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

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

(a.) $S^{\circ }T$ is the identity map of ${\mathbb{R}}^4$.
(b.) $S^{\circ }T$ is one-one, but not onto.
(c.) $S^{\circ }T$ is onto, but not one-one.
(d.) $S^{\circ }T$ is neither one-one nor onto.

Explanation for Question 11:
Same as Question 44.

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

(a.) $S$ is always $\mathbb{C}$-linear, that is $S(z_1+z_2)=S\left(z_1\right)+S\left(z_2\right)$ for all $z_1,z_2\in \mathbb{C}$ and $S(\alpha z)=\alpha S(z)$ for all $\alpha \in \mathbb{C}$ and $z\in \mathbb{C}$.
(b.) $S$ is $\mathbb{C}$-linear if $b=-c$ and $d=a$.
(c.) $S$ is $\mathbb{C}$-linear only if $b=-c$ and $d=a$.
(d.) $S$ is $\mathbb{C}$-linear if and only if $T$ is the identity transformation.

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 $\{e_1,e_2,\dots ,e_{n+1}\}$ is a basis of $V$ and $T:V\to V$ is the linear transformation satisfying $T\left(e_i\right)=e_{i+1}$ for $i=1,2,\dots ,n$ and $T\left(e_{n+1}\right)=0$. Then

(a.) Trace of $T$ is non-zero.
(b.) Rank of $T$ is $n$.
(c.) Nullity of $T$ is 1.
(d.) $T^n=T\circ T\circ \dots \circ T$ ($n$ times) is the zero map.

Explanation for Question 13:
We are given the matrix representation of $T$:
$$[T]={\left[\begin{array}{ccccc}0& 0& \cdots & 0& 0\\ 1& 0& \cdots & 0& 0\\ 0& 1& \cdots & 0& 0\\ 0& 0& \cdots & 0& 0\\ ⋮& ⋮& ⋮& ⋮& ⋮\\ 0& 0& \cdots & 1& 0\end{array}\right]}_{(n+1)\times (n+1)}$$ 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 $S:V\to V$ by $S(p(x))=Q(x)$, $\mathrm{\forall }p(x)\in V$, where $Q(x)=p(x+1)$. Then the matrix of $S$ in the basis $\{1,x,x^2,x^3\}$, considered as column vectors, is given by:

(a.) $\left[\begin{array}{llll}1& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 3& 0\\ 0& 0& 0& 4\end{array}\right]$
(b.) $\left[\begin{array}{llll}1& 1& 1& 1\\ 0& 1& 2& 3\\ 0& 0& 1& 3\\ 0& 0& 0& 1\end{array}\right]$
(c.) $\left[\begin{array}{llll}1& 1& 2& 3\\ 1& 1& 2& 3\\ 2& 2& 2& 3\\ 3& 3& 3& 3\end{array}\right]$
(d.) $\left[\begin{array}{llll}0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\end{array}\right]$

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. $p(x)\le n$, and let $B_n$ denote the standard basis of $P_n$ given by $B_n=\{1,x,x^2,\dots ,x^n\}$. If $T:P_3\to P_4$ is the linear transformation defined by $T(p(x))=x^2{p}^{\prime }(x)+{\int }_0^{x}p(t)dt$ and $A=\left[a_{ij}\right]$ is the $5\times 4$ matrix of $T$ with respect to standard bases $B_3$ and $B_4$, then

(a.) $a_{32}=\frac{3}{2}$ and $a_{33}=\frac{7}{3}$.
(b.) $a_{32}=\frac{3}{2}$ and $a_{33}=0$.
(c.) $a_{32}=0$ and $a_{33}=\frac{7}{3}$.
(d.) $a_{32}=0$ and $a_{33}=0$.

Explanation for Question 15:
We have a transformation $T$ defined as follows:
$T(p)=x^2{p}^{\prime }+{\int }_0^{x}p(t)dt$
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 $T:{\mathbb{R}}^7\to {\mathbb{R}}^7$ defined by

$$T(x_1,x_2,\dots ,x_6,x_7)=(x_7,x_6,\dots ,x_2,x_1)$$

Which of the following statements are true?

