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I am trying to prove that Hermitian Matrices are diagonalizable.

I have already proven that Hermitian Matrices have real roots and any two eigenvectors associated with two distinct eigen values are orthogonal. If $A=A^H;\;\;\lambda_1,\lambda_2$ be two distinct eigenvalues and $v_1,v_2$ be two eigenvectors associated with them. $$Av_1=\lambda_1v_1\Rightarrow v_1^HAv_1=\lambda_1v_1^Hv_1\Rightarrow \lambda_1=\lambda_1^*.$$ Similarly, $$Av_1=\lambda_1v_1\Rightarrow v_2^HAv_1=\lambda_1v_2^Hv_1\Rightarrow v_1^H\lambda_2v_2=\lambda_1v_1^Hv_2\Rightarrow v_1^Hv_2=0$$

However, I am not aware of Spectral Theorem. Given this circumstances, how can I prove that Hermitian Matrices are diagonalizable? It should follow from above but the only sufficient condition of a matrix being diagonalizable is to have $dim(A)$ distinct eigenvalues, or existance of $P$ such that $A=P^{-1}DP$. I am not sure how this follows from above conditions.

glS
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    Induction. Note that in general a subspace $F$ is $A$ invariant if and only if its orthogonal $F^\perp$ is $A^*$-invariant. – Julien Apr 16 '13 at 14:28
  • Induce on number of distinct eigenvalues? How would be the induction step? –  Apr 16 '13 at 14:38
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    Induction on $n$ the size of the matrix. See $A$ as a self-adjoint operator. Prove that every self-adjoint operator is diagonalizable in an orthonormal basis. Trivial for $n=1$. Assume true for $n-1$. Then take a $n\times n$ hermitian matrix. Take an eigenvalue $\lambda$ and a corresponding eignevector $x$ of norm $1$. Then $\mathbb{C}^n=\mathbb{C}x\oplus x^\perp$ is an orthogonal decomposition which is $A$-invariant. Apply induction hypothesis to the restriction of $A$ to $x^\perp$, which is still self-adjoint. – Julien Apr 16 '13 at 14:46
  • @julien Could you may be, check my solution? –  Apr 16 '13 at 16:55

2 Answers2

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The proof of this property is not so easy as those of the basic properties of eigenvalues and eigenvectors. It can be shown by induction, or by explicit construction (see eg here)

I like to visualize the property in this way:

We know that an hermitian matrix with $n$ distict eigenvalues has $n$ eigenvectors that are not only LI (as in general matrices) but, more than that, orthogonal. We also know that this matrix is diagonalizable, with unitary $U$ (both properties are easy to prove).

Now, if our hermitian matrix happens to have repeated (degenerate) eigenvalues, we can regard it as a perturbation of some another hermitian matrix with distinct eigenvalues. By a continuity argument, we should see that the matrix perturbation than transforms different (but perhaps close) eigenvalues into coincident ones, cannot make the orthogonal eigenvectors linearly dependent.

Put in other way: an hermitian matrix $A$ with repeated eigenvalues can be expressed as the limit of a sequence of hermitian matrices with distinct eigenvalues. Because all members of the sequence have $n$ orthogonal eigenvectors, by a continuity argument, they cannot end in LD eigenvectors.

This approach leads to a nice intuition, IMO, and it can be formalized. But for a formal proof the other methods are to be preferred.

