Something about hermitian matrixes

[twoflower]

Hi all,

I don't understand to one part of proof of this theorem:

All eigenvalues of each hermitian matrix A are real numbers and, moreover, there exists unitary matrix R such, that

$$R^{-1}AR$$

is diagonal

Proof: By induction with respect to n (order of matrix A)

For n = 1 it's obvious.
Suppose that the theorem holds for 1, 2, ..., n-1
We know that $\exists$ eigenvalue $\lambda$ and appropriate eigenvector $x \in \mathbb{C}$.
Using Steinitz's theorem, we can extend $x$ to orthonormal base of $\mathbb{C}^{n}$.
Suppose that $||x|| = 1$ and construct matrix $P_n$ from vectors of this base ($P_n$ will have these vectors in its columns).

$P_n$ is unitary $\Leftarrow P_{n}^{H}P_n = I$, because standard inner product of two different vectors in the orthonormal base is zero and inner product of two identical vectors is 1.

This holds:

$$\left(P_{n}^{H}A_{n}P_{n}\right)^{H} = P_{n}^{H}A_{n}^{H}\left(P_{n}^{H}\right)^{H} = P_{n}^{H}A_{n}P_{n}$$

Last line is what I don't understand, probably it's trivial but I can't see that

$$\left(P_{n}^{H}A_{n}P_{n}\right)^{H} = \left(P_{n}^{H}\right)^{H}A_{n}^{H}P_{n}^{H} = P_{n}^{H}A_{n}^{H}\left(P_{n}^{H}\right)^{H}$$

(the second equality)

Thank you for the explanation.

 

[matt grime]

because you're forgetting that taking the daggeer reverses the order of the matrices i'l; use start instead, but (AB)*=B*A*

the second equality as you have it is wrong, but then it isn't supposed to be true.

 

[twoflower]

because you're forgetting that taking the daggeer reverses the order of the matrices i'l; use start instead, but (AB)*=B*A*

the second equality as you have it is wrong, but then it isn't supposed to be true.

Thank you a lot Matt, I was looking at it for ten minutes and it's as simple as normal transposition

 

[dextercioby]

Well,the first part (the real values of eigenvalues of hermitean operators) can be proven quite easily for a hermitean linear operator defined on a dense everywhere subset of a separable Hilbert space.

Daniel.

 

[dextercioby]

The key to the second part is to remark that that matrix

$$M^{\dagger}AM \ ,\ M\in U(n,\mathbb{C})$$

is hermitean,which means that the linear operator associated is hermitean.A hermitean linear operator in a finite dimensional complex Hilbert space admits a spectral decomposition (moreover,the spectrum is purely discrete),which means that the operator $M^{\dagger}AM$ has zero off-diagonal matrix elements.

Daniel.

 

[twoflower]

The key to the second part is to remark that that matrix

$$M^{\dagger}AM \ ,\ M\in U(n,\mathbb{C})$$

is hermitean,which means that the linear operator associated is hermitean.A hermitean linear operator in a finite dimensional complex Hilbert space admits a spectral decomposition (moreover,the spectrum is purely discrete),which means that the operator $M^{\dagger}AM$ has zero off-diagonal matrix elements.

Daniel.

Thank you Daniel for this explanation, but I don't have a clue what Hilbert space is (I only heard of it) and what hermitean linear operator is.

However, I've been studying the proof on and I again encountered place I don't understand to.

If I continue from where I finished my initial post:

...

And thus $P_{n}^{H}A_{n}P_{n}$ is hermitian matrix.

Next,

$$\left( \begin{array}{cc} \lambda & 0 ... 0 \\ 0 & A_{n-1} \\ 0 \end{array} \right)$$

Because this matrix is equal to its hermitian transposition, $\lambda \in \mathbb{R}$

// I'm not sure why this matrix is here and whether it should mean that it is the matrix $P_{n}^{H}A_{n}P_{n}$, I really don't know...anyway, let's continue

From induction presumption $\exists$ unitary matrix $R_{n-1}$ such, that

$$R_{n-1}^{-1}A_{n-1}R_{n-1} = D_{n-1}$$

Let's take

$$S = \left(\begin{array}{cc} 1 & 0 ... 0 \\ 0 & R_{n-1} \\ 0 \end{array} \right)$$

$$R_n = P_{n}S$$

S is unitary, as well as $P_{n}$. Is also their product unitary? (In another words, is product of two unitary matrixes unitary matrix?) Let's see.

$$R_{n}^{H}R_{n} = \left(P_{n}S\right)^{H}P_{n}S = S^{H}P_{n}^{H}P_{n}S = I$$

So, it holds that $R_{n}$ is also unitary. Is $R_{n}$ the matrix we're looking for?

$$R_{n}^{-1}A_{n}R_{n} = \left(P_{n}S\right)^{H}AP_{n}S = S^{H}P_{n}^{H}AP_{n}S = \left(\begin{array}{cc} 1 & 0 ... 0 \\ 0 & R_{n-1}^{H} \\ 0 \end{array} \right) \left(\begin{array}{cc} \lambda & 0 ... 0 \\ 0 & A_{n-1} \\ 0 \end{array} \right) \left(\begin{array}{cc} 1 & 0 ... 0 \\ 0 & R_{n-1} \\ 0 \end{array} \right) = \left(\begin{array}{cc} \lambda & 0 ... 0 \\ 0 & D_{n-1} \\ 0 \end{array} \right) = D$$

Q.E.D

What I don't understand is that according to this,

$$P_{n}^{H}AP_{n} = \left(\begin{array}{cc} \lambda & 0 ... 0 \\ 0 & A_{n-1} \\ 0 \end{array} \right)$$

Why?

Thank you.