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

11. Let \(S=\left\{A: A\left[a_{i j}\right]_{5 \times 5}, a_{i j}=0\right.\) or \(1 \forall i, j\),

\(\sum_{i} a_{i j}=1 \forall j\) and \(\left.\sum_{j} a_{i j}=1 \forall i\right\}.\)

Then the number of elements is \(S\) is

(a.) \(5^{2}\)
(b.) \(5^{5}\)
(c.) \(5!\)
(d.) 55

  Explanation for Question 11:
  (c) Number of choices to place '1' in row 1 = 5
  Number of choices to place '1' in row 2 = 4
  Number of choices to place '1' in row 3 = 3
  Number of choices to place '1' in row 4 = 2
  Number of choices to place '1' in row 5 = 1
  Therefore, the number of possibilities for the matrix \(A\) in set \(S\) is \(5 \times 4 \times 3 \times 2 \times 1 = 5!\).

12. The dimension of the vector space of all symmetric matrices of order \(n \times n\) (\(n \geq 2\)) with real entries and trace equal to zero is

(a.) \(\frac{\left(n^{2}-n\right)}{2}-1\)
(b.) \(\frac{\left(n^{2}-2n\right)}{2}-1\)
(c.) \(\frac{\left(n^{2}+n\right)}{2}-1\)
(d.) \(\frac{\left(n^{2}+2n\right)}{2}-1\)

Explanation for Question 12:
(a) Number of places with the freedom to choose in the 1st row \(=n\)
Number of places with freedom to choose in the 2nd row \(=n-1\)
\(\vdots\)
Number of places with freedom to choose in the \(n\)th row \(=1\)
\(\Rightarrow\) Total number of places with freedom to choose \(=n+(n-1)+(n-2)+\ldots+1=\frac{n(n+1)}{2}\)
Now, the trace needs to be zero.
\(\Rightarrow\) Dimension \(=\frac{n(n+1)}{2}-1=\frac{n^2+n}{2}-1\)

13. Let \(M\) be the vector space of all \(3 \times 3\) real matrices and let

\(A=\left(\begin{array}{lll} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{array}\right)\)

Which of the following are subspaces of \(M\)?

(a.) \(\{X \in M: X A=A X\}\)
(b.) \(\{X \in M: X+A=A+X\}\)
(c.) \(\{X \in M: \operatorname{trace}(A X)=0\}\)
(d.) \(\{X \in M: \operatorname{det}(A X)=0\}\)

Explanation for Question 13:
For option (a): Let \(X_{1}, X_{2} \in V_{1}, \alpha \in \mathbb{R}\)
\(\Rightarrow X_{1} A = A X_{1} \quad \& \quad X_{2} A = A X_{2}\)
Consider \(X = X_{1} + X_{2}\)
\(\begin{aligned} & XA = \left(X_{1}+X_{2}\right) A = X_{1} A + X_{2} A = A X_{1} + A X_{2} = A X \end{aligned}\)
Consider \(\alpha X_{1}\),
\((\alpha X_{1}) A = \alpha(X_{1}A)\)
\(\Rightarrow V_{1}\) is a vector space
For option (b): Let \(X_{1}, X_{2} \in V_{2}, \alpha \in \mathbb{R}\)
\(\Rightarrow X_{1}+A=A+X_{2} \quad \& \quad X_{2}+A=A+X_{2}\)
Observe \(\left(X_{1}+X_{2}\right)+A=A+\left(X_{1}+X_{2}\right) \quad \& \quad \alpha\left(X_{1}\right)+A=A+\alpha\left(X_{1}\right)\)
\(\Rightarrow V_{2}\) is a vector space
For option (c): Let \(X_{1}, X_{2} \in V_{3}, \alpha \in \mathbb{R}\)
We know \(\text {trace} \left(A\left(X_{1}+X_{2}\right)\right) = \text {trace}\left(A X_{1}+A X_{2}\right)\)
\(=\text {trace}\left(A X_{1}\right) + \text {trace}\left(A X_{2}\right)\)
\(\text {trace} \left(A\left(\alpha, X_{1}\right)\right) = \text {trace} (\alpha A X_{1}) = \text {trace} \left(A X_{1}\right)\)
\(\Rightarrow V_{3}\) is a vector space

14. Let \(W=\{p(B): p\) is a polynomial with real coefficients\(\}\), where

\(B=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right)\)

The dimension \(d\) of the vector space \(W\) satisfies

(a.) \(4 \leq d \leq 6\)
(b.) \(6 \leq d \leq 9\)
(c.) \(3 \leq d \leq 8\)
(d.) \(3 \leq d \leq 4\)

Explanation for Question 14:
Because \(B^3 = I\),
\(\begin{aligned} \therefore \quad W & =\left\{a_0 I+a_1 B+a_2 B^2 \mid a_0, a_1, a_2 \in \mathbf{R}\right\} \\ & =\operatorname{span}\left\{I, B, B^2\right\} \end{aligned}\)
As \(\left\{I, B, B^2\right\}\) is linearly independent.
Hence, \(\operatorname{dim} W=3\)

15. Let \(V\) be a real vector space and let \(\left\{x_{1}, x_{2}, x_{3}\right\}\) be a basis for \(V\). Then

(a.) \(\left\{x_{1}+x_{2}, x_{2}, x_{3}\right\}\) is a basis for \(V\)
(b.) The dimension of \(V\) is 3
(c.) \(\left\{x_{1}-x_{2}, x_{2}-x_{3}, x_{1}-x_{3}\right\}\) is a basis for \(V\)
(d.) None of these

