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Explanation & Answer
Let A be a Hermitian matrix.
Then, by definition:
- A=A∗
where A∗ denotes the conjugate transpose of A.
Let λ be an eigenvalue of A.
Let v be an eigenvector corresponding to the eigenvalue λ.
By definition of eigenvector:
- Av=λv
Left-multiplying both sides by v∗, we obtain:
- (1):v∗Av=v∗λv=λv∗v
Firstly, note that both v∗Av and v∗v are 1×1-matrices.
Now observe that, using Conjugate Transpose of Matrix Product: General Case:
- (v∗Av)∗=v∗A∗(v∗)∗
As A is Hermitian, and (v∗)∗=v by Double Conjugate Transpose is Itself, it follows that:
- v∗A∗(v∗)∗=v∗Av
That is, v∗Av is also Hermitian.
By Product with Conjugate Transpose Matrix is Hermitian, v∗v is Hermitian.
So both v∗Av and v∗v are Hermitian 1×1 matrices.
Now suppose that we have for some a,b∈C:
- v∗Av=[a]
- v∗v=[b]
Note that b≠0 as an Definition:Eigenvector is non-zero.
By definition of a Hermitian matrix:
- a=a¯ and b=b¯
where a¯ denotes the complex conjugate of a.
By Complex Number equals Conjugate iff Wholly Real, it follows that a,b∈R, that is, are real.
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