PinkyWay
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leonbloy
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    Any matrix is a perturbation of a matrix with pairwise distinct eigenvalues. But not every matrix is diagonalizable. You need to be more precise if you want your "intuitive" argument to make sense. But maybe I just don't understand your "by a continuity argument..." – Julien Apr 16 '13 at 18:24
  • @julien: For general (non-hermitian) matrices, the small perturbation that (say) makes $\lambda_1$ and $\lambda_2$ equal can well make $p_1$ and $p_2$ (originally LI) colineal (LD); the unperturbed eigenvectors were LI but "almost LD", as the unperturbed eigenvalues were distict but "almost equal". This cannot happen in the hermitian case. You can, continuosly go from LI to LD, you cannot go continously from orthogonal to LD. – leonbloy Apr 16 '13 at 18:29
  • How do you prove the existence of the perturbation with pairwise distinct eigenvalues without diagonalizing the original matrix first? – Julien Apr 16 '13 at 18:44
  • @julien I would not into the way of formally proving it (I mentioned that in the answer), but it seems quite convincing to me; repeated eigenvalues correspond to degenerate roots of the c. polynomial, which should be in a zero-measure set in the space of polynomial coefficients. – leonbloy Apr 16 '13 at 19:03
  • Yes, that's a nice heuristic argument anyway. For your continuity argument, compactness of orthogonal matrices is crucial. Maybe it would be helpful to mention it. Indeed, fix $M$ orthogonal. Take a sequence of orthogonal matrices $M_n$ with pairwise distinct eigenvalues which converge to $M$. Diagonalize them $M_n=P_nD_nP_n^$ in an orthonormal basis. Up to an extraction, this converges to $PDP^$ by compactness. So you're not far to have a formal proof. – Julien Apr 16 '13 at 19:11
  • Unfortunately the reference you link to for a proof has several errors, e.g. it says "the eigenvectors of a symmetric matrix are mutually orthogonal" which isn't true (e.g. (1,0) and (1,1) are both eigenvectors of the zero matrix and aren't orthogonal), and sloppiness / details like this is exactly what matters in this context. – Max Sep 25 '20 at 14:07
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    @Max Bro, people automatically exclude the trivial case of zero matrix... And in fact that the eigenvectors of a non-zero symmetric matrix are mutually orthogonal is true... See here. – Sam Wong Apr 27 '21 at 09:25
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Let $A$ be defective (= non diagonalizable). Consequently, it has a defective eigenvalue $\lambda$, with an algebraic multiplicity bigger than its geometric multiplicity. Then, there is a generalized eigenvector $\vec{u}_{m}$ of $A$, of rank $m>1$, such that $\left(A-\lambda I\right)^{m}\vec{u}_{m}=\vec{0}$ (proof). Furthermore, $\vec{u}_{m}$ generates a Jordan chain of generalized eigenvectors of $A$: For every natural number $n$, $0<n<m$, there is a generalized eigenvector $\vec{u}_{n}$, belonging to the chain, such that $\vec{u}_{n}=\left(A-\lambda I\right)^{m-n}\vec{u}_{m}\neq\vec{0}$.

For $n=2$, we have: $\vec{u}_{2}=\left(A-\lambda I\right)^{m-2}\vec{u}_{m}$. Then: $$\left(A-\lambda I\right)^{m}\vec{u}_{m}=\vec{0}\Rightarrow\left(A-\lambda I\right)^{2}\left(A-\lambda I\right)^{m-2}\vec{u}_{m}=\vec{0}\Rightarrow\left(A-\lambda I\right)^{2}\vec{u}_{2}=\vec{0}$$ However, if $A$ is hermitian, then $A=A^{H}$ και $\left(A-\lambda I\right)^{H}=A^{H}-(\lambda I)^{H}=A-\lambda^{*}I^{H}$ (where $\lambda^{*}$ is the complex conjugate of $\lambda$). But, since hermitian matrices have real eigenvalues, $\lambda^{*}=\lambda$ and $I^{H}=I$, because the identity matrix is hermitian. So, $\left(A-\lambda I\right)$ is also hermitian: $\left(A-\lambda I\right)=\left(A-\lambda I\right)^{H}$. Thus: $$\left(A-\lambda I\right)^{2}\vec{u}_{2}=\left(A-\lambda I\right)\left[\left(A-\lambda I\right)\vec{u}_{2}\right]=\left(A-\lambda I\right)^{H}\left[\left(A-\lambda I\right)\vec{u}_{2}\right]=\vec{0}$$ Multiplying from the left, with $\vec{u}_{2}^{H}$, both sides of the last equality, we get: $$\left[\vec{u}_{2}^{H}\left(A-\lambda I\right)^{H}\right]\left[\left(A-\lambda I\right)\vec{u}_{2}\right]=\left[\left(A-\lambda I\right)\vec{u}_{2}\right]^{H}\left[\left(A-\lambda I\right)\vec{u}_{2}\right]=\left|\left(A-\lambda I\right)\vec{u}_{2}\right|^{2}=\vec{u}_{2}^{H}\vec{0}=0$$ This means that $\left(A-\lambda I\right)\vec{u}_{2}=\vec{u}_{1}=\vec{0}$, which contradicts the condition stated above, that, for every $n\in\mathbb{N}$, $1\leq n<m$, $\vec{u}_{n}=\left(A-\lambda I\right)^{m-n}\vec{u}_{m}\neq\vec{0}$. So, either $A$ is not defective, or is not hermitian.

Aris Makrides
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