Linear Algebra MCQs and MSQs with Solutions for CSIR NET : Linear Transformation – II

11. Let \(S: \mathbb{R}^{3} \rightarrow \mathbb{R}^{4}\) and \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{3}\) be linear transformations such that \(T^{\circ} S\) is the identity map of \(\mathbb{R}^{3}\). Then

Explanation for Question 11:
Same as Question 44.

12. Let \(a, b, c, d \in \mathbb{R}\) and let \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be the linear transformation defined by \(T\left(\left[\begin{array}{l}x \\ y\end{array}\right]\right)=\left[\begin{array}{l}a x+b y \\ c x+d y\end{array}\right]\) for \(\left[\begin{array}{l}x \\ y\end{array}\right] \in \mathbb{R}^{2}\). Let \(S: \mathbb{C} \rightarrow \mathbb{C}\) be the corresponding map defined by \(S(x+i y)=(a x+b y)+i(c x+d y)\) for \(x, y \in \mathbb{R}\). Then

Explanation for Question 12:
We can easily verify that \(S(z_1 + z_2) = S(z_1) + S(z_2)\) for all \(z_1, z_2 \in \mathbb{C}\).
Let \(\alpha = \alpha_1 + i\alpha_2\). On comparing \(S(\alpha z) = \alpha S(z)\), we find that \(b = -c\).

13. Let \(n\) be a positive integer and \(V\) be an \((n+1)\)-dimensional vector space over \(\mathbb{R}\). If \(\left\{e_{1}, e_{2}, \ldots, e_{n+1}\right\}\) is a basis of \(V\) and \(T: V \rightarrow V\) is the linear transformation satisfying \(T\left(e_{i}\right)=e_{i+1}\) for \(i=1,2, \ldots, n\) and \(T\left(e_{n+1}\right)=0\). Then

Explanation for Question 13:
We are given the matrix representation of \(T\):
\[ [T] = \left[\begin{array}{ccccc} 0 & 0 & \cdots & 0 & 0 \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ 0 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \end{array}\right]_{(n+1) \times (n+1)} \] For \(n = 1\), the matrix becomes:
\[ [T] = \left[\begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right] \] Now, we observe that \(T^{n+1} = 0\) and \(T^n\) is not the zero transformation. Therefore, option \(d\) is not true. This completes the explanation for Question 13.

14. Let \(V\) be the vector space of all real polynomials of degree at most 3. Define \(S: V \rightarrow V\) by \(S(p(x))=Q(x)\), \(\forall p(x) \in V\), where \(Q(x)=p(x+1)\). Then the matrix of \(S\) in the basis \(\left\{1, x, x^{2}, x^{3}\right\}\), considered as column vectors, is given by:

Explanation for Question 14:
Same as Question 28.

15. For a positive integer \(n\), let \(P_{n}\) denote the space of all polynomials \(p(x)\) with coefficients in \(\mathbb{R}\) such that deg. \(p(x) \leq n\), and let \(B_{n}\) denote the standard basis of \(P_{n}\) given by \(B_{n}=\left\{1, x, x^{2}, \ldots, x^{n}\right\}\). If \(T: P_{3} \rightarrow P_{4}\) is the linear transformation defined by \(T(p(x))=x^{2} p^{\prime}(x)+\int_{0}^{x} p(t) d t\) and \(A=\left[a_{i j}\right]\) is the \(5 \times 4\) matrix of \(T\) with respect to standard bases \(B_{3}\) and \(B_{4}\), then

Explanation for Question 15:
We have a transformation \(T\) defined as follows:
\(T(p) = x^2 p’ + \int_{0}^{x} p(t) dt\)
Using this transformation, we find that:
\(1 \mapsto 0\)
\(x \mapsto x^2 + \frac{x^2}{2}\)
\(x^2 \mapsto 2x^3 + \frac{x^3}{3}\)
\(x^3 \mapsto 3x^4 + \frac{x^4}{4}\)
Therefore, the matrix representation of \(T\) is:
\[ [T] = \left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & \frac{3}{2} & 0 & 0 \\ 0 & 0 & \frac{7}{3} & 0 \\ 0 & 0 & 0 & \frac{13}{4} \end{array}\right] \] This completes the explanation for Question 15.

16. Consider the linear transformation \(T: \mathbb{R}^{7} \rightarrow \mathbb{R}^{7}\) defined by

\[T\left(x_{1}, x_{2}, \ldots, x_{6}, x_{7}\right)=\left(x_{7}, x_{6}, \ldots, x_{2}, x_{1}\right)\]

Which of the following statements are true?