(a.) The determinant of $T$ is 1.
(b.) There is a basis of ${\mathbb{R}}^7$ with respect to which $T$ is a diagonal matrix.
(c.) $T^2=I$.
(d.) The smallest $n$ such that $T^n=I$ is even.

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 $T:W\to W$ by $(T(p(x))=p^{\prime }(x)$ where $p^{\prime }$ is the derivative of $p$. The matrix of $T$ in the basis $\{1,x,x^2,x^3\}$, considered as column vectors, is given by

(a.) $\left(\begin{array}{llll}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3\end{array}\right)$
(b.) $\left(\begin{array}{llll}0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 3& 0\end{array}\right)$
(c.) $\left(\begin{array}{llll}0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3\\ 0& 0& 0& 0\end{array}\right)$
(d.) $\left(\begin{array}{llll}0& 1& 2& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$

Explanation for Question 17:
We have a set $W$ consisting of polynomials $p(x)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3$ where $a_i\in \mathbb{R}$.
The transformation $T$ maps a polynomial $p$ to its derivative $p^{\prime }$.
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 $T:{\mathbb{R}}^2\to {\mathbb{R}}^2$ is a linear transformation such that $Ty=ax$ and $Tx=0$. Then with respect to some basis in ${\mathbb{R}}^2,T$ is of the form

(a.) $\left(\begin{array}{ll}a& 0\\ 0& a\end{array}\right),a>0$
(b.) $\left(\begin{array}{ll}a& 0\\ 0& b\end{array}\right),a,b>0,a\ne b$
(c.) $\left(\begin{array}{ll}0& a\\ 0& 0\end{array}\right)$
(d.) $\left(\begin{array}{ll}0& 0\\ 0& 0\end{array}\right)$

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 $a\ne 0$, 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

(a.) $\mathrm{V}$ is a vector space of dimension 5.
(b.) $\mathrm{V}$ is a vector space of dimension 6.
(c.) $\mathrm{V}$ is a vector space of dimension 8.
(d.) $\mathrm{V}$ is a vector space of dimension 9.

Explanation for Question 19:
The set of linear transformations from ${\mathbb{R}}^3\to {\mathbb{R}}^2$ 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 $A:{\mathbb{R}}^6\to {\mathbb{R}}^5$ and $B:{\mathbb{R}}^5\to {\mathbb{R}}^7$ be two linear transformations. Then which of the following can be true?

(a.) $A$ and $B$ are both non one-one.
(b.) $A$ is one-one and $B$ is not one-one.
(c.) $A$ is onto and $B$ is one-one.
(d.) $A$ and $B$ both are onto.

Explanation for Question 20:
Define $A_1:{\mathbb{R}}^6\to {\mathbb{R}}^5$ and $B_1:{\mathbb{R}}^5\to {\mathbb{R}}^7$ 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 $A_2:{\mathbb{R}}^6\to {\mathbb{R}}^5$. If Rank$(A_2)\le 5$, then by the Rank-Nullity Theorem, $\eta (A_2)\ge 6-5=1$.
This implies that $A_2$ cannot be one-to-one ($1-1$), so option $b$ is not true.
Next, define $A_3:{\mathbb{R}}^6\to {\mathbb{R}}^5$ as follows:
$A_3:(x_1,x_2,x_3,x_4,x_5,x_6)\mapsto (x_1,x_2,x_3,x_4,x_5)$, and $B_3:{\mathbb{R}}^5\to {\mathbb{R}}^7$ as:
$B_3:(y_1,y_2,y_3,y_4,y_5)\mapsto (y_1,y_2,y_3,y_1,y_5,0,0)$.
It's clear that $A_3$ is onto and $B_3$ is one-to-one ($1-1$), so options $c$ is true.
Now, for $B:{\mathbb{R}}^5\to {\mathbb{R}}^7$, if $\eta (B)\ge 0$, then by the Rank-Nullity Theorem, $\text{Rank}(B)\le 5-0=5$.
This implies that $B$ cannot be onto, so option $d$ is not true.