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

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:

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

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

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

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

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

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

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

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

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

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

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 \(\frac{\mathbb{R}^{2}}{W}\)

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?

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