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

31. Let \(M\) be the real vector space of \(2 \times 3\) matrices with real entries. Let \(T: M \rightarrow M\) be defined by

\(T\left(\begin{bmatrix} x_{1} & x_{2} & x_{3} \\ x_{4} & x_{5} & x_{6} \end{bmatrix}\right) = \begin{bmatrix}-x_{6} & x_{4} & x_{1} \\ x_{3} & x_{5} & x_{2}\end{bmatrix}\).

Then the determinant of \(T\) is:

(a.) 1
(b.) -1
(c.) 0
(d.) Cannot be determined

Explanation for Question 31:
Let \(B=\left\{\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 0\end{array}\right],\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 0\end{array}\right], \cdots,\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 1\end{array}\right]\right\}\) be the basis of \(M\) \[ \begin{gathered} \Rightarrow T=\left[\begin{array}{cccccc} 0 & 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{array}\right] \\ |T|=1(-1(1(-1(-1))))=-1 \end{gathered} \] This completes the explanation for Question 31.

32. Let the linear transformation \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) be defined by

\(T(x_{1}, x_{2}) = (x_{1}, x_{1}+x_{2}, x_{2})\).

Then the nullity of \(T\) is:

(a.) 0
(b.) 1
(c.) 2
(d.) 3

Explanation for Question 32:
\(T\left(x_{1}, x_{2}\right)=\left(x_{1}, x_{1}+x_{2}, x_{2}\right)\) \[ \Rightarrow[T]=\left[\begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{array}\right] \] Rank \(T=2 \Rightarrow \eta(T)=2-2=0/).

33. Let \(P_{3}=\{p(x) \mid p(x)\) be a polynomial with real coefficients and degree at most 3\} and \(T: P_{3} \rightarrow P_{3}\) be the map given by

\(T(p(x)) = \int_{1}^{x} p^{\prime}(t) dt\).

If the matrix of \(T\) relative to the standard bases \(B_{1}=B_{2}=\{1, x, x^{2}, x^{3}\}\) is \(M\) and \(M^{\prime}\) denotes the transpose of the matrix \(M\), then \(M+M^{\prime}\) is:

(a.) \(\begin{bmatrix}0 & -1 & -1 & -1 \\ -1 & 2 & 0 & 0 \\ -1 & 0 & 2 & 0 \\ -1 & 0 & 0 & 2\end{bmatrix}\)
(b.) \(\begin{bmatrix}-1 & 0 & 0 & 2 \\ 0 & -1 & 1 & 0 \\ 0 & 1 & -1 & 0 \\ 2 & 0 & 2 & -1\end{bmatrix}\)
(c.) \(\begin{bmatrix}2 & 0 & 0 & -1 \\ 0 & 2 & 1 & 0 \\ 0 & 1 & 2 & -1 \\ -1 & 0 & -1 & 0\end{bmatrix}\)
(d.) \(\begin{bmatrix}0 & 2 & 2 & 2 \\ 2 & -1 & 0 & 0 \\ 2 & 0 & -1 & 0 \\ 2 & 0 & 0 & -1\end{bmatrix}\)

Explanation for Question 33:
\(T: P_{3} \rightarrow P_{3}\) defined as \(T(p)=\int_{1}^{x} p^{\prime}(t) d t\) \[ \begin{aligned} & \Rightarrow T(1)=\int_{1}^{x}0 \ d t=0 \\ & T(x)=\int_{1}^{x} 1 d t=x-1 \\ & T\left(x^{2}\right)=\int_{1}^{x} 2 t d t=x^{2}-1 \\ & T\left(x^{3}\right)=\int_{1}^{x} 3 t^{2} d t=t^{3}-1 \\ & \Rightarrow M {\left[\begin{array}{cccc} 0 & -1 & -1 & -1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] } \\ & \Rightarrow M + M^{\prime}=\left[\begin{array}{cccc} 0 & -1 & -1 & -1 \\ -1 & 2 & 0 & 0 \\ -1 & 0 & 2 & 0 \\ -1 & 0 & 0 & 2 \end{array}\right] \end{aligned} \] This completes the explanation for Question 33.

