Answer to Question #152789 in Linear Algebra for Sourav Mondal

Question

True or false and give reason for your answer


i) If all the Eigen values of a matrix are real, then the
matrix must be symmetric.
ii) If V and W are real vector spaces, with T : V --> W
being a linear transformation then V/kerT isomorphic to W
iii) There are at least three different unitary matrices of order 2.
iv) There are two subspaces U and W of R3 such that U(intersection) W is empty.
v) If the determinant of a matrix is zero, then the matrix cannot be diagonalised.

Answer 1

i) False

Reason:-

Let a matrix $A = \begin{bmatrix}1&2\\0&0\end{bmatrix}$

Clearly A is not symmetric matrix.

Now let $\lambda$ be an eigen value of A.

So, $det(\lambda I-A)=0$ $\Rightarrow \lambda=0$ or, 1

$\therefore$ Eigen values of A are real but A is not symmetric.

ii) False

Reason:-

Let $V=W=R^2$

And $T: V\to W$ such that $T(\begin{bmatrix}a\\b\end{bmatrix} ^T)=\begin{bmatrix}a\\0\end{bmatrix} ^T$

It is easy to see that $T$ is well-defined and $T$ is a Linear Transformation.

[As $v_1=v_2 \Rightarrow T(v_1)=T(v_2)$ and $T(v_1+cv_2)=T(v_1)+cT(v_2)$]

Now $Im(T) = \{T(v) : v\in V\} = \{\begin{bmatrix}a\\0\end{bmatrix}^T : a$ is a real number$\}$

Clearly $Im(T) \subset W \Rightarrow |Im(T)|<|W|$

Now, From $1st$ isomorphism theorem,

$V/ker(T) \cong Im(T)$

$\Rightarrow |V/ker(T)| = |Im(T)| < |W|$

$\Rightarrow |V/ker(T)| < |W|$

$\exists$ no bijection between $V/ker(T)$ and $W$

So, $V/ker(T)$ is not isomorphic to $W$

iii) True

Reason :-

Let 3 matrices of order 2 are A, B and C, where

$A = \begin{bmatrix}1&0\\0&1\end{bmatrix}$ , $B = \begin{bmatrix}0&i\\i&0\end{bmatrix}$ and $C = \begin{bmatrix}i/2&\sqrt{3}/2\\\sqrt{3}/2&i/2\end{bmatrix}$

It is easy to check, $A(A^\theta)^T=B(B^\theta)^T=C(C^\theta)^T=I$

Where $M^\theta$ is conjugate of M.

$\therefore A,B,C$ are Unitary.

So, A,B,C are 3 different unitary matrices of order 2.

$\therefore \exists$ at least 3 different unitary matrices of order 2.

iv) False

Reason :-

Let $U, W$ are two subspaces of $\R^3$ .

let $u\in U$ and c be a scalar. $\Rightarrow cu\in U$

Take $c=0$ $\therefore 0\in U$

Similarly $0\in W$

So, $0\in U\cap W$

$\therefore$ $U\cap W$ is not empty

v) False

Reason :-

Let $A = \begin{bmatrix}1&2\\0&0\end{bmatrix}$ be a matrix of order 2.

Clearly $det(A)=0$

Let $P = \begin{bmatrix}0&1\\1&2\end{bmatrix}$

$\therefore P^{-1} = \begin{bmatrix}-2&1\\1&0\end{bmatrix}$

Now $PAP^{-1}= \begin{bmatrix}0&1\\1&2\end{bmatrix}\begin{bmatrix}1&2\\0&0\end{bmatrix}\begin{bmatrix}-2&1\\1&0\end{bmatrix}=\begin{bmatrix}0&0\\0&1\end{bmatrix}$

$\therefore PAP^{-1}$ is a diagonal matrix. $\Rightarrow A$ is diagonalizable.

So, the Determinant of $A$ is but $A$ is diagonalizable.