Linear Algebra MCQs and MSQs with Solutions : Linear Transformation – I

1. Let \(V\) be the space of twice differentiable functions on \(\mathbb{R}\) satisfying \(f^{\prime \prime}-2 f^{\prime}+f=0\). Define \(T: V \rightarrow \mathbb{R}^{2}\) by \(T(f)=(f^{\prime}(0), f(0))\). Then \(T\) is

NA

2. Given a \(4 \times 4\) matrix, let \(T: \mathbb{R^4}\rightarrow \mathbb{R^4}\) be a linear transformation defined by \(T v=A v\), where we think of \(\mathbb{R}^{4}\) as the set of real \(4 \times 1\) matrices. For which choices of \(A\) given below, do Image \((T)\) and Image \((T^{2})\) have respective dimensions 2 and 1 ? (* denotes a nonzero entry)

Explanation for Option a:
Let \(a, b, c, d, e \in \mathbb{R}\). For option a: \[ A=\left[\begin{array}{llll} 0 & 0 & a & b \\ 0 & 0 & c & d \\ 0 & 0 & 0 & e \\ 0 & 0 & 0 & 0 \end{array}\right] \Rightarrow A^2=\left[\begin{array}{llll} 0 & 0 & 0 & a e \\ 0 & 0 & 0 & b \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \] \(\Rightarrow \rho(A)=2\) and \(\rho\left(A^2\right)=1\).
Explanation for Option b:
For option b: \[ A=\left[\begin{array}{llll} 0 & 0 & a & 0 \\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & c \\ 0 & 0 & 0 & d \end{array}\right] \Rightarrow A^2=\left[\begin{array}{llll} 0 & 0 & 0 & a c \\ 0 & 0 & 0 & b c \\ 0 & 0 & 0 & c d \\ 0 & 0 & 0 & d^2 \end{array}\right] \] \(\rho(A)=2\) and \(\rho\left(A^2\right)=1\).

Explanation for Option c:
For option c: \[ A=\left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & a \\ 0 & 0 & b & 0 \end{array}\right] \Rightarrow A^2=\left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & a b & 0 \\ 0 & 0 & 0 & a b \end{array}\right] \] \(\rho(A)=2\) and \(\rho\left(A^2\right)=2\).

Explanation for Option d:
For option d: \[ A=\left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & a & b \\ 0 & 0 & c & d \end{array}\right] \Rightarrow A^2=\left[\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & a^2+bc & ab+bd \\ 0 & 0 & ac+cd & b^2+cd \end{array}\right] \] \(\rho(A)=2\) and \(\rho\left(A^2\right)=2\).

3. Which of the following is a linear transformation from \(\mathbb{R}\) to \(\mathbb{R^2}\)
A. \(f\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}4 \\ x+y\end{array}\right)\)
B. \(g\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}xy \\ x+y\end{array}\right)\)
C. \(h\left(\begin{array}{l}x \\ y \\ z\end{array}\right)=\left(\begin{array}{l}x-y \\ x+y\end{array}\right)\)

Explanation for Question 3:
Consider linear transformations \(A\) and \(B\). We have:
\(A: \left(\begin{array}{l}0 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{l}4 \\ 0\end{array}\right) \neq \left(\begin{array}{l}0 \\ 0\end{array}\right)\)
\(B: g\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)+g\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}0 \\ 1\end{array}\right)+\left(\begin{array}{l}0 \\ 1\end{array}\right)=\left(\begin{array}{l}0 \\ 2\end{array}\right)\)
\(g\left(\left(\begin{array}{l}1 \\ 0 \\ 0\end{array}\right)+\left(\begin{array}{l}0 \\ 1 \\ 0\end{array}\right)\right)=g\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right)=\left(\begin{array}{l}1 \\ 2\end{array}\right)\neq\left(\begin{array}{l}0 \\ 2\end{array}\right)\). Therefore, options \(a\), \(b\), and \(d\) are false.