34. If the nullity of the matrix

\(\begin{bmatrix}k & 1 & 2 \\ 1 & -1 & -2 \\ 1 & 1 & 4\end{bmatrix}\)

is 1, then the value of \(k\) is:

(a.) -1
(b.) 0
(c.) 1
(d.) 2

Explanation for Question 34:
\[ \begin{aligned} & A=\left[\begin{array}{rrr} k & 1 & 2 \\ 1 & -1 & -2 \\ 1 & 1 & 4 \end{array}\right] \\ & \Rightarrow \quad|A|=-2 k-2 \\ & \text { nullity of } A>0 \Rightarrow|A|=0 \\ & \Rightarrow-2 k-2=0 \\ & \Rightarrow k=1 \end{aligned} \] This completes the explanation for Question 34.

35. Let \(T: P_{3}[0,1] \rightarrow P_{2}[0,1]\) be defined by

\(T(p(x))=p^{\prime \prime}(x)+p^{\prime}(x)\).

Then the matrix representation of \(T\) with respect to the bases \(\{1, x, x^{2}, x^{3}\}\) and \(\{1, x, x^{2}\}\) of \(P_{3}[0,1]\) and \(P_{2}[0,1]\) respectively is:

(a.) \(\begin{bmatrix}0 & 0 & 0 \\ 1 & 0 & 0 \\ 2 & 2 & 0 \\ 0 & 6 & 3\end{bmatrix}\)
(b.) \(\begin{bmatrix}0 & 1 & 2 & 0 \\ 0 & 0 & 2 & 6 \\ 0 & 0 & 0 & 3\end{bmatrix}\)
(c.) \(\begin{bmatrix}0 & 2 & 1 & 0 \\ 6 & 2 & 0 & 0 \\ 3 & 0 & 0 & 0\end{bmatrix}\)
(d.) \(\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 2 & 2 \\ 3 & 6 & 0\end{bmatrix}\)

Explanation for Question 35:
\[ \begin{aligned} T(p) & =p^{\prime \prime}+p^{\prime} \\ \Rightarrow \quad 1 & \mapsto 0 \\ x & \mapsto 1 \\ x^{2} & \mapsto 2 x+2 \\ \Rightarrow T & x^{3} \longmapsto 3 x^{2}+6 x \\ & =\left[\begin{array}{llll} 0 & 1 & 2 & 0 \\ 0 & 0 & 2 & 6 \\ 0 & 0 & 0 & 3 \end{array}\right] \end{aligned} \] This completes the explanation for Question 35.

36. Let \(V\) be the column space of the matrix

\(\begin{bmatrix}1 & -1 \\ 1 & 2 \\ 1 & -1 \end{bmatrix}\).

Then the projection of \(\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}\) on \(V\) is:

(a.) \(\begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}\)
(b.) \(\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\)
(c.) \(\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\)
(d.) \(\begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}\)

Explanation for Question 36:
\(\mathrm{Coll}_{A}=\{\alpha(1,1,1)+\beta(-1,2,-1) / \alpha, \beta \in \mathbb{R}\}\) Observe for \(\alpha=1=\beta; (1,1,1)+(-1,2,-1)=(0,1,0)\) \(\Rightarrow\) projection of \((0,1,0)\) on \(\mathrm{Coll}_{A}=(0,1,0)\) This completes the explanation for Question 36.

37. Let \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be the linear transformation defined by

\(T\left(x_{1}, x_{2}, x_{3}\right)=(x_{1}+3 x_{2}+2 x_{3}, 3 x_{1}+4 x_{2}+x_{3}, 2 x_{1}+x_{2}-x_{3})\).

I. The dimension of the range space of \(T^{2}\) is:

(a.) 0
(b.) 1
(c.) 2
(d.) 3

Explanation for Question 37:
\[ \begin{aligned} {[T] } & =\left[\begin{array}{ccc} 1 & 3 & 2 \\ 3 & 4 & 1 \\ 2 & 1 & -1 \end{array}\right] \Rightarrow T^{2}=\left[\begin{array}{ccc} 14 & 17 & 3 \\ 17 & 26 & 9 \\ 3 & 9 & 6 \end{array}\right] \\ \& & T^{3}=\left[\begin{array}{ccc} 71 & 113 & 42 \\ 113 & 164 & 51 \\ 42 & 51 & 9 \end{array}\right] \end{aligned} \] \[ \begin{aligned} \operatorname{Rank}\left(T^{2}\right) & =2 \ \text { & } \operatorname{Rank}\left(T^{3}\right)=2 \\ & \Rightarrow N\left(T^{3}\right)=3-2=1 \end{aligned} \] This completes the explanation for Question 37.