Explanation for Question 16:
The matrix representation of \(T\) is given as:
\[ [T] = \left[\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \] We can see that \(|T| = -1\) and \(T^2 \equiv I\), where \(I\) is the identity matrix.
Therefore, the minimal polynomial of \(T\) divides \(x^2 – 1 = 0\), which factors as \((x + 1)(x – 1) = 0\).
This implies that \(T\) is diagonalizable, and with respect to some basis, \(T\) is a diagonal matrix.
This completes the explanation for Question 16.

17. Let \(W\) be the vector space of all real polynomials of degree at most 3. Define \(T: W \rightarrow W\) by \(\left(T(p(x))=p^{\prime}(x)\right.\) where \(p^{\prime}\) is the derivative of \(p\). The matrix of \(T\) in the basis \(\left\{1, x, x^{2}, x^{3}\right\}\), considered as column vectors, is given by

Explanation for Question 17:
We have a set \(W\) consisting of polynomials \(p(x) = a_0 + a_1x + a_2x^2 + a_3x^3\) where \(a_i \in \mathbb{R}\).
The transformation \(T\) maps a polynomial \(p\) to its derivative \(p’\).
Using this transformation, we find that:
\(1 \mapsto 0\)
\(x \mapsto 1\)
\(x^2 \mapsto 2x\)
\(x^3 \mapsto 3x^2\)
Therefore, the matrix representation of \(T\) is:
\[ [T] = \left[\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right] \]

18. Let \(x, y\) be linearly independent vectors in \(\mathbb{R}^{2}\) suppose \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) is a linear transformation such that \(T y=a x\) and \(T x=0\). Then with respect to some basis in \(\mathbb{R}^{2}, T\) is of the form

Explanation for Question 18:
We have \(T(x) = 0\) and \(T(y) = ax\).
Therefore, \(T^2(x) = 0\) and \(T^2(y) = T(T(y)) = T(ax) = aT(x) = a \cdot 0 = 0\).
This implies that \([T]^2 = 0\). However, \(T\) is not the zero transformation because \(T(y) = ax\) for some \(a\) where \(a \neq 0\), making T a nilpotent operator of index 2.
Therefore, option \(c\) is true.

19. Let \(\mathrm{V}\) be the space of all linear transformations from \(\mathbb{R}^{3}\) to \(\mathbb{R}^{2}\) under usual addition and scalar multiplication. Then

Explanation for Question 19:
The set of linear transformations from \(\mathbb{R}^3 \rightarrow \mathbb{R}^2\) is equivalent to the set of matrices of order \(2 \times 3\), denoted as \(M_{2 \times 3}(\mathbb{R})\).
The dimension of \(M_{2 \times 3}(\mathbb{R})\) is 6.
This completes the explanation for Question 19.

20. Let \(A: \mathbb{R}^{6} \rightarrow \mathbb{R}^{5}\) and \(B: \mathbb{R}^{5} \rightarrow \mathbb{R}^{7}\) be two linear transformations. Then which of the following can be true?

Explanation for Question 20:
Define \(A_1: \mathbb{R}^6 \rightarrow \mathbb{R}^5\) and \(B_1: \mathbb{R}^5 \rightarrow \mathbb{R}^7\) as:
\(A_1(x) = 0\) for all \(x \in \mathbb{R}^6\), and \(B_1(y) = 0\) for all \(y \in \mathbb{R}^5\).
Therefore, neither \(A_1\) nor \(B_1\) are one-to-one (\(1-1\)), so option \(a\) is true.
Now, consider \(A_2: \mathbb{R}^6 \rightarrow \mathbb{R}^5\). If Rank\((A_2) \leq 5\), then by the Rank-Nullity Theorem, \(\eta(A_2) \geq 6 – 5 = 1\).
This implies that \(A_2\) cannot be one-to-one (\(1-1\)), so option \(b\) is not true.
Next, define \(A_3: \mathbb{R}^6 \rightarrow \mathbb{R}^5\) as follows:
\(A_3: \left(x_1, x_2, x_3, x_4, x_5, x_6\right) \mapsto \left(x_1, x_2, x_3, x_4, x_5\right)\), and \(B_3: \mathbb{R}^5 \rightarrow \mathbb{R}^7\) as:
\(B_3: \left(y_1, y_2, y_3, y_4, y_5\right) \mapsto \left(y_1, y_2, y_3, y_1, y_5, 0, 0\right)\).
It’s clear that \(A_3\) is onto and \(B_3\) is one-to-one (\(1-1\)), so options \(c\) is true.
Now, for \(B: \mathbb{R}^5 \rightarrow \mathbb{R}^7\), if \(\eta(B) \geq 0\), then by the Rank-Nullity Theorem, \(\text{Rank}(B) \leq 5 – 0 = 5\).
This implies that \(B\) cannot be onto, so option \(d\) is not true.