Span of an orthogonal basis of an hermitian matrix constructed using eigenvectors

Question

If $A \in \mathbb{C}^{n \times n}$ is hermitian, then all it's eigenvalues are real and eigenvectors of different eigenspaces are orthogonal. Left to prove is the fact that there exists an orthonormal Basis $\{ \ v_1 \ v_2 \ \dots \ v_n \ \}$ of eigenvectors of $A$.

In every eigenspace, I can construct an orthonormal basis using the Gram-Schmidt algorithm. Since the eigenvectors of different eigenspaces are orthogonal, all left to prove is that this basis spans $\mathbb{C}^{n \times n}$ - the geometric multiplicity of every eigenvalue must be equal to it's algebraic multiplicity.

Let's try induction over $n$:

$n = 1: \xi_1 v_1 = 0 \implies \xi_1 = 0 \quad \text{since, per Definition} \quad v_1 \neq 0$

This is where I am stuck - I don't know how to prove that adding an eigenvector from an orthogonal basis from one of the eigenspaces still implies that $\Sigma_{i = 1}^{n+1} \xi_i v_i = o\implies \forall i: \xi_i = 0$. It is easy enough when for $1 \leq i \leq n: \lambda_{n+1} \neq \lambda_i$. Can somebody point me in the right direction for the other case?

Answer 1

Here is a standard argument.

Let $\lambda$ be an eigenvalue of $A$, and $v$ be an associated eigenvector. Set $E=span\left\{ v\right\} $ and $E^{\perp}$ be the orthogonal complement, i.e., \begin{alignat*}{1} E^{\perp} & =\left\{ u\in\mathbb{C}^{n}\mid\left\langle u,v\right\rangle =0\right\} . \end{alignat*} $E$ and $E^{\perp}$ are both invariant under $A$. To see that $E^{\perp}$ is $A$-invariant, let $u\in E^{\perp}$ then \begin{align*} \left\langle Au,v\right\rangle & =\left\langle u,A^{*}v\right\rangle =\left\langle u,Av\right\rangle =\left\langle u,\lambda v\right\rangle =0. \end{align*}

Now the restriction of $A$ to $E^{\perp}$ is also Hermitian, so by repeating the above argument, there exist $\lambda_{1},\cdots,\lambda_{n}$ and $v_{1},\cdots,v_{n}$ such that $Av_{j}=\lambda_{j}v_{j}$.

The $\lambda_{j}$'s are not necessarily distinct. By construction, $v_{j}$'s are mutually orthogonal.

Answer 2

This is based on Schur triangularization theorem: every square matrix is unitarily similar to a triangular matrix. More explicitly, if $A$ is a square matrix (over the complex numbers), there exist $U$ unitary (that is, $UU^H=I$) and $T$ upper triangular such that $$ A=UTU^H $$ (Here $A^H$ denotes the Hermitian transpose.)

Suppose $A$ is normal, that is, $AA^H=A^HA$; then we have $$ UTU^HUT^HU^H=UT^HU^HUTU^H $$ so $UTT^HU^H=UT^HTU^H$ and, finally, $$ TT^H=T^HT $$ Actually, $T$ must be diagonal: write $T$ as a block matrix $$ T=\begin{bmatrix}t & v^H \\ 0 & T_1\end{bmatrix} $$ where the upper left block is $1\times 1$ and the lower right block is $(n-1)\times(n-1)$; then $$ TT^H= \begin{bmatrix}t & v^H \\ 0 & T_1\end{bmatrix} \begin{bmatrix}\bar{t} & 0^H \\ v & T_1^H\end{bmatrix}= \begin{bmatrix} t\bar{t}+v^Hv & v^HT_1 \\ T_1v & T_1T_1^H \end{bmatrix} $$ whereas $$ T^HT= \begin{bmatrix}\bar{t} & 0^H \\ v & T_1^H\end{bmatrix} \begin{bmatrix}t & v^H \\ 0 & T_1\end{bmatrix}= \begin{bmatrix} \bar{t}t & \bar{t}v^H \\ tv & vv^H+T_1^HT_1 \end{bmatrix} $$ whence $v^Hv=0$, which implies $v=0$; therefore $T_1T^H=T_1^HT_1$ and induction ends the proof.

A Hermitian matrix is obviously normal.

Answer 3

You have, that the eigenvectors and generalized eigenvectors form a invariant subspace $U$, that is there exists a matrix $X$, s.t. $AU=UX$. Since there is definiatly a eigenvector in there, you have the first collumn of $X$ as a multiple of $e_1$.

It is definiatly possible to form a orthogonal-basis of $U$, so that you can determine $X=U^HAU$. If $A=A^H$ you can conjugate-transpose that equation and get $X^H=(U^HAU)^H = U^HAU = X$. So $X$ has to be itself hermitian.

Since the first collumn of $X$ has only one nonzero component in the diagonal, the same goes for the first collumn, meaning ever other vector of $U$ (forming $\underline{U}$) will be mapped into $\underline{U}$.

From there on, you can use induction to prove by this, that $X$ is diagonal and every vector in $U$ is as eigenvector.