Complex spectral theorem proof

Question

So I need to prove the complex spectral theorem without using matrices, The complex spectral theorem says that for a complex inner product space $U$ and some linear operator $T$ on $U$, $T$ is normal if and only if there exists an orthonormal $T$ eigenbasis.

I'm not entirely sure where to start as the proofs I've seen before use matrices?!

Any help would be appreciated

Answer 1

Let me outline how to prove that if $T$ is normal then you can find an orthonormal basis of eigenvectors for $T$.

1. Using the fact that $T$ is normal, prove that $\|Tv\| = \|T^{*}v\|$ for all $v \in V$.
2. Deduce from the previous item that if $v$ is an eigenvector of $T$ with eigenvalue $\lambda$ then $v$ is also an eigenvector of $T^{*}$ with eigenvalue $\overline{\lambda}$.
3. Deduce from the previous item that if $v$ is an eigenvector of $T$ then $W = \operatorname{span} \{ v \}^{\perp}$ is $T$-invariant.
4. Show that if $W$ is a $T$-invariant subspace then $W^{\perp}$ is $T^{*}$-invariant. Show in addition that if $W$ is both $T$ and $T^{*}$-invariant then $\left( T|_{W} \right)^{*} = T^{*}|_{W}$ when we consider $T|_{W} \colon W \rightarrow W$ as an operator on $W$ with the inner product $\left< \cdot, \cdot \right>|_{W}$ which is the restriction of the original inner product to $W$. This holds for all (not neccesarily normal) operators. If $T$ is normal and $W$ is both $T$ and $T^{*}$-invariant then we have $$ T|_{W} \circ T^{*}|_{W} = \left( T \circ T^{*} \right)|_{W} = \left( T^{*} \circ T \right)|_{W} = T^{*}|_{W} \circ T|_{W} $$ so $T|_{W}$ is also normal.
5. Finally, prove that $T$ is orthogonally diagonalizable by induction on $\dim U$. If $\dim U = 1$, any non-zero vector in $U$ is an eigenvector of $T$ so just choose a unit length vector and it will form a basis for $U$. For the general case, choose an eigenvector $v_1$ of $T$ such that $\| v_1 \| = 1$. This is always possible because we are working over the complex numbers. Set $W = \operatorname{span} \{ v \}^{\perp}$. Then $\dim W < \dim U$ and by $(3)$ the space $W$ is $T$-invariant. By $(4)$, $T|_{W}$ is also normal. By the induction hypothesis, we can find an orthonormal basis $v_2,\dots,v_n$ of $W$ of eigenvectors for $T|_{W}$. Then $(v_1,\dots,v_n)$ is an orthonormal basis of eigenvectors of $U$ for $T$.

For the other direction, assume that $v_1,\dots,v_n$ is an orthonormal basis of $U$ consisting of eigenvectors of $T$ and write $Tv_i = \lambda_i v_i$. Show using the defining property of $T^{*}$ that $T^{*}v_i = \overline{\lambda_i} v_i$ and then $$ (T^{*}T)(v_i) = |\lambda_i|^2 = (TT^{*})(v_i) $$ for all $1 \leq i \leq n$ showing that $TT^{*} = T^{*}T$ so $T$ is normal.

Answer 2

This isn't what you're looking for, but I really like the analytic proof of the Complex Spectral Theorem. So:

Let $T$ be a self-adjoint operator in $\mathbb{C}^d$, and let $q(x) =\langle Tx, x \rangle$ be its associated quadratic form. Before proceeding with the proof of the Spectral Theorem, we show that $q$ is real valued and has real eigenvalues. Via properties of inner products and that $T$ is self-adjoint, we have $q(x) = \langle Tx, x \rangle = \langle x, Tx \rangle = \overline{\langle Tx, x \rangle}$. Thus $\langle Tx, x \rangle$ equals its conjugate, and is therefore real-valued. Since this holds for all $x$, it follows that $q$ is real-valued. Notice we have proved that the associated quadratic form of self-adjoint operators is always real.

To see that $T$ has real eigenvalues, let $v$ be an eigenvector and $\lambda$ its corresponding eigenvalue. Then $\langle Tv, v \rangle = \langle \lambda v, v \rangle = \lambda \|v\|^2$. Now, since $\|v\|^2$ is real, and $q(v) = \langle Tv, v \rangle$ is real, it follows that $\lambda$ is real as well.

Now, restrict $q$ to the unit sphere in $\mathbb{C}^d$, which we denote by $\mathbb{S}_{\mathbb{C}}^{d-1}$. Then $q$ is a continuous function on a compact set, so its maximum is attained. Let $x_0 \in \mathbb{S}_{\mathbb{C}}^{d-1}$ be such that $q(x_0)$ is a maximum. Now with the constraint equation $g(x) = \|x\|^2 -1$, we have by Lagrange multipliers $Dq(x_0) = \lambda Dg(x_0)$.