Explanation for Question 15:
\(\begin{aligned} & \text{Let } \alpha (x_1 + x_2) + \beta x_2 + \gamma x_3 = 0 \\ & \Rightarrow \alpha x_1 + (\alpha + \beta) x_2 + \gamma x_3 = 0 \\ & \Rightarrow \alpha = 0 = \gamma \quad \text{and} \quad \beta = -\alpha = 0 \end{aligned}\)
\(\Rightarrow \{x_1 + x_2, x_2, x_3\}\) are 3 linearly independent and \(\operatorname{dim}(V) = 3\) (because the cardinality of the basis is 3).
\(\Rightarrow \{x_1 + x_2, x_2, x_3\}\) is the basis of \(V\).
Observe \(x_1 - x_3 = (x_1 - x_2) + (x_2 - x_3)\),
\(\Rightarrow \{x_1 - x_2, x_2 - x_3, x_1 - x_3\}\) is not linearly independent.
\(\Rightarrow\) not a basis

16. Let \(V\) be the set of all real \(n \times n\) matrices \(A=\left(a_{i j}\right)\) with the property that \(a_{i j}=-a_{j i}\) for all \(i, j=1,2, \ldots, n\). Then

(a.) \(V\) is a vector space of dimension \(n^{2}-n\)
(b.) For every \(A\) in \(V, a_{i i}=0\) for all \(i=1,2, \ldots, n\)
(c.) \(V\) consists of only diagonal matrices
(d.) \(V\) is a vector space of dimension \(\frac{n^{2}-n}{2}\)

Explanation for Question 16 (d):
In the 1st row, there are:
1st place with freedom = \(n - 1\)
2nd place with freedom = \(n - 2\)
\(\vdots\)
\(n\)th place with freedom = 1
\((n + 1)\)th place with freedom = 0
\(\Rightarrow\) Total places with freedom = \((n - 1) + (n - 2) + \cdots + 1 + 0 = \frac{n(n - 1)}{2}\) Also note that diagonal entries of a skew symm. matrix are always zero.

17. Let \(V=\left\{\left(x_{1}, \ldots, x_{100}\right) \in \mathbb{R}^{100}: x_{1}=\ldots=x_{50}\right.\) and \(\left.x_{51}+x_{52}+\ldots+x_{100}=0\right\}\). Then

(a.) \(\operatorname{dim} V=98\)
(b.) \(\operatorname{dim} V=51\)
(c.) \(\operatorname{dim} V=49\)
(d.) \(\operatorname{dim} V=50\)

Explanation for Question 17 (d):
Given \(V = \left\{(x_1, \ldots, x_{100}) \in \mathbb{R}^{100} \mid x_1 = \ldots = x_{50}, x_{51} + \cdots + x_{100} = 0\right\}\), we can see that: \[ \begin{aligned} \Rightarrow \operatorname{dim} V & = \left|\left\{x_1, x_{51}, x_{52}, \ldots, x_{99}\right\}\right| \\ & = 50 \end{aligned} \]

18. Let \(P(3)=\left\{a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3} \mid a_{i} \in \mathbb{R}\right.\) (\(i=0,1,2,3\))

Under the standard operation of addition (+) and scalar multiplication (.)P(3) is

(a.) Not a vector space
(b.) A vector space of infinite dimension
(c.) A vector space of dimension 3
(d.) A vector space of dimension 4

Explanation for Question 18 (d):
Let \( p, q \in P(3) \) and \( \alpha \in \mathbb{R} \). Then, \[ \begin{aligned} & p = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} \\ & q = b_{0} + b_{1} x + b_{2} x^{2} + b_{3} x^{3} \\ \end{aligned} \] \[ \begin{aligned} p+q & =\left(a_{0}+b_{0}\right)+\left(a_{1}+b_{1}\right) x+\left(a_{2}+b_{2}\right) x^{2}+\left(a_{3}+b_{3}\right) x^{3} \\ & \in P(3) \end{aligned} \] \[ \alpha p=\alpha a_{0}+\alpha a_{1} x+\alpha a_{2} x^{2}+\alpha a_{3} x^{3} \in P(3) \] Therefore, \( P(3) \) is a vector space with \[ \begin{aligned} \operatorname{dim}(P(3)) & = \left|\left\{a_{0}, a_{1}, a_{2}, a_{3}\right\}\right| \\ & = 4 \end{aligned} \]

19. Let \(V=\left\{\left(x_{1}, \ldots ; x_{100}\right) \in R^{100} \mid x_{1}=2 x_{2}=3 x_{3}\right.\) and \(\left.x_{51}-x_{52}-\ldots-x_{100}=0\right\}\). Then

(a.) \(\operatorname{dim} V=98\)
(b.) \(\operatorname{dim} V=49\)
(c.) \(\operatorname{dim} V=99\)
(d.) \(\operatorname{dim} V=97\)

Explanation for Question 19 (d):
For \( V = \left\{\left(x_{1}, \ldots, x_{100}\right) \in \mathbb{R}^{100} \mid x_{1} = 2 x_{2} = 3 x_{3}, x_{51} - x_{52}, \ldots, x_{100} = 0\right\} \), we can see that: \[ \begin{aligned} \Rightarrow \operatorname{dim} V & = \left|\left\{x_{11}, x_{4}, x_{52}, \ldots, x_{50}, x_{52}, x_{53}, \ldots, x_{100}\right\}\right| \\ & = 97 \end{aligned} \]

20. Consider the linear differential equation

\(y^{(3)}+3 x y^{(2)}+4 y^{(1)}+2 x^{2} y=\sin x,\) (\(x \in[0,1])\).

Then the set of solutions of the above equation

(a.) Is a linear space of infinite dimension
(b.) Is a linear space of dimension 3
(c.) Is not a linear space
(d.) Is a linear space of dimension less than 3

Explanation for Question 20 (c):
Solution of a non-homogeneous differential equation do not form a vector space.