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

41. Let \(V\) be the vector space of all real polynomials. Consider the subspace \(W\) spanned by \(t^{2}+t+2\), \(t^{2}+2 t+5\), \(5 t^{2}+3 t+4\), and \(2 t^{2}+2 t+4\). Then the dimension of \(W\) is

(a.) 4
(b.) 3
(c.) 2
(d.) 1

Explanation for Question 41 (c):
Consider the matrix \(A\) of the coefficients of the given vectors: \[ A=\left[\begin{array}{lll} 1 & 1 & 2 \\ 1 & 2 & 5 \\ 5 & 3 & 4 \\ 2 & 2 & 4 \end{array}\right] \] Note that \(\text{Rank}(A)=2\). One can use row echelon form to get the rank.
\(\Rightarrow\) Out of the given 4 vectors, 2 are linearly independent.
\(\Rightarrow\) The space generated by the 4 vectors is of dimension 2.

42. The set of all \(x \in \mathbb{R}\) for which the vectors \((1, x, 0)\), \((0, x^{2}, 1)\), and \((0,1, x)\) are linearly independent in \(\mathbb{R}^{3}\) is

(a.) \(\{x \in \mathbb{R}: x=0\}\)
(b.) \(\{x \in \mathbb{R}: x \neq 0\}\)
(c.) \(\{x \in \mathbb{R}: x \neq 1\}\)
(d.) \(\{x \in \mathbb{R}: x \neq-1\}\)

Explanation for Question 42 (c):
Consider the matrix \(A\): \[ A=\left[\begin{array}{lll} 1 & x & 0 \\ 0 & x^2 & 1 \\ 0 & 1 & x \end{array}\right] \] \((1, x, 0), (0, x^2, 1), (0, 1, x)\) are linearly independent if and only if \(|A| \neq 0\).
\(|A|=x^3-1 \neq 0\) if and only if \(x \neq 1\).

43. Let \(S\) and \(T\) be two subspaces of \(\mathbb{R}^{24}\) such that \(\operatorname{dim}(S)=19\) and \(\operatorname{dim}(T)=17\). Then, the

(a.) Smallest possible value of \(\operatorname{dim}(S \cap T)\) is 2
(b.) Largest possible value of \(\operatorname{dim}(S \cap T)\) is 18
(c.) Smallest possible value of \(\operatorname{dim}(S+T)\) is 19
(d.) Largest possible value of \(\operatorname{dim}(S+T)\) is 22

Explanation for Question 43 (c):
We know \(\operatorname{S+T} \subseteq \mathbb{R}^{14}\) and \[ \operatorname{dim}(S+T)=\operatorname{dim} S+\operatorname{dim} T-\operatorname{dim} (S \cap T) \] \[ \begin{aligned} & S+T \subseteq R^{14} \\ & \Rightarrow \operatorname{dim}(S+T) \leqslant 24 \\ & \Rightarrow 24 \geqslant 19+17-\operatorname{dim} (S \cap T) \\ & \Rightarrow \operatorname{dim}(S \cap T) \geqslant 12 \\ & \operatorname{dim}(S \cap T) \leqslant \operatorname{dim} T=17 \\ & \Rightarrow 17 \geqslant \operatorname{dim} S \cap T \geqslant 12 \\ & \Rightarrow 17 \geqslant \operatorname{dim} S+\operatorname{dim} T-\operatorname{dim}(S+T) \geqslant 12 \\ & \Rightarrow 19 \leqslant \operatorname{dim}(S+T) \leqslant 24 \end{aligned} \]

44. Let \(v_{1}=(1,2,0,3,0)\), \(v_{2}=(1,2,-1,-1,0)\), \(v_{3}=(0,0,1,4,0)\), \(v_{4}=(2,4,1,10,1)\), and \(v_{5}=(0,0,0,0,1)\). The dimension of the linear span of \(\left(v_{1}, v_{2}, v_{3}, v_{4}, v_{5}\right)\) is

(a.) 2
(b.) 3
(c.) 4
(d.) 5

Explanation for Question 44 (b):
\(\operatorname{Rank} A=3\) where \[ A=\left[\begin{array}{ccccc} 1 & 2 & 0 & 3 & 0 \\ 1 & 2 & -1 & -1 & 0 \\ 0 & 0 & 1 & 4 & 0 \\ 2 & 4 & 1 & 10 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right] \] Use row echelon form to determine the rank.

