CSIR -NET Linear algebra Solved question : part B : June 2016

1. Given a $n\times n$ matrix $B$ defined by $e^B$ by 
$$e^B=\sum_{j=0}^\infty \frac{B^j}{j!}$$
Let p be the characteristic polynomial of $B$ Then the matrix $e^{P(B)}$ is

1. $I_{n\times n}$
2. $0_{n\times n}$
3. $e \times I_{n\times n}$
4. $\pi \times I_{n\times n}$

Solution:

Since every characteristic polynomial is satisfied by the matrix. (caley hamilton Theorem)

So, $P(B)=0$ So , $e^0=I_{n\times n}$

So, 1st option is correct.

2. Let $x=(x_1,x_2,x_3), y=(y_1,y_2,y_3)\in \mathbb{R}^3$ be linearly independent. 
Let 
$$\delta_1 =x_2y_3-y_2x_3$$  
$$\delta_2 =x_1y_3-y_1x_3$$  
$$\delta_3 =x_1y_2-y_1x_2$$  
 If V is the span of $x,y$ then,

1. $V=\{(u,v,w):\delta_1u-\delta_2v+\delta_3w=0\}$
2. $V=\{(u,v,w):-\delta_1u+\delta_2v+\delta_3w=0\}$
3. $V=\{(u,v,w):\delta_1u+\delta_2v-\delta_3w=0\}$
4. $V=\{(u,v,w):\delta_1u+\delta_2v+\delta_3w=0\}$

Solution:

Consider this determinant,
$$\begin{vmatrix} u & v & w  \\ x_1 &x_2&x_3 \\ y_1&y_2&y_3 \\   \notag \end{vmatrix}$$
If $\{u,v,w\}$ spans $\mathbb{R}^3$ thus also spans $x,y$ and thus the value of this determinant is zero.
The expression written in the first option is the value of this determinant
so, 1st option is correct.


3. Let $A=\begin{bmatrix} 1&0\\0&-1\end{bmatrix}$ .Let $f:\mathbb{R}^2\times \mathbb{R}^2\to \mathbb{R}$ be defined by $f(v,w)=w^TAv$,
Pick the correct statement from below:

1. There exists an eigenvector $v$ of $A$ such that $Av$ is perpendicular to $v$
2. The set $\{v\in \mathbb{R}^2 |f(u,v)=0\}$ is a nonzero subspace of $\mathbb{R}^2$
3. If $u,v\in \mathbb{R}^2$, are non-zero vectors such that $f(v,v)=0=f(w,w)$, then $v$ is a scala  multiple of $w$
4. For every $v\in \mathbb{R}^2$, there exists a nonzero $w\in \mathbb{R}^2$ such that $f(v,w)=0$

Solution: 

For every $(u,v)\in \mathbb{R}$ there will exists $(v,u)\in \mathbb{R}^2$ such that

$\begin{bmatrix}u&v\end{bmatrix}\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ $\begin{bmatrix}v\\u\end{bmatrix}=0$

So,

4th option is correct.

4. Let $A$ be a $n\times m$  matrix and $b$ be $n\times 1$ vector (with real entries), suppose the equation $Ax=b, x\in \mathbb{R}^m$ admits a unique solution, Then we can conclude that

1. $m\geq n$
2. $n\geq m$
3. $n=m$
4. $n>m$

Solution:Let $A_{n\times m}$ matrix has rank $r$ then there exists unique solution in two cases

• $n=m=r$
• $r=m<n$

Combining two cases, We get $n\geq m$

So, 2nd option is correct.

5.Let $V$ be the vactor space of all real polynomials of degree $\leq 10$. Let $Tp(x)=p'(x)$ for $p\in V$ be a linear transformation from $V$ to $V$. Consider the basis $\{1,x,x^2,\cdots x^{10}\}$ of $V$. Let $A$ be the matrix of $T$ with respect to this basis. Then 
Trace$A$=1
Det $A$=0
There is no $m\in \mathbb{N}$ Such that $A^m=0$
$A$ has a nonzero eigenvalue

Solution:
The matrix of above linear transformation is
$\begin{bmatrix}0&1&0&0&\cdots&0\\0&0&2&0&\cdots&0\\0&0&0&3&\cdots&0\\ \cdots&\cdots&\cdots&\cdots&\cdots\\0&0&0&0&\cdots&0\end{bmatrix}$
This is an upper triangular matrix so, It's determinant is equal to zero.
So, 2nd option is correct.

6. Let $A$ be a $n\times n$ real symmetric non-singular matrix. Suppose there exists $x\in \mathbb{R}^n$ such that 
$x'Ax<0$
Then we can conclude that

1. $det(A)<0$
2. $B=-A$ is positive definite
3. $\exists y\in \mathbb{R}^n \quad y'A^{-1}y<0$
4. $\forall y\in \mathbb{R}^n \quad y'A^{-1}y<0$

Solution:

Option 1:
Consider,
 $A=\begin{pmatrix}-1&0\\0&-1\end{pmatrix}$ Clearly $\det(A)>0$
But $\exists x=\begin{bmatrix}1&0\end{bmatrix}$ such that $\begin{bmatrix}1&0\end{bmatrix} \begin{bmatrix}-1&0\\0&-1\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix}<0$
So, 1st option is not correct.

Option 2:
We can not coclude that $A$ is negative definite since criteria for being negative definite is:
$\forall x\in \mathbb{R}^n$ such that
$x'Ax<0$
But here this condition is not given.

Option 3:
Take now $\;v:=Ax\;$ , then

$$v^tA^{-1}v=x^tA^tA^{-1}Ax=x^tA^tx=x^tAx<0$$

option 4: 
$A:=\begin{pmatrix}-1&0\\0&1\end{pmatrix}\;\;\text{is a counterexample to 4th option  since}\;\; (0\;1)A\binom01>0$

$\text{but}\;\;(1\;0)A\binom10<0\;\;\text{and}\;\;A^{-1}=A$

So, 3rd option is correct.