# Problem of the Week #85 - January 13th, 2014

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#### Chris L T521

##### Well-known member
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Here's this week's problem!

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Problem: Let $E$ be a real vector space with an inner product $\langle\cdot,\cdot\rangle$. Show that any self-adjoint operator on $E$ can be diagonalized.

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#### Chris L T521

##### Well-known member
Staff member
This week's problem was correctly answered by caffeinemachine. You can find his very detailed solution below.

Notation
Given an operator $T\in\mathcal L(V)$, where $V$ is a finite dimensional inner product space, we write $T^*$ to denote the adjoint of $T$.
We use $U^\perp$ to denote the orthogonal complement of a subspace $U$ of an inner product space $V$.

Theorem
Let $V$ be a finite dimensional real vector space. Define set $U=V\times V$; addition on $U$ occurs component-wise.
For any member $a+ib$ of $\mathbb C$ and any member $(v,w)\in U$, define $(a+ib)(v,w)=(av-bw,aw+bv)$.
Then $U$ is a finite dimensional vector space over the complex number field.
Proof.
The fact that $U$ is a vector space over the field of complex numbers is routinely proved.
We show that $U$ is finite dimensional.
Let $\{x_1,\ldots,x_n\}$ be any basis of $V$.
Let $(u,v)$ be any vector in $U$.
Write $\displaystyle\frac{u+v}{2}=\sum_{j=1}^{n}a_jx_j$ and $\displaystyle\frac{v-u}{2}=\sum_{j=1}^{n}b_jx_j$.
Thus $\displaystyle u=\sum_{j=1}^{n}a_jx_j-\sum_{j=1}^{n}b_jx_j$ and $\displaystyle v=\sum_{j=1}^{n}a_jx_j+\sum_{j=1}^{n}b_jx_j$.
This shows that $\displaystyle (u,v)=\sum_{j=1}^{n}(a_j+ib_j)(x_j,x_j)$.
So we see that $\displaystyle \{(x_j,x_j)\}_{j=1}^n$ spans $U$.
Therefore $U$ is a finite dimensional vector space over $\mathbb C$.

Theorem.
Let $V$ be a finite dimensional real vector space and $T$ be any operator on $V$.
Then there exists $\lambda,\mu\in\mathbb R$ such that $Tu=\lambda u-\mu v$ and $Tv=\lambda v + \mu u$ for some $u,v\in V$, at least one of which is a non-zero vector.
Proof.
Define a vector space $U=V\times V$ as done in the theorem above.
Now define $S:U\to U$ as $S(v,w)=(Tv,Tw)$.
It's easy to see that $S$ is a linear transformation.
Since $U$ is a finite dimensional complex vector space, $S$ has an eigenvalue.
Say $\lambda+\mu i$ is an eigenvalue of $S$.
Then there exists $(u,v)\neq(0,0)$ such that $S(u,v)=(\lambda+\mu i)(u,v)$.
This gives the desired result and we are done.

Definition .
An operator $T$ on $V$ is said to be self-adjoint if $T=T^*$.

Theorem.
Every self-adjoint operator has an eigenvalue.
Proof.
Let $T$ be a self-adjoint operator on a vector space $V$.
If $V$ is a complex vector space then the theorem is true even without the hypothesis that $T$ is self-adjoint.
So we may assume that $V$ is a real vector space.
From the above theorem, we know that there exist vectors $u,v\in V$, not both zero, such that $Tu=\lambda u+\mu v$ and $Tv=-\mu u+\lambda v$, for some $\lambda,\mu\in \mathbb R$.
Now, by using the fact that $T$ is self-adjoint, we can write $\langle{Tu,v}\rangle=\langle{u,Tv}\rangle$.
This gives $\langle{\lambda u+\mu v,v}\rangle=\langle{u,-\mu u+\lambda v}\rangle$.
Thus $\lambda\langle{u,v}\rangle+\mu|v|^2=-\mu|u|^2+\lambda\langle{u,v}\rangle$, from where we get $\mu(|u|^2+|v|^2)=0$.
This forces $\mu=0$ and hence $Tu=\lambda u$ and $Tv=\lambda v$. Since at least one of $u$ and $v$ is not zero, we conclude that $T$ has an eigenvalue.

Theorem.
Real Spectral Theorem. Let $T$ be a linear operator on a finite dimensional real inner-product space $V$.
Then $V$ has an orthonormal basis consisting only of eigenvectors if and only if $T$ is self-adjoint.
Proof.
One direction of the proof is trivial.
We do the proof in the other direction.
Suppose $T$ is self-adjoint over a finite dimensional real inner product space $V$.
We prove the theorem by induction on the dimension of $V$.
The base case, $\dim V=1$, is trivial.
Assume that $\dim V>1$ and that the theorem holds for all vector spaces of strictly smaller dimension.
We know that $T$ has an eigenvalue, say $\lambda$, and let $u$ be the corresponding eigenvector of norm $1$.
Let $U=\{au:a\in\mathbb R\}$ and write $S=T|_{U^\perp}$.
We now show that $U^\perp$ is invariant under $S$.
Let $v\in S$.
Now $\langle{u,Sv}\rangle=\langle{u,Tv}\rangle=\langle{Tu,v}\rangle=\lambda\langle{u,v}\rangle=0$ for all $u\in U$.
Thus $Sv\in U^\perp$.
Now we show that $S$ is self adjoint.
Let $v,w\in U^\perp$.
Now, $\langle{Sv,w}\rangle=\langle{Tv,w}\rangle=\langle{v,Tw}\rangle=\langle{v,Sw}\rangle$.
Thus $S$ is self-adjoint.
This provides us with the necessary tools to apply the induction hypothesis on $U^\perp$ as the dimension of $U^\perp$ is one less than the dimension of $V$.
Using induction, there is an orthonormal basis of $U^\perp$ consisting only of eigenvectors of $S$.
Since every eigenvector of $S$ is an eigenvector of $T$, adjoining $u$ to this orthonormal basis of $U^\perp$ given us an orthonormal basis of $V$ consisting only of eigenvectors and we are done.

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