45. The set \(V=\left\{(x, y) \in \mathbb{R}^{2}: x y \geq 0\right\}\) is

(a.) A vector subspace of \(\mathbb{R}^{2}\)
(b.) Not a vector subspace of \(\mathbb{R}^{2}\) since every element does not have an inverse in \(V\)
(c.) Not a vector subspace of \(\mathbb{R}^{2}\) since it is not closed under scalar multiplication
(d.) Not a vector subspace of \(\mathbb{R}^{2}\) since it is not closed under vector addition

Explanation for Question 45 (d):
We check for vector addition first.
Let \(v_1=(0,5)\) and \(v_2=(-5,0) \in V\).
\(v_1=(0,5), v_2=(-5,0)\) but \(v_1+v_2=(-5,5) \notin V\).

46. Let \(\left\{v_{1}, v_{2}, v_{3}, v_{4}\right\}\) be a basis of \(\mathbb{R}^{4}\) and \(v=a_{1} v_{1}+a_{2} v_{2}+a_{3} v_{3}+a_{4} v_{4}\) where \(a_{i} \in \mathbb{R}\), \(i=1,2,3,4\).

Then \(\left\{v_{1}-v, v_{2}-v, v_{3}-v, v_{4}-v\right\}\) is a basis of \(\mathbb{R}^{4}\) if and only if

(a.) \(a_{1}=a_{2}=a_{3}=a_{4}\)
(b.) \(a_{1} a_{2} a_{3} a_{4}=-1\)
(c.) \(a_{1}+a_{2}+a_{3}+a_{4} \neq 1\)
(d.) \(a_{1}+a_{2}+a_{3}+a_{4}=0\)

Explanation for Question 46 :
NA

47. The dimension of the vector space of all \(3 \times 3\) real symmetric matrices is

(a.) 3
(b.) 9
(c.) 6
(d.) 4

Explanation for Question 47 (c):
The first row has 3 places of choice.
The second row has 2 places of choice.
The third row has 1 place of choice.
\(\Rightarrow\) Total places for choice \(=3+2+1=6\).

48. Let \(W\) be the space spanned by \(f=\sin x\) and \(g=\cos x\). Then for any real value of \(\theta\), \(f_{1}=\sin (x+\theta)\) and \(g_{1}=\cos (x+\theta)\).

(a.) Are vectors in \(W\)
(b.) Are linearly independent
(c.) Do not form a basis for \(W\)
(d.) Form a basis for \(W\)

Explanation for Question 48 (abd):
Consider the equations: \[ \begin{aligned} \sin (x+\theta) & =\sin x \cos \theta+\cos x \sin \theta \\ & =\alpha_1 \sin x+\beta \cos x \\ \cos (x+\theta) & =\cos x \cos \theta-\sin x \sin \theta \\ & =\beta \cos x-\alpha \sin x \end{aligned} \] \((\theta \in \mathbb{R}\) is fixed)\)
\(\cos (x+\theta)=\cos x \cos \theta-\sin x \sin \theta\).
\(\Rightarrow \sin (x+\theta)\) and \(\cos (x+\theta)\) are in \(W\) as they are linear combinations of \(f\) and \(g\).
Also, we have \(\left|\begin{array}{cc}\alpha & \beta \\ \beta & -\alpha\end{array}\right|=-\alpha^2-\beta^2=-\left(\cos ^2 \theta+\sin ^2 \theta\right)=-1\).
\(\Rightarrow \sin (x+\theta)\) and \(\cos (x+\theta)\) are linearly independent and form a basis for \(W\).

49. Let \(W\) be an \(m\)-dimensional subspace of an \(n\)-dimensional vector space \(V\), where \(m < n\). Then the dimension of \(V / W\) is

(a.) \(\frac{n}{m}\)
(b.) \(n-m\)
(c.) \(n+m\)
(d.) 0

Explanation for Question 49 (b):
One can use the second theorem on homomorphism to prove this.