II. The dimension of the null space of \(T^{3}\) is:

(a.) 0
(b.) 1
(c.) 2
(d.) 3

Explanation for Question 37:
\[ \begin{aligned} {[T] } & =\left[\begin{array}{ccc} 1 & 3 & 2 \\ 3 & 4 & 1 \\ 2 & 1 & -1 \end{array}\right] \Rightarrow T^{2}=\left[\begin{array}{ccc} 14 & 17 & 3 \\ 17 & 26 & 9 \\ 3 & 9 & 6 \end{array}\right] \\ \& & T^{3}=\left[\begin{array}{ccc} 71 & 113 & 42 \\ 113 & 164 & 51 \\ 42 & 51 & 9 \end{array}\right] \end{aligned} \] \[ \begin{aligned} \operatorname{Rank}\left(T^{2}\right) & =2 \ \text { & } \operatorname{Rank}\left(T^{3}\right)=2 \\ & \Rightarrow N\left(T^{3}\right)=3-2=1 \end{aligned} \] This completes the explanation for Question 37.

38. Let \(T: \mathbb{R}^{4} \rightarrow \mathbb{R}^{4}\) be the linear map satisfying

\(T(e_{1})=e_{2}, T(e_{2})=e_{3}, T(e_{3})=0, T(e_{4})=e_{3}\)

Where \(\{e_{1}, e_{2}, e_{3}, e_{4}\}\) is the standard basis of \(\mathbb{R}^{4}\). Then:

(a) \(T\) is idempotent
(b) \(T\) is invertible
(c) Rank \(T=3\)
(d) \(T\) is nilpotent

Explanation for Question 38:
\[ [T]=\left[\begin{array}{llll} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right] \] Observe \(T^{3}=0\). This completes the explanation for Question 38.

39. For any \(n \in \mathbb{N}\), let \(P_{n}\) denote the vector space of all polynomials with real coefficients and of degree at most \(n\). Define \(T: P_{n} \rightarrow P_{n+1}\) by

\(T(p)(x)=p^{\prime}(x)-\int_{0}^{x} p(t) dt\).

The dimension of the null space of \(T\) is:

(a) 0
(b) 1
(c) \(n\)
(d) \(n+1\)

Explanation for Question 39:
\[ \begin{aligned} & T(p)=p^{\prime}(x)-\int_{0}^{x} p(t) d t \\ & \Rightarrow 1 \mapsto 0-x \\ & x \mapsto 1-\frac{x^{2}}{2} \\ & x^{2} \mapsto 2 x-\frac{x^{3}}{3} \\ & \vdots \\ & x^{n} \mapsto n x^{n-1}-\frac{x^{n+1}}{n+1} \\ & \\ & \Rightarrow[T]=\left[\begin{array}{ccccc} 0 & 1 & 0 & & 0 \\ -1 & 0 & 2 & & \vdots \\ 0 & -1 / 2 & 0 & \ddots & n \\ \vdots & \vdots & \ddots & & 0 \\ 0 & 0 & & & -1 / (n+1) \end{array}\right]_{(n+2) \times(n+1)} \\ & \Rightarrow \operatorname{Rank} T=n+1 \end{aligned} \] (observe by substituting different values of \(n\)) \(T\) goes from \((n+1)\)-dim space to \((n+2)\)-dim space with rank \((n+1)\), and therefore, by the rank-nullity theorem, \(\eta(T)=0\). This completes the explanation for Question 39.

40. Let \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) be a linear transformation defined by \(T((x, y, z))=(x+y-z, x+y+z, y-z)\). Then the matrix of the linear transformation \(T\) with respect to the ordered basis \(B=\{(0,1,0),(0,0,1),(1,0,0)\}\) of \(\mathbb{R}^{3}\) is:

(a.) \(\begin{bmatrix}1 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & 1\end{bmatrix}\)
(b.) \(\begin{bmatrix}1 & 1 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & -1\end{bmatrix}\)
(c.) \(\begin{bmatrix}1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & -1 & 1\end{bmatrix}\)
(d .) \(\begin{bmatrix}1 & -1 & 1 \\ 1 & 1 & 1 \\ 1 & -1 & 0\end{bmatrix}\)

Explanation for Question 40:
\[ \begin{aligned} & T(0,1,0)=(1,1,1)=1 v_{1}+1 v_{2}+1 v_{2} \\ & T(0,0,1)=(-1,1,-1)=1 v_{1}+(-1) v_{2}+(-1) v_{3} \end{aligned} \] \[ \Rightarrow a, b, d \text { are false } \] This completes the explanation for Question 40.