# Does an n by n Hermitian matrix always has n independent eigenvectors?

I am learning the MIT ocw 18.06 Linear Algebra, and I have learnt: an arbitrary $n×n$ matrix A with n independent eigenvectors can be written as $A=SΛS^{-1}$, and then for the Hermitian matrices, because the eigenvectors can be chosen orthonormal, it can be written as $A=QΛQ^H$ further.

I wonder does every $n×n$ Hermitian matrix has n independent eigenvectors and why? Thank you!

P.S. MIT 18.06 Linear Algebra Lecture 25: Symmetric Matrices and Positive Definiteness. You may wish to start from 4:20. From the course, I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why?

• You mean independent eigenvectors, not independent eigenvalues. – uniquesolution Feb 8 '18 at 12:04
• Oh, thanks, I have edited it. – Thomas Feb 9 '18 at 2:28
• It's on the Wikipedia page: en.wikipedia.org/wiki/Hermitian_matrix. – Robert Wolfe Feb 9 '18 at 4:25
• @Robert Thank you for the link, I had read the page though, and there wasn't an answer to my question. – Thomas Feb 9 '18 at 10:34
• Look here for a simple proof. – user Feb 11 '18 at 20:15

This is a theorem with a name: it is called the Spectral Theorem for Hermitian (or self-adjoint) matrices. As pointed out by Jose' Carlos Santos, it is a special case of the Spectral Theorem for normal matrices, which is just a little bit harder to prove.

Actually we can prove the spectral theorem for Hermitian matrices right here in a few lines.

We are going to have to think about linear operators rather than matrices. If $T$ is a linear operator on a finite dimensional complex inner product space $V$, its adjoint $T^*$ is another linear operator determined by $\langle T v, w\rangle = \langle v, T^* w \rangle$ for all $v, w \in V$. (Note this is a basis-free description.) $T$ is called Hermitian or self-adjoint if $T = T^*$.

Let $B$ be an $n$-by-$n$ complex matrix and $B^*$ the conjugate transpose matrix. Let $T_B$ and $T_{B^*}$ be the corresponding linear operators. Then $(T_B)^* = T_{B^*}$, so a a matrix is Hermitian if and only if the corresponding linear operator is Hermitian.

Let $A$ be a Hermitian linear operator on a complex inner product space $V$ of dimension $n$. We need to consider $A$--invariant subspaces of $V$, that is linear subspaces $W$ such that $A W \subseteq W$. We should think about such a subspace as on an equal footing as our original space $V$. In particular, any such subspace is itself an inner product space, $A_{|W} : W \to W$ is a linear operator on $W$, and $A_{|W}$ is also Hermitian. If $\dim W \ge 1$, $A_{|W}$ has an least one eigenvector $w \in W$ -- because any linear operator at all acting on a (non-zero) finite dimensional complex vector space has at least one eigenvector.

The basic phenomenon is this: Let $W$ be any invariant subspace for $A$. Then $W^\perp$ is also invariant under $A$. The reason is that if $w \in W$ and $x \in W^\perp$, then $$\langle w, A x\rangle = \langle A^* w , x \rangle = \langle A w, x \rangle = 0,$$ because $Aw \in W$ and $x \in W^\perp$. Thus $A x \in W^\perp$.

Write $V = V_1$. Take one eigenvector $v_1$ for $A$ in $V_1$. Then $\mathbb C v_1$ is $A$--invariant. Hence $V_2 = (\mathbb C v_1)^\perp$ is also $A$ invariant. Now just apply the same argument to $V_2$: the restriction of $A$ to $V_2$ has an eigenvector $v_2$ and the perpendicular complement $V_3$ to $\mathbb C v_2$ in $V_2$ is $A$--invariant. Continuing in this way, one gets a sequence of mutually orthogonal eigenvectors and a decreasing sequence of invariant subpsaces, $V = V_1 \supset V_2 \supset V_3 \dots$ such that $V_k$ has dimension $n - k + 1$. The process will only stop when we get to $V_n$ which has dimension 1.

• You really hit the nail on the head as I'm afraid my question is too unusual to be understand. May I repeat you answer in my imprecise words? – Thomas Feb 9 '18 at 12:42
• Regard A as an operator on V, its invariant subspaces contains at least one eigenvector, their orthogonal complement are invariant subspaces too because A is a Hermitian matrix, we start from $\mathbb C^n$, every time we take a eigenvector from the space, it on a line which is a invariant subspace, and the space left which like a plane perpendicular to the line is also a invariant subspace, and we can take n times (i.e. n orthogonal/independent eigenvectors). Do I understand it correctly? – Thomas Feb 9 '18 at 12:47
• I still have some trouble with this, what if I take an eigenvector from the last space? As we know "Let $W$ be any invariant subspace for $A$. Then $W^⊥$ is also invariant under $A$", I take the eigenvector out, then there will be only zero vector left, and I think it really orthogonal to that eigenvector we take, is zero vector an invariant subspace? It may be, but "invariant space has at least one eigenvector" would be false. This "Let $W$ be any invariant subspace for $A$. Then $W^⊥$ is also invariant under $A$" makes me confused. – Thomas Feb 9 '18 at 13:43
• When you have reached $V_n$, you already have your collection of $n$ mutually orthogonal eigenvectors. You can't go further and you don't need to go further. I corrected the answer at one point to say that a linear operator on a non-zero finite dimensional complex vector space has an eigenvector. – fredgoodman Feb 9 '18 at 14:19
• It's important in the second paragraph that "acting on" means "mapping it into itself". – Ian Feb 9 '18 at 15:26

Yes, it has. That is due to the spectral theorem: every normal $n\times n$ matrix is diagonalizable. And Hermitian matrices are normal.

• From my understanding, the factorization $A=QΛQ^H$ (i.e., the spectral theorem) is the special form of $A=SΛS^{-1}$ when A is a Hermitian matrix, and $A=SΛS^{-1}$ exists under the situation that there is a full set of independent eigenvectors. I wonder, if it is, why a Hermitian matrix is always diagonalizable? – Thomas Feb 9 '18 at 2:58
• @Thomas Because the spectral theorem says so. – José Carlos Santos Feb 9 '18 at 6:30
• From the course (I add the link into the question now), I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why? – Thomas Feb 9 '18 at 10:57
• @Thomas Every Heritian matrix is normal and every normal matrix is diagonalizable. – José Carlos Santos Feb 9 '18 at 11:08
• I had not learnt the concept of normal matrix from the course. I search for it now, and I realize that all matrices can be written as $A=QTQ^H$ with triangular T, so $A^HA=AA^H$ would be $T^HT=TT^H$, for a 2 by 2 matrix, it's easy to prove the upper right corner entry of such T is 0, extend it to size n, and we prove that T is diagonal. Is that a right poof? Thank you for your patience. – Thomas Feb 9 '18 at 11:58