50. Consider the set

\begin{aligned}& V=\left\{\left(x_{1}-x_{2}+x_{3}, x_{1}+x_{2}-x_{3}\right):\right. \\& \left.\left(x_{1}, x_{2}, x_{3}\right) \in \mathbb{R}^{3}\right\}\end{aligned} Then

(a.) \(V\) is not a vector subspace of \(\mathbb{R}^{2}\)
(b.) \(V\) is a vector subspace of \(\mathbb{R}^{2}\) of dimension 0
(c.) \(V\) is a vector subspace of \(\mathbb{R}^{2}\) of dimension 1
(d.) \(V=\mathbb{R}^{2}\)

Explanation for Question 50 (d):
Consider the linear transformation \(T: \mathbb{R}^3 \rightarrow \mathbb{R}^2\) defined as \(T=\left[\begin{array}{ccc}1 & -1 & 1 \\ 1 & 1 & -1\end{array}\right]\).
Observe that \(V=\operatorname{Img}(T)\).
\(\Rightarrow V\) is a subspace and \(\operatorname{Rank}(T)=2\).
\(\Rightarrow \operatorname{dim}(V)=2\).
\(V \subseteq \mathbb{R}^2\) of dimension 2, so \(V=\mathbb{R}^2\).

51. Which of the following subsets of \(\mathbb{R}^{4}\) is a basis of \(\mathbb{R}^{4}\) ? \begin{aligned} & B_{1}=\{(1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)\} \\ & B_{2}=\{(1,0,0,0),(1,2,0,0),(1,2,3,0),(1,2,3,4)\} \\ & B_{3}=\{(1,2,0,0),(0,0,1,1),(2,1,0,0),(-5,5,0,0)\} \end{aligned}

(a.) \(B_{1}\) and \(B_{2}\) but not \(B_{3}\)
(b.) \(B_{1}, B_{2}\) and \(B_{3}\)
(c.) \(B_{1}\) and \(B_{3}\) but not \(B_{2}\)
(d.) Only \(B_{1}\)

Explanation for Question 51 (a):
Let \(A_i\) be the matrix with rows as vectors of \(B_i\) for \(i=1,2,3\).
\(\operatorname{Rank}\left(A_1\right)=4\), \(\operatorname{Rank}\left(A_2\right)=4\), \(\operatorname{Rank}\left(A_3\right)=3\).
\(\Rightarrow B_1\) and \(B_2\) are a basis but not \(B_3\).

52. Which of the following sets of functions from \(\mathbb{R}\) to \(\mathbb{R}\) is a vector space over \(\mathbb{R}\)

\begin{aligned} & S_{1}=\left\{f \mid \lim _{x \rightarrow 3} f(x)=0\right\} \\ & S_{2}=\left\{g \mid \lim _{x \rightarrow 3} g(x)=1\right\} & S_{3}=\left\{h \mid \lim _{x \rightarrow 3} h(x) \text { exists }\right\} \end{aligned}

(a.) Only \(S_{1}\)
(b.) Only \(S_{2}\)
(c.) \(S_{1}\) and \(S_{3}\) but not \(S_{2}\)
(d.) All the three are vector spaces

Explanation for Question 52 (c):
Let \(f, g \in S\), and \(\alpha, \beta \in \mathbb{R}\). Then \[ \begin{aligned} \lim _{x \rightarrow 3}(\alpha f+\beta g)(x) & = \alpha \lim _{x \rightarrow 3} f(x) + \beta \lim _{x \rightarrow 3} g(x) \\ & = \alpha \cdot 0 + \beta \cdot 0 = 0 \end{aligned} \] \(\therefore \quad \alpha f+\beta g \in S_1\).
Similarly, \(S_3\) is a vector space because if \(f, g \in S_3, \alpha, \beta \in \mathbb{R}\), then \[ \lim _{x \rightarrow 3}(\alpha f+\beta g) = \alpha \lim _{x \rightarrow 3} f(x) + \beta \lim _{x \rightarrow 3} g(x) \] exists.
\(S_2\) is not a vector space since the zero function does not belong to \(S_2\), and if \(f, g \in S_2\), then \[ \lim _{x \rightarrow 3}(f+g)(x) = \lim _{x \rightarrow 3} f(x) + \lim _{x \rightarrow 3} g(x) = 1 + 1 = 2 \] \(\therefore \quad f+g \notin S_2\).

53. Let \(M_{n}\) denote the vector space of all \(n \times n\) real matrices.

Among the following subsets of \(M_{n}\) decide which are linear subspaces.

