right) \\ &=\alpha^{3}-2 \alpha \\ \Rightarrow \alpha & \neq 0 \text { and } \alpha^{2} \neq 2 \end{aligned} \]

26. Let \(S\) be the set of all \(n \times n\) matrices over \(\mathbb{R}\) with zero trace. Then

(a.) \(S\) is not a vector space
(b.) \(S\) is a vector space of dimension \(n-1\)
(c.) \(S\), together with the identity matrix, form a vector space
(d.) \(S\) is a vector space and it has a basis consisting of \(n^{2}-1\) matrices

Explanation for Question 26 (d):
We know \( \text{trace} \left(X_{1}+X_{2}\right)= \text{trace}X_{1}+ \text{trace} X_{2} \) and \( \text{trace} (\alpha x)=\alpha \operatorname{trace}(x) \) \( \Rightarrow V \) is a vector space

27. Let \(A\) and \(B\) be \(n \times n\) matrices. Then

(a.) Every column of \(A B\) can be expressed as a linear combination of columns of \(A\)
(b.) Every column of \(A B\) can be expressed as a linear combination of rows of \(A\)
(c.) Every column of \(A B\) can be expressed as a linear combination of columns of \(B\)
(d.) Every row of \(A B\) can be expressed as a linear combination of rows of \(A\)

Explanation for Question 27 (a):
Let \( A=\left[\begin{array}{ll} a_{1} & a_{2} \\ a_{3} & a_{4} \end{array}\right], B=\left[\begin{array}{ll} b_{1} & b_{2} \\ b_{3} & b_{4} \end{array}\right] \) \[ \begin{aligned} & \Rightarrow A B=\left[\begin{array}{ll} a_{1} & a_{2} \\ a_{3} & a_{4} \end{array}\right]\left[\begin{array}{ll} b_{1} & b_{2} \\ b_{3} & b_{4} \end{array}\right]=\left[\begin{array}{ll} a_{1} b_{1}+a_{2} b_{3} & a_{1} b_{2}+a_{2} b_{4} \\ a_{3} b_{1}+a_{4} b_{3} & a_{3} b_{2}+a_{4} b_{3} \end{array}\right] \end{aligned} \] Consider the first column of \( A B=\left[\begin{array}{l}a_{1} b_{1}+a_{2} b_{3} \\ a_{3} b_{1}+a_{4} b_{3}\end{array}\right\) \[ =b_{1}\left[\begin{array}{l} a_{1} \\ a_{3} \end{array}\right]+b_{3}\left[\begin{array}{l} a_{2} \\ a_{4} \end{array}\right] \] But it cannot be written as a linear combination of rows of \( A \) or columns of \( B \) \( \Rightarrow \) options \( b \) and \( c \) are not true Consider the first row of \( A B=\left[\begin{array}{lll}a_{1} b_{1}+a_{2} b_{3} & a_{1} b_{2}+a_{2} b_{4}\end{array}\right] \) which might not be written as a linear combination of rows of \( A \) \( \Rightarrow \) option \( d \) is not true

28. Let \(V\) be the vector space of real polynomials of degree not exceeding 2.

Let

\(f(x)=x-1, g(x)=x+1, h(x)=x^{2}-1, j(x)=x^{2}+1\)

Then the set \(\{f, g, h, j\}\) is

(a.) Linearly independent
(b.) Linearly dependent because \(f g=h\)
(c.) Linearly dependent \(f+g-h=0\)
(d.) Linearly dependent \(f-g-h+j=0\)

Explanation for Question 28 (d):
\[ \begin{aligned} & V=P_{2}(x) \\ & f(x)=x-1, g(x)=x+1, h(x)=x^{2}-1, j(x)=x^{2}+1 \\ & g-f=2=j-h \\ & \text{i.e.} \quad f-g-h+j=0 \end{aligned} \]

29. Let \(W\) be the subspace of \(\mathbb{R}^{2}\) spanned by \((1,2)\). Which of the following pairs represent the same element of the quotient space \(\frac{\mathbb{R}^{2}}{W}\)

(a.) \(W,\left(\frac{1}{2}, 3\right)+W\)
(b.) \(\left(1, \frac{1}{2}\right)+W,\left(6 ,\frac{21}{2}\right)+W\)
(c.) \(\left(1, \frac{1}{2}\right)+W,\left(2 ,\frac{3}{2}\right)+W\)
(d.) \((\sqrt{2}, 1)+W,(\sqrt{2}, 2)+W\)

Explanation for Question 29 (b):
\[ \begin{aligned} & W=\langle(1,2)\rangle=\{(x, 2x) / x \in \mathbb{R}\} \\ & \text{Observe } (1, \frac{1}{2})+(5,10)=(6, \frac{21}{2}) \\ & \Rightarrow (1, \frac{1}{2})+W=(6, \frac{21}{2})+W \end{aligned} \]

30. Suppose \(V_{1}, V_{2}, V_{3}, V_{4}\) are linearly independent vectors of a real vector space. Consider the two sets of vectors

\(S_{1}=\left\{V_{1}+V_{2}, V_{1}+V_{3}, V_{1}+V_{4}\right\}\)

\(S_{2}=\left\{V_{1}+V_{2}, V_{1}+V_{3}, V_{1}+V_{4}, V_{4}+V_{1}\right\}\)

Which of the following is true?

(a.) Both \(S_{1}\) and \(S_{2}\) are linearly independent
(b.) \(S_{2}\) is linearly independent but not \(S_{1}\)
(c.) \(S_{1}\) is linearly independent but not \(S_{2}\)
(d.) Neither \(S_{1}\) nor \(S_{2}\) is linearly independent

Explanation for Question 30 (c):
\[ \begin{aligned} & \text{Let } \alpha(v_{1}+v_{2})+\beta(v_{1}+v_{3})+\gamma(v_{1}+v_{4})=0 \\ & \Rightarrow (\alpha+\beta+\gamma) v_{1}+\alpha v_{2}+\beta v_{3}+\gamma v_{4}=0 \\ & \Rightarrow \alpha=0=\beta=\gamma \\ & S_{1} \text{ is an independent set } \end{aligned} \] \(\Rightarrow S_{1}\) is a linearly independent set Clearly, \(S_{2}\) is a linearly dependent set.

Explanation for Question 31 (a):
Consider \( \alpha = \sqrt{2} \) and \( v = (1,0,0) \) \[ \alpha v = (\sqrt{2}, 0, 0) \notin V_{1} \] \(\Rightarrow V_{1}\) is not a vector space