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I have read different definitions, or properties, of a Hermitian matrix, and still am not sure if I have a sufficient number of properties to define a Hermitian matrix.

Suppose the following is true about Hermitian Matrices.

$M\vec{a} = \lambda\vec{a}$

where all scalars $\lambda$s (eigenvalues) are real.

Lets just use a 2D matrix.

If $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, then $det\begin{bmatrix} a-\lambda & b \\ c & d-\lambda \end{bmatrix}=0$.

This means that (by quadratic solution to characteristic equation) $$\lambda = \frac{(a+d)\pm\sqrt{(a+d)^2-4*(a^2-cb)}}{2}$$

For lambda to be real (in both + and - cases), there are certain restrictions that must be obeyed. However they are not as restrictive as the definition of a Hermitian matrix.

One restriction is that $$Im(a+d)=0$$ (imaginary components cancel). Another is $$(a+d)^2-4*(a^2-cb) >= 0 $$ (>= or just >?) and simultaneously $$Im((a+d)^2-4*(a^2-cb))=0$$

Even though Hermitian matrices do not allow imaginary diagonal elements, I'm pretty sure I could think of a matrix that would satisfy these conditions with imaginary $a$ and $d$ elements.

I'm guessing I'm not understanding the properties of a Hermitian matrix. What am I missing?

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  • $\begingroup$ These are just necessary conditions for a matrix to be Hermitian. If they also were sufficient, then these properties would define a Hermitian matrix. $\endgroup$ – Svetoslav Sep 10 '15 at 16:40
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By definition, a complex square matrix $A$ is hermitian if and only if $A^H = A$. Where $A^H = \bar A^T$ denotes the conjugate transposed of $A$.

It is the result of the spectral theorem, that all eigenvalues of a hermitian matrix are real and there exists eigenvector basis.

The matrix $$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$ has clearly only real eigenvalues, but is not hermitian.

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  • $\begingroup$ Your matrix only has one double-degenerate eigenvalue, and it is zero. Maybe Hermitian matrices need to have certain conditions on the eigenvalues? I'm guessing it wouldn't help. What defines if "an eigenvector basis exists"? $\endgroup$ – OrangeSherbet Sep 10 '15 at 17:52
  • $\begingroup$ @OrangeSherbet you could change the diagonal to what ever you want. the point remains, real eigenvalues does not imply hermitian $\endgroup$ – user251257 Sep 10 '15 at 17:53
  • $\begingroup$ Sounds good, I just don't understand what defines an eigenvector basis existing or not, which is probably sad. $\endgroup$ – OrangeSherbet Sep 10 '15 at 17:56
  • $\begingroup$ @OrangeSherbet: There is a basis of eigenvectors if and only if $A$ is diagonalizable if and only if the dimensions of the eigenspaces agree with multiplicities of the eigenvalues in the characteristic polynomial. $\endgroup$ – user251257 Sep 10 '15 at 18:05
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After looking in my Griffiths quantum mechanics book, I think I now realize that the Hermitian Operator defines what a Hermitian Matrix is. Anyways, I wonder if the following properties is enough to define what a Hermitian Matrix is.

1) All eigenvalues are real

2) The eigenvectors are mutually orthogonal (Inner product between any of them yields $0$).

3) Eigenvectors span all of the space

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  • $\begingroup$ It would be better if you combine 2) und 3) to "there is an orthonormal basis of eigenvectors". Yes, under these conditions $A$ is hermitian. Also, it would also be better if you modify your question or ask a new question instead of posting a question as answer. $\endgroup$ – user251257 Sep 10 '15 at 20:09

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