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.