Linear Algebra MCQs and MSQs with Solutions for CSIR NET, IIT JAM : Linear Transformation - III

21. The transformation \((x, y, z) \rightarrow(x+y, y+z): \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}\) is

(a.) Linear and has zero kernel
(b.) Linear and has a proper subspace as kernel
(c.) Neither linear nor 1-1
(d.) Neither linear nor onto

Explanation for Question 21:
The transformation \(T(x, y, z) = (x + y, y + z)\) corresponds to the matrix \(\left[\begin{array}{lll}1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right]\).
With \(\operatorname{Rank}(T) = 2\), we have \(\eta(T) = 3 - 2 = 1\). Thus, \(T\) is not \(1-1\) and \(\text{Ker } T \neq \{0\}\). The calculations for \(T(x_1 + x_2, y_1 + y_2, z_1 + z_2)\) and \(T(x_1, y_1, z_1) + T(x_2, y_2, z_2)\) are shown, confirming that \(T\) is a linear transformation. This completes the explanation for Question 21.

22. Let \(T: \mathbb{R}^{3} \rightarrow W\) be the orthogonal projection of \(\mathbb{R}^{3}\) onto the \(x z\) plane \(W\). Then

(a.) \(T(x, y, z)=(x+y, 0, y+z)\)
(b.) \(T(x, y, z)=(x-y, 0, y-z)\)
(c.) \(T(x, y, z)=(x+y+z, 0, z)\)
(d.) \(T(x, y, z)=(x, 0, z) \)

Explanation for Question 22:
The transformation \(T(x, y, z) = (x ,0, z)\) represent the projection on xz plane.

23. \(\left\{v_{1}, v_{2}, v_{3}\right\}\) is a basis of \(V=\mathbb{R}^{3}\) and a linear transformation \(T: V \rightarrow V\) is defined by \(T(v_{1})=v_{1}+v_{2}, T(v_{2})=v_{2}+v_{3}, T(v_{3})=v_{3}+v_{1}\), then

(a.) \(T\) is \(1-1\)
(b.) \(T(v_{1}+v_{2}+v_{3})=0\)
(c.) \(T(v_{1}+2 v_{2}+v_{3})=T(v_{2}+2 v_{3})\)
(d.) \(T(v_{1}-v_{3})=v_{1}-v_{3}\)

Explanation for Question 23:
Given the matrix representation of the linear transformation \(T\) as: \[ [T]=\left[\begin{array}{lll}1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{array}\right] \] We can determine the rank of \(T\), which is the number of linearly independent rows or columns in the matrix. Here, \(\text{Rank}(T)=3\). Since the rank is equal to the number of rows (which is 3), we conclude that \(\eta(T)=0\), meaning \(T\) is a one-to-one (injective) transformation. Next, let's calculate the result of applying \(T\) to various vectors: \begin{align*} T(v_1+v_1+v_3) &= T(v_1)+T(v_2)+T(v_3) \\ &= 2v_1+2v_2+2v_3 \end{align*} \begin{align*} T(v_1+2v_2+v_3) &= v_1+v_2+2v_2+2v_3+v_1+v_3 \\ &= 2v_1+3v_2+3v_3 \end{align*} \begin{align*} T(v_2+2v_3) &= v_2+v_3+2v_1+2v_3 \\ &= 2v_1+v_2+3v_3 \end{align*} \begin{align*} T(v_1-v_3) &= v_1+v_2-v_1-v_3 \\ &= v_2-v_3 \end{align*} This concludes the explanation for Question 23.

24. For the standard basis \(\{(1,0,0),(0,1,0),(0,0,1)\}\) of \(\mathbb{R}^{3}\), a linear transformation \(T\) from \(\mathbb{R}^{3}\) to \(\mathbb{R}^{3}\) has the matrix representation \(\left[\begin{array}{ccc}2 & 1 & -1 \\ 1 & -1 & 1 \\ 3 & 1 & -2\end{array}\right]\). The image under \(T\) of \((2,1,2)\) is

(a.) \((11,0,-1)\)
(b.) \((11,3,-5)\)
(c.) \((7,3,-1)\)
(d.) None of these

Explanation for Question 24:
To calculate \(T(2,1,2)\), we use the matrix representation of \(T\) as \(\left[\begin{array}{ccc}2 & 1 & -1 \\ 1 & -1 & 1 \\ 3 & 1 & -2\end{array}\right]\) and the input vector \(\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right]\).
Performing matrix-vector multiplication, we get: \[ T(2,1,2) = \left[\begin{array}{ccc}2 & 1 & -1 \\ 1 & -1 & 1 \\ 3 & 1 & -2\end{array}\right]\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right] = \left[\begin{array}{l}3 \\ 3 \\ 1\end{array}\right] \] This completes the explanation for Question 24.

25. For a linear transformation \(T: \mathbb{R}^{10} \rightarrow \mathbb{R}^{6}\), the kernel is having dimension 5. Then the dimension of the range of \(T\) is

(a.) Five
(b.) Six
(c.) Four
(d.) Two

Explanation for Question 25:
Using the Rank-Nullity theorem, we have \(\operatorname{dim}(\mathbb{R}^{10}) = \operatorname{Rank}(T) + \eta(T)\).

