Question

If A is hermitian matrix and B skew hermitian matrix , prove that (B + iA) is skew hermitian matrix , full proof​

Answer 1

Step-by-step explanation:

If HH be a Hermitian matrix, prove that detHdetH is real number.

If SS be a skew Hermitian matrix of order nn, prove that

(i). if nn be even, then detSdetS is real number;

(ii). if nn be odd, then detSdetS is a purely imaginary number or zero.

Attempt: 1. Let H=P+iQH=P+iQ be a Hermitian matrix, where P,QP,Q are real matrices. Then H¯t=H⟹Pt−iQt=P+iQ⟹Pt=PH¯t=H⟹Pt−iQt=P+iQ⟹Pt=P and Qt−QQt−Q. How can I show that detHdetH is real

Answer 2

Answer:

The matrix $(B+iA)$ is a Skew-Hermitian matrix.

Step-by-step explanation:

Hermitian matrices: A square matrix $A=[a_{ij}]$ is said to be Hermitian if $A^{\theta}=A$.

Skew-Hermitian matrices: A square matrix $A=[a_{ij}]$ is said to be Skew-Hermitian if $A^{\theta}=-A$.

Here, $A^{\theta}$ is the $transposed$ $conjugate$ of the matrix $A$.

According to the question,

Since $A$ is a Hermitian matrix.

⇒ $A^{\theta}=A$ . . . . . (1)

Also, $B$ is a Skew-Hermitian matrix.

⇒ $B^{\theta}=-B$ . . . . . (2)

To prove: $(B+iA)$ is a Skew-Hermitian matrix, i.e.,

$(B+iA)^{\theta}=-(B+iA)$

Consider the matrix as follows:

$(B+iA)$

Taking the conjugate as follows:

$\overline{(B+iA)}$

Simplify as follows:

$\overline{(B+iA)}=\overline{B}-i \overline{A}$

Take transpose both the sides.

⇒ $(\overline{B+iA})^T=(\overline{B})^T-(i \overline{A})^T$

⇒ $({B+iA})^{\theta}={B}^{\theta}-i{A}^{\theta}$ . . . . . (3) (Transpose conjugate)

Now, substitute the value of $B^{\theta}$ and $A^{\theta}$ in (3) as follows:

⇒ $({B+iA})^{\theta}=-B-iA$ (From (1) and (2))

⇒ $({B+iA})^{\theta}=-(B+iA)$

Thus, $(B+iA)$ is Skew-Hermitian matrix.

Hence proved.

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