4. Consider non-zero vector spaces \(V_{1}, V_{2}, V_{3}, V_{4}\) and linear transformations \(\phi_{1}: V_{1} \rightarrow V_{2}\), \(\phi_{2}: V_{2} \rightarrow V_{3}\), \(\phi_{3}: V_{3} \rightarrow V_{4}\) such that \(\operatorname{ker}\left(\phi_{1}\right)=\{0\}\), Range \(\left(\phi_{1}\right)=\operatorname{ker}\left(\phi_{2}\right)\), Range \(\left(\phi_{2}\right)=\operatorname{ker}\left(\phi_{3}\right)\), Range \(\left(\phi_{3}\right)=V_{4}\). Then

Explanation for Question 4:
Given linear transformations \(\phi_1: V_1 \rightarrow V_2\), \(\phi_2: V_2 \rightarrow V_3\), and \(\phi_3: V_3 \rightarrow V_4\), we can analyze their kernel (null space) and range (image) properties:
\(\bullet\) Ker\(\left(\phi_1\right) = \{0\}\), which implies \(\text{Rank}\left(\phi_1\right) = \text{dim} V_1\).
\(\bullet\) Range\(\left(\phi_1\right)\) is the kernel of \(\phi_2\), so \(\text{Rank}\left(\phi_1\right) = \text{Nullity}\left(\phi_2\right)\).
\(\bullet\) Range\(\left(\phi_2\right)\) is the kernel of \(\phi_3\), so \(\text{Rank}\left(\phi_2\right) = \text{Nullity}\left(\phi_3\right)\).
\(\bullet\) Range\(\left(\phi_3\right) = V_4\), which implies \(\text{Rank}\left(\phi_3\right) = \text{dim} V_4\).
Now, \[ \begin{align*} &\sum_{i=1}^4 (-1)^i \text{dim} V_i \\ &= -\text{dim} V_1 + \text{dim} V_2 – \text{dim} V_3 + \text{dim} V_4 \\ &= -\rho(\phi_1) + (\rho(\phi_2) + \eta(\phi_2)) – (\rho(\phi_3) + \eta(\phi_3)) + \rho(\phi_3) \\ &= -\rho(\phi_1) + \rho(\phi_2) – \rho(\phi_3) – \eta(\phi_2) + \eta(\phi_3) + \rho(\phi_3) \\ &= 0 \end{align*} \] Additionally,
\[ \begin{align*} &\sum_{i=2}^4 (-1)^i \text{dim} V_i \\ &= \sum_{i=1}^4 (-1)^i \text{dim} V_i + \text{dim} V_1 \\ &= 0 + \text{dim} V_1 \geq 0 \end{align*} \] Therefore, the given conditions hold, and the explanation for Question 4 is complete.

5. Let \(M_{n}(K)\) denote the space of all \(n \times n\) matrices with entries from a field \(K\). Fix a non-singular matrix \(A=\left(A_{i j}\right) \in M_{n}(K)\) and consider the linear map \(T: M_{n}(K) \rightarrow M_{n}(K)\) given by: \(T(X)=A X\). Then

Hint:
With respect to some basis, the matrix of T will be a matrix with matrix A on its block and rest of the entries will be zero. what can you conclude from that ?

6. Let \(M_{m \times n}(\mathbb{R})\) be the set of all \(m \times n\) matrices with real entries. Which of the following statements is correct?

Explanation for Question 6:
For option a,b and d Let’s consider linear transformations \(A\) in the context of \(A: \mathbb{R}^5 \longrightarrow \mathbb{R}^2\). We are given the conditions:
\(\rho(A) \leqslant 2\) and \(\eta(A) \geqslant 3\).
Now, for option \(C\), we have matrices \(A\) and \(B\) defined as follows:
\[ A = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \end{bmatrix} \] \[ B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} \] When we perform matrix multiplication \(A B\), we get the result:
\[ A B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \] Therefore, for option \(C\), the matrix \(A B\) is a 2×2 identity matrix.
In summary, for option \(C\), matrix \(A B\) is the 2×2 identity matrix, which implies that \(\rho(A B) = 2\). Since \(\rho(A B) \leqslant \min(\rho(A), \rho(B))\), we have \(\rho(A B) \leqslant 2\).
Additionally, since \(\rho(A B) \leqslant 2\), it follows that \(\eta(A B) \geqslant 3\) because \(\eta(A B) = 5 – \rho(A B)\).
Therefore, the given conditions hold for option \(C\), and the explanation for Question 6 is complete.

