4,493 views

1 Answer

Best answer
1 votes
1 votes

Given that $M=\begin{bmatrix} 2&3+2i &-4 \\ 3-2i&5 &6i \\-4 &-6i &3 \end{bmatrix}$

If $M$ is Harmition Matrix then $M=M^{\theta}$

find the $M^{\theta}:$

         first, find the conjugate of $M:$

                                                   $\overline{M}=\begin{bmatrix} 2&3-2i &-4 \\ 3+2i&5 &-6i \\-4 &6i &3 \end{bmatrix}$

        after that find the transpose of $\overline{M}:$

                                                   $M^{\theta}=\begin{bmatrix} 2&3+2i &-4 \\ 3-2i&5 &6i \\-4 &-6i &3 \end{bmatrix}$

Where $M^{\theta}$ is called Transposed conjugate matrix.

                              Here $M=M^{\theta}$

                           So,$M$ is called Harmition Matrix.

$iM=\begin{bmatrix} 2i&3i-2 &-4i \\ 3i+2&5i &-6 \\-4i &6 &3i \end{bmatrix}$

$\overline{iM}=\begin{bmatrix} -2i&-3i-2 &-4i \\ -3i+2&-5i &-6 \\4i &6 &-3i \end{bmatrix}$

$iM^{\theta}=\begin{bmatrix} -2i&-3i+2 &4i \\ -3i-2&-5i &6 \\4i &-6 &-3i \end{bmatrix}$

$iM^{\theta}=-\begin{bmatrix} 2i&3i-2 &-4i \\ 3i+2&5i &-6 \\-4i &6 &3i \end{bmatrix}$

Here $iM=-iM^{\theta}$

 So,$iM$ is called Skew-Harmition Matrix.

$\Rightarrow$ Every real symmetric matrix $(A=A^{T})$ is Hermitian Matrix$(A=A^{\theta})$

$\Rightarrow$ Eigenvalues of a real symmetric matrix are real numbers.

$\Rightarrow$ All the eigenvalues of the Hermitian matrix are real numbers.

$\Rightarrow$ The Eigen value of skew-symmetric matrx$(A=-A^{T})$ are either purely imaginary or zeros.

$\Rightarrow$ The eigenvalues of skew-Hermitian matrix$(A=-A^{\theta})$ are either pure imaginary or zeros.

So $(Q)$ and $(R)$ is the right choice.

selected by

Related questions

0 votes
0 votes
0 answers
1
Balaji Jegan asked Dec 11, 2018
181 views
1 votes
1 votes
0 answers
2
Balaji Jegan asked Dec 11, 2018
204 views
1 votes
1 votes
1 answer
3
Balaji Jegan asked Dec 11, 2018
3,605 views
0 votes
0 votes
0 answers
4
Balaji Jegan asked Dec 11, 2018
180 views
The number of 5-Sylow subgroup(s) in a group of order 45 is ____