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

  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

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 \(M\)?

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

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

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

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

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

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

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

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