(a.) \(V_{1}=\left\{A \in M_{n}: A\right.\) is non-singular \(\}\)
(b.) \(V_{2}=\left\{A \in M_{n}: \operatorname{det}(A)=0\right\}\)
(c.) \(V_{3}=\left\{A \in M_{n}: \operatorname{trace}(A)=0\right\}\)
(d.) \(V_4=\left\{B A: A \in M_n\right\}\), where \(B\) is some fixed matrix in \(M_{n}\).

Explanation for Question 53 (cd):
Let \(M_n\) be the vector space of all \(n \times n\) real matrices.
\(V_1=\left\{A \in M_n: A\right.\) is non-singular \(\}\) is not a subspace.
Since \(0 \in V_1\).
\(V_2=\left\{A \in M_n: \operatorname{det}(A)=0\right\}\) is not a subspace.
Since \(\operatorname{det}(A+B) \neq \operatorname{det}(A)+\operatorname{det}(B)\).
Hence, vector addition fails.
\(V_3=\left\{A \in M_n: \operatorname{Trace}(A)=0\right\}\) is a subspace.
\(\because \quad 0 \in V_3\) and \(\operatorname{Tr}(\alpha A+\beta B)=\operatorname{Tr}(\alpha A)+T(\beta B)\)
\(=\alpha \operatorname{Tr}(A)+\beta \operatorname{Tr}(B)=0\) where \(\alpha, \beta \in F\).
We can easily verify that \(V_4=\left\{B A: A \in M_n\right\}\) where \(B\) is some fixed matrix in \(M_n\) also forms a subspace.

54. Let \(W_{1}, W_{2}, W_{3}\) be three distinct subspaces of \(\mathbb{R}^{10}\) such that each \(W_{i}\) has dimension 9. Let \(W=W_{1} \cap W_{2} \cap W_{3}\). Then we can conclude that

(a.) \(W\) may not be a subspace of \(\mathbb{R}^{10}\)
(b.) \(\operatorname{dim} W \leq 8\)
(c.) \(\operatorname{dim} W \geq 7\)
(d.) \(\operatorname{dim} W \leq 3\)

Explanation for Question 54 (bc):
\(\because \quad\) Intersection of subspaces is also a subspace.
Hence, option (a) is incorrect.
Since each \(W_i\)'s are distinct and \(\operatorname{dim}\left(W_i\right)=9\).
Hence, L.I. constraint is imposed on each \(W_i\). \(W_1, W_2, W_3<\mathbb{R}^{10}: \operatorname{dim} W_i=9\).
\(W=W_1 \cap W_2 \cap W_3\).
Number of L.I. constraints imposed on \(W=W_1 \cap W_2 \cap W_3\) is 3.
Hence, \(\operatorname{dim}(W)=10-3=7\).

55. Let \(x=\left(x_{1}, x_{2}, x_{3}\right), y=\left(y_{1}, y_{2}, y_{3}\right) \in \mathbb{R}^{3}\) be linearly independent. Let \(\delta_{1}=x_{2} y_{3}-y_{2} x_{3}, \delta_{2}=x_{1} y_{3}-y_{1} x_{3}, \delta_{3}=x_{1} y_{2}-y_{1} x_{2}\). If \(V\) is the span of \(x, y\), then

(a.) \(V=\left\{(u, v, w): \delta_{1} u-\delta_{2} v+\delta_{3} w=0\right\}\)
(b.) \(V=\left\{(u, v, w):-\delta_{1} u+\delta_{2} v+\delta_{3} w=0\right\}\)
(c.) \(V=\left\{(u, v, w): \delta_{1} u+\delta_{2} v-\delta_{3} w=0\right\}\)
(d.) \(V=\left\{(u, v, w): \delta_{1} u+\delta_{2} v+\delta_{3} w=0\right\}\)

Explanation for Question 55 (a):
Given \(x=\left(x_1, x_2, x_3\right)\) and \(y=\left(y_1, y_2, y_3\right)\) are linearly independent.
That is if \((u, v, w) \in V\) where \(V\) is the span of \(x\) and \(y\), then \[ \left|\begin{array}{lll} u & v & w \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{array}\right|=0 \] \(u(x_2 y_3-y_2 x_3)-v(x_1 y_3-y_1 x_3)+w(x_1 y_2-y_1 x_2)=0\). ie \(u \delta_1-v \delta_2+w \delta_3=0\).