Math, asked by sallosahrma821, 12 hours ago

9.
Which of the following matrices are Hermitian and Skew Hermitian?
TI 2+1 3-1
4 2-15 +21
2 + 1 2
(b)
2+1 1 2-5i
3+1 4+ 13
15-21 2+51 2
21 -3 4
1 - 1 2 +371
-i
3 31 - 5
-- 0 61
4 5
1-2 +316i o
(c
)
(d
)​

Answers

Answered by nirmalkumar60
1

Answer:

number field, see Skew-symmetric matrix.

In linear algebra, a square matrix with complex entries is said to be skew-Hermitian or anti-Hermitian if its conjugate transpose is the negative of the original matrix.[1] That is, the matrix {\displaystyle A}A is skew-Hermitian if it satisfies the relation

{\displaystyle A{\text{ skew-Hermitian}}\quad \iff \quad A^{\mathsf {H}}=-A}{\displaystyle A{\text{ skew-Hermitian}}\quad \iff \quad A^{\mathsf {H}}=-A}

where {\displaystyle A^{\textsf {H}}}{\displaystyle A^{\textsf {H}}} denotes the conjugate transpose of the matrix {\displaystyle A}A. In component form, this means that

{\displaystyle A{\text{ skew-Hermitian}}\quad \iff \quad a_{ij}=-{\overline {a_{ji}}}}{\displaystyle A{\text{ skew-Hermitian}}\quad \iff \quad a_{ij}=-{\overline {a_{ji}}}}

for all indices {\displaystyle i}i and {\displaystyle j}j, where {\displaystyle a_{ij}}a_{ij} is the element in the {\displaystyle j}j-th row and {\displaystyle i}i-th column of {\displaystyle A}A, and the overline denotes complex conjugation.

Skew-Hermitian matrices can be understood as the complex versions of real skew-symmetric matrices, or as the matrix analogue of the purely imaginary numbers.[2] The set of all skew-Hermitian {\displaystyle n\times n}n\times n matrices forms the {\displaystyle u(n)}u(n) Lie algebra, which corresponds to the Lie group U(n). The concept can be generalized to include linear transformations of any complex vector space with a sesquilinear norm.

Note that the adjoint of an operator depends on the scalar product considered on the {\displaystyle n}n dimensional complex or real space {\displaystyle K^{n}}K^{n}. If {\displaystyle (\cdot \mid \cdot )}{\displaystyle (\cdot \mid \cdot )} denotes the scalar product on {\displaystyle K^{n}}{\displaystyle K^{n}}, then saying {\displaystyle A}A is skew-adjoint means that for all {\displaystyle \mathbf {u} ,\mathbf {v} \in K^{n}}{\displaystyle \mathbf {u} ,\mathbf {v} \in K^{n}} one has {\displaystyle (A\mathbf {u} \mid \mathbf {v} )=-(\mathbf {u} \mid A\mathbf {v} )}{\displaystyle (A\mathbf {u} \mid \mathbf {v} )=-(\mathbf {u} \mid A\mathbf {v} )}.

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