7. Let \(V\) be the vector space of polynomials over \(\mathbb{R}\) of degree less than or equal to \(n\). For \(p(x)=a_{0}+a_{1} x+\ldots+a_{n} x^{n}\) in \(V\), define a linear transformation \(T: V \rightarrow V\) by \((T p)(x)=a_{0}-a_{1} x+a_{2} x^{2}-\ldots .+(-1)^{n} a_{n} x^{n}\). Then which of the following are correct?

Explanation for Question 7:
The linear transformation \(T\) is defined as follows:
\[ T\left(a_0 + a_1 x + \cdots + a_n x^n\right) = a_0 – a_1 x + \cdots + (-1)^n a_n x^n \] We can represent \(T\) using a matrix \([T]\) as follows:
\[ [T] = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & -1 & \cdots & 0 \\ 0 & 0 & \ddots & 0 \\ \vdots & \vdots & & \vdots \\ 0 & 0 & & (-1)^n \end{bmatrix}_{(n+1) \times (n+1)} \] Now, to determine whether \(T\) is invertible, we need to check if the determinant \(\det(T)\) is non-zero. If \(\det(T) \neq 0\), then \(T\) is invertible.
Since \(\det(T) = (-1)^n\) (the determinant of a diagonal matrix is the product of its diagonal entries), we have \(\det(T) \neq 0\) for all values of \(n\). Therefore, \([T]\) is invertible for all \(n\), and by extension, \(T\) is invertible.
An invertible linear transformation is both one-to-one (injective) and onto (surjective). Therefore, \(T\) is one-to-one and onto.
This completes the explanation for Question 7.

8. A linear transformation \(T\) rotates each vector in \(\mathbb{R}^{2}\) clockwise through \(90^{\circ}\). The matrix \(T\) relative to the standard ordered basis \(\left(\left[\begin{array}{l}1 \\ 0\end{array}\right],\left[\begin{array}{l}0 \\ 1\end{array}\right]\right)\) is

Explanation for Question 8:
We are given a rotation matrix for clockwise rotation by an angle \(\theta\):
\[ \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \] This matrix represents a transformation in a 2D Euclidean space where points are rotated clockwise by an angle of \(\theta\) about the origin.
Note that for anticlockwise rotation, the matrix is:
\[ \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \] The rotation matrices are commonly used in computer graphics, physics, and engineering to perform 2D rotations.
This completes the explanation for Question 8.

9. Let \(T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}\) be a linear transformation. Which of the following statements implies that \(T\) is bijective?

Explanation for Question 9:
We know that \(\operatorname{Rank}(T) + \operatorname{Nullity}(T) = \operatorname{dim}(\mathbb{R}^n) = n\).
If \(\operatorname{Rank}(T) – \operatorname{Nullity}(T) = n\), then we have: \[ \begin{align*} \operatorname{Rank}(T) – \operatorname{Nullity}(T) &= n \\ \Rightarrow \operatorname{Rank}(T) &= n + \operatorname{Nullity}(T) \end{align*} \] Simplifying, we get \(\operatorname{Nullity}(T) = 0\), and therefore, \(\operatorname{Rank}(T) = n\).
When the rank of a linear transformation is equal to the dimension of the codomain, and the nullity is zero, the linear transformation is bijective (both one-to-one and onto).

10. Let \(n\) be a positive integer and let \(M_{n}(\mathbb{R})\) denote the space of all \(n \times n\) real matrices. If \(T: M_{n}(\mathbb{R}) \rightarrow M_{n}(\mathbb{R})\) is a linear transformation such that \(T(A)=0\) whenever \(A \in M_{n}(\mathbb{R})\) is symmetric or skew symmetric, then the rank of \(T\) is

Explanation for Question 10:
Let \(A\) be an arbitrary matrix. We have the equation: \[ A = \frac{A + A^{\top}}{2} + \frac{A – A^{\top}}{2} \] Observe that \(\frac{A + A^{\top}}{2}\) is a symmetric matrix, and \(\frac{A – A^{\top}}{2}\) is a skew-symmetric matrix.
Now, let’s consider the transformation \(T\) defined as \(T(A) = T\left(\frac{A + A^{\top}}{2}\right) + T\left(\frac{A – A^{\top}}{2}\right)=0\). Since \(T\) is linear, we can split it into two parts as shown.
So, for arbitrary \(A \in M_n(\mathbb{R})\), we have \(T(A) = 0\), which implies that the transformation \(T\) is identically zero (\(T \equiv 0\)).