Let $v$ be a unit vector, so we can take a directional derivative in its direction. So, oriented in the direction of $v$, we have $D_{v}q(x_0) = \displaystyle \lim_{h \to 0} \frac{q(x_0+hv) - q(x_0)}{h} = \lim_{h \to 0} \frac{\langle Tx_0 + hTv, x_0 + hv \rangle - \langle Tx_0, x_0 \rangle}{h} =\lim_{h \to 0}\frac{\langle Tx_0, x_0 \rangle + h\langle Tx_0, v \rangle + h\langle Tv, x_0 \rangle + h^2 \langle Tv, v \rangle - \langle Tx_0, x_0 \rangle}{h} = \langle x_0, Tv \rangle + \langle Tv, x_0 \rangle = 2\operatorname{Re}(\langle Tx_0, v \rangle).$

Now we have

$$D_{v}q(x_0) = Dq(x_0)v = \lambda Dg(x_0)v = \lambda D_{v}g(x_0) = \lambda \displaystyle \lim_{h \to 0} \frac{g(x_0 +hv) - g(x_0)}{h} = \lim_{h \to 0} \frac{\langle x_0 + hv, x_0 + hv \rangle -1 - \langle x_0, x_0 \rangle +1}{h} = \lim_{h \to 0} \frac{\langle x_0, x_0 \rangle + h\langle x_0, v \rangle + h\langle v, x_0 \rangle +h^2\langle v,v \rangle - \langle x_0, x_0 \rangle}{h} = 2\operatorname{Re}(\langle x_0, v \rangle).$$

And so $$2\operatorname{Re}(\langle Tx_0, v \rangle) = 2\lambda\operatorname{Re}(\langle x_0, v\rangle).$$ Our claim now is that $Tx_0 = \lambda x_0$.

To see, this, remembering that our quadratic form is real, we have $\langle Tx_0 - \lambda x_0, Tx_0 - \lambda x_0 \rangle = \langle Tx_0, Tx_0 \rangle -2\lambda \langle Tx_0, x_0 \rangle + \lambda^2 \langle x_0, x_0 \rangle$. Now, we have shown that $2$Re$(\langle Tx_0, v \rangle) = 2\lambda$Re$(\langle x_0, v\rangle)$, so with $x_0$ in place of $\lambda$, $\langle Tx_0, x_0 \rangle = \lambda \langle x_0, x_0 \rangle$, and with $Tx_0$ in place of $v$, we have $\langle Tx_0, Tx_0 \rangle = \lambda \langle x_0, Tx_0 \rangle = \langle Tx_0, x_0 \rangle$. Then putting it all together, we have $\langle Tx_0 - \lambda x_0, Tx_0 - \lambda x_0 \rangle = \lambda \langle Tx_0, x_0 \rangle - 2\lambda \langle Tx_0, x_0 \rangle + \lambda \langle Tx_0, x_0 \rangle = 0$. By the positive-definiteness of the norm, this implies $Tx_0 = \lambda x_0$. Thus our Lagrange multiplier is an eigenvalue, and $x_0$ a corresponding eigenvector!

We then have $x_0$ as one extremum. Now consider $x_0^\bot = \{x \in \mathbb{S}_{\mathbb{C}}^{d-1} : \langle x, x_0 \rangle = 0 \}$. First it is pertinent to show that $x_0^\bot$ is invariant under $T$. So take $x \in x_0^\bot$, and consider $\langle Tx, x_0 \rangle = \langle x, Tx_0 \rangle = \langle x, \lambda x_0 \rangle = \lambda \langle x, x_0 \rangle = 0$ (note that $\lambda = \bar{\lambda}$ because $T$ has real eigenvalues). Hence $Tx \in x_0^\bot$, so $x_0^\bot$ is invariant under $T$. It follows that $x_0^\bot \cap \mathbb{S}_{\mathbb{C}}^{d-1}$ is a space of dimension $d-1$, on which we perform the same operations to find a unit eigenvector perpendicular to $x_0$. Continuing inductively $d-1$ steps, we wish to find a unit eigenvector of $T$ on intersection of $\mathbb{S}^0_{\mathbb{C}}$ and all of the orthogonal complements of the other $d-1$ eigenvectors, where this space is invariant. But this is trivial, for if $T$ is an operator on a $1$ dimensional invariant space, let $w$ be any nonzero vector which therefore spans this space and take $\frac{w}{\|w\|}$ as the eigenvector.

We have thus selected $d$ orthonormal and hence linearly independent eigenvectors of $T$, and therefore $\mathbb{C}^d$ has an orthonormal basis of eigenvectors of $T$. Therefore $T$ is diagonalizable.

Apologies for my poor formatting, I wrote this up on the first assignment I TeXed and don't feel like fixing it all.

Answer 3

For step #4 in the forward direction, how do you go from "Now restrict the inner product and $T$ to $W$" to "verify that $T|W$ is normal." ? I see that you've commented "Regarding the restriction, show that $(T|W)_∗ = (T_∗)|W$ using the defining property of $T_∗$ and this will imply that $T|W$ is normal from the fact that $T$ is normal." How does $(T|W)_∗ = (T_∗)|W$ become incorporated in the proof? I feel like there's something to be said before that, but I'm not sure what. Thank you!