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

31. Which of the following is not a subspace of \(\mathbb{R}^{3}\)?

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.
Rest can be proved using one step test for subspaces.

32. Let \(V\) denote the vector space \(C^{5}[a, b]\) over \(\mathbb{R}\) and \(W=\left\{f \in V: \frac{d^{4} f}{d t^{4}}+2 \frac{d^{2} f}{d t^{2}}-f=0\right\}\). Then

Explanation for Question 32 (b):
Solution space of the homogeneous differential equation of \(n\)th order forms a vector space of dimension \(n\).
Again, we have an ifinite LI subset in V. Namely \[ \begin{aligned} & \{ e^{nx} | n \in \mathbb{N}, a\leq x \leq b\} \subset C^{5}[a, b] \\ & \Rightarrow \operatorname{dim}\left(C^{5}\{0, b]\right)=\infty \end{aligned} \]

33. Let \(F_{3}\) be the field of 3 elements and let \(F_{3} \times F_{3}\) be the vector space over \(F_{3}\). The number of distinct linearly dependent sets of the form \(\{u, v\}\), where \(u, v \in F_{3} \times F_{3} /\{(0,0)\}\) and \(u \neq v\) is __.

Explanation for Question 32 (b):
NA

34. Let \(M\) be the space of all \(4 \times 3\) matrices with entries in the finite field of three elements. Then the number of matrices of rank three in \(M\) is

Explanation for Question 32 (b):
NA

35. The dimension of the vector space \(V=\left\{A=\left[a_{i j}\right]_{n \times n} \mid a_{i j} \in \mathbb{C}, a_{i j}=-a_{j i}\right\} \quad\) over field \(\mathbb{R}\) is

Explanation for Question 35 (c):
(Referred to as Q16) The same as Question 16.
\(\operatorname{dim} V\) over field \(\mathbb{C}=\frac{n(n+1)}{2}\)
\(\operatorname{dim} \mathbb{C}\) over field \(\mathbb{R}=2\)
\(\Rightarrow \operatorname{dim} V\) over field \(\mathbb{R}=\frac{n(n+1)}{2} \times 2=n(n+1)\)

36. Consider the subspace

\(W=\{\left[a_{i j}\right]: a_{i j}=0\) if \(i\) is even \(\}\) of all \(10 \times 10\)

real matrices. Then the dimension of \(W\) is

Explanation for Question 36 (b):
This is subspace of all matrices with 2,4,6,8,10 rows as 0.

37. Let \(M=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta\end{array}\right]\), where \(0<\theta<\frac{\pi}{2}\). Let \(V=\left\{u \in \mathbb{R}^{3}: M u^{t}=u^{t}\right\}\). Then the dimension of \(V\) is

Explanation for Question 37 (b):
Linear transformation corresponding to \(M\) is rotation about the X-axis in \(\mathbb{R}^3\) using the right-hand thumb rule. This implies that every vector \(Mv=v\) lies on the X-axis, and vice-versa.
\(\Rightarrow V=\{(x, 0, 0) \mid x \in \mathbb{R}\}\)
\(\Rightarrow \operatorname{dim} V=1\)
Another way to look at it is that the geometric multiplicity of eigenvalue 1 is 1.

38. Let \(A=\left[\begin{array}{lll}1 & 1 & 1 \\ 2 & 2 & 3 \\ x & y & z\end{array}\right]\) and let \(V=\left\{(x, y, z) \in \mathbb{R}^{3}: \operatorname{det}(A)=0\right\}\). Then the dimension of \(V\) equals

Explanation for Question 38 (a):
\(A=\left[\begin{array}{lll} 1 & 1 & 1 \\ 2 & 2 & 3 \\ x & y & z \end{array}\right]\)
\(|A|=(2z-3y)-(2z-3x)+1(2y+2x)=5x-y\)
\(\Rightarrow V=\{(x, y, z) \ | \ 5x-y=0\}\)
\(=\{(x, 5x, z) | \ \ x,z \in \mathbb{R}\}\)
Thus, the number of unknowns is 2.
\(\Rightarrow \operatorname{dim} V=2\)

39. A basis of

\(V=\{(x, y, z, w) \in \mathbb{R}^{4}: x+y-z=0\), \(y+z+w=0,2 x+y-3 z-w=0\}\)

Explanation for Question 39 (d):
For option a) \(1+1-(-1) \neq 0 \Rightarrow x+y-z \neq 0\) $\Rightarrow\{(1,1,-1,0),(0,1,1,1),(2,1,-3,1)\}$ is not a basis of \(V\).
For option b) & c) the vectors \((1,-1,0,1)\) & \((1,0,1,-1)\) both are linearly independent and satisfy conditions given for \(V\). Therefore, the basis set contains at least 2 elements. Hence, they are incorrect.
For d), it can be proved that the dimension of \(V\) is 2.

40. The dimension of the subspace \(\left\{\left(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right): 3 x_{1}-x_{2}+x_{3}=0\right\}\) of \(\mathbb{R}^{5}\) is

Explanation for Question 40:
Correct answer is d.
\[ \begin{aligned} V & =\left\{\left(x_1, x_2, x_3, x_4, x_5\right) \ x_2 =3 x_1+x_3\right\} \\ & =\left\{\left(x_1, 3 x_1+x_3, x_3, x_4, x_5\right) |\ \ x_i \in \mathbb{R}\right\} \\ \text{That is V can be generated by 4 vectors.} \Rightarrow \operatorname{dim} V & =4 \end{aligned} \]