26. Consider the \(3 \times 3\) matrix \(T=\left(\begin{array}{ccc}1 & -1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1\end{array}\right)\). Which of the following is false?

(a.) There is a non-zero vector which is not in the image of \(T\)
(b.) There is a non-zero vector which is in the kernel of \(T\)
(c.) There is a non-zero vector which is not in the kernel of \(T\)
(d.) The matrix is invertible

Explanation for Question 26:
Given the matrix \(T\) as \(\left[\begin{array}{ccc}1 & -1 & 0 \\ 1 & -2 & 1 \\ 0 & 1 & -1\end{array}\right]\), we find that \(\operatorname{Rank}(T) = 2\) implies d is our answer. Since \(\operatorname{Rank}(T) \neq 3\), \(T\) is not onto.
Also, \(\eta(T) = 1\), which means \(\operatorname{ker}(T) \neq \{0\}\) and \(\operatorname{ker}(T) \neq \mathbb{R}^{3}\). This implies that there is a nonzero vector in \(\operatorname{ker}(T)\) and there is a vector not in \(\operatorname{ker}(T)\), proving a,b and c. This completes the explanation for Question 26.

27. Let \(V\) be the vector space of \(2 \times 2\) matrices over \(\mathbb{R}\). Which of the following is/are not linear transformations?

(a.) \(T\left(\left[a_{i j}\right]\right)=a_{11}\)
(b.) \(T(A)=A+1\)
(c.) \(T(A) = \text{Trace } A\)
(d.) \(T\left(\left[a_{i j}\right]\right)=a_{11}+a_{12}+a_{21}+a_{22}\)

Explanation for Question 27:
For b: - \(T(0) = 1\), which implies \(T\) does not preserve the zero vector, violating the linearity property. - Thus, \(T\) is not a linear transformation. Rest can be proved using the definition of LT.

28. Let \(V\) be the real vector space of real polynomials of degree < 3 and let \(T: V \rightarrow V\) be the linear transformation defined by \(P(t)\) and \(Q(t)\) where \(Q(t)=p(a t+b)\). Then the matrix of \(T\) with respect to the basis \(1, t, t^{2}\) of \(V\) is:

(a.) \(\left(\begin{array}{ccc}1 & b & b^{2} \\ 0 & a & 2 a b \\ 0 & 0 & a^{2}\end{array}\right)\)
(b.) \(\left(\begin{array}{ccc}0 & a & a^{2} \\ 0 & b & 2 a b \\ 0 & 0 & b^{2}\end{array}\right)\)
(c.) \(\left(\begin{array}{ccc}b & b & b^{2} \\ a & a & 0 \\ 0 & b & a^{2}\end{array}\right)\)
(d.) \(\left(\begin{array}{ccc}a & a & a^{2} \\ b & b & 0 \\ 0 & a & b^{2}\end{array}\right)\)

Explanation for Question 28:
We have a vector space \(V\) of polynomials \(P_2(x)\) and a linear transformation \(T\) defined as \(T(p) = p(ax + b)\). Here's how we determine the matrix representation of \(T\):
- \(1 \longmapsto 1\) - \(x \longmapsto (ax + b)\) - \(x^2 \longmapsto (ax + b)^2 = a^2x^2 + 2abx + b^2\) Thus, \(T\) can be represented as the matrix \(\left[\begin{array}{lll}1 & b & b^2 \\ 0 & a & 2ab \\ 0 & 0 & a^2\end{array}\right]\).

29. Let \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}\) be a linear map defined by \(T(x, y, z, w)=(x+z, 2x+y+3z, 2y+2z, w)\). Then the rank of \(T\) is equal to

(a.) Four
(b.) Three
(c.) Two
(d.) One

Explanation for Question 29:
The matrix representation of \(T\) is \(\left[\begin{array}{llll}1 & 0 & 1 & 0 \\ 2 & 1 & 3 & 0 \\ 0 & 2 & 2 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]\). Here's the analysis:
- \(\operatorname{Rank}(T) = 3\) - \(\eta(T) = 4 - 3 = 1\) Since the rank of \(T\) is 3, it implies that \(\eta(T)\) is the difference between the number of rows and the rank of \(T\).

30. Let \(T_{1}, T_{2}: \mathbb{R}^{5} \rightarrow \mathbb{R}^{3}\) be linear transformations such that rank \(T_{1}=3\) and nullity \(T_{2}=3\). Let \(T_{3}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be a linear transformation such that \(T_{3} \circ T_{1}=T_{2}\). Then rank \(T_{3}\) is

(a.) Two
(b.) Three
(c.) Four
(d.) Five

Explanation for Question 30:
We are given three linear transformations \(T_1\), \(T_2\), and \(T_3\). We have the following information:
- \(\operatorname{Rank}(T_1) = 3\), which implies \(\eta(T_3) = 2\).
- \(\eta(T_2) = 3\), which implies \(\operatorname{Rank}(T_3) = 2\).
- \(T_3 \circ T_1 = T_2\), and we know that \(\eta(T_3 \circ T_1) = \eta(T_1) + \eta(T_3) = \eta(T_2) = 3\).
From the above information, we can deduce that \(\eta(T_3) = 1\), which, in turn, implies \(\operatorname{Rank}(T_3) = 2\). This completes the explanation for Question 30.