Real Spectral Theorem Induction Proof I am currently reading Linear Algebra Done right and was having trouble with the proof of the Real Spectral Theorem.
Statement:

7.13 Real Spectral Theorem: Suppose that V is a real inner-product space and T ∈ L(V). Then V has an orthonormal basis consisting of
eigenvectors of T if and only if T is self-adjoint.

The theorem he is refering to in the proof is:

7.12 Lemma: Suppose T ∈ L(V) is self-adjoint. Then T has an eigenvalue.

I am only interested in the proof from one side. Here it is:

“To prove the other direction, now suppose that $T$ is self-adjoint. We will prove that $V$ has an orthonormal basis consisting of eigenvectors of $T$ by induction on the dimension of $V$. To get started, note that our desired result clearly holds if $\operatorname{dim} V=1$. Now assume that $\operatorname{dim} V>1$ and that the desired result holds on vector spaces of smaller dimension.
The idea of the proof is to take any eigenvector $u$ of $T$ with norm 1 then adjoin to it an orthonormal basis of eigenvectors of $\left.T\right|_{\{u\}^{\perp}}$. Now
for the details, the most important of which is verifying that $\left.T\right|_{\{u\}^{\perp}}$ is self-adjoint (this allows us to apply our induction hypothesis).
Let $\lambda$ be any eigenvalue of $T$ (because $T$ is self-adjoint, we know from the previous lemma that it has an eigenvalue) and let $u \in V$ denote a corresponding eigenvector with $\|u\|=1$. Let $U$ denote the one-dimensional subspace of $v$ consisting of all scalar multiples of $u$. Note that a vector $v \in V$ is in $U^{\perp}$ if and only if $\langle u, v\rangle=0$.
Suppose $v \in U^{\perp}$. Then because $T$ is self-adjoint, we have
$$
\langle u, T v\rangle=\langle T u, v\rangle=\langle\lambda u, v\rangle=\lambda\langle u, v\rangle=0
$$
and hence $T v \in U^{\perp}$. Thus $Tv \in U^{\perp}$ whenever $v \in U^{\perp}$. In other words, $U^{\perp}$ is invariant under $T$. Thus we can define an operator $S \in \mathcal{L}\left(U^{\perp}\right)$ by $S=\left.T\right|_{U^{\perp}}$. If $v, w \in U^{\perp}$, then
$$
\langle S v, w\rangle=\langle T v, w\rangle=\langle v, T w\rangle=\langle v, S w\rangle
$$
which shows that $S$ is self-adjoint (note that in the middle equality above we used the self-adjointness of $T$ ). Thus, by our induction hypothesis, there is an orthonormal basis of $U^{\perp}$ consisting of eigenvectors of $S$. Clearly every eigenvector of $S$ is an eigenvector of $T$ (because $S v=T v$ for every $v \in U^{\perp}$ ). Thus adjoining $u$ to an orthonormal basis of $U^{\perp}$ consisting of eigenvectors of $S$ gives an orthonormal basis of $V$ consisting of eigenvectors of $T$, as desired.”

It isn't fully clicking to me. He defined an artificial operator himself and is saying this must have an orthogonal basis even though he never showed it for matrices greater than dimension 1. Dimension 1 is such an unrealistic case for matrices. Can anyone clear out his thinking or approach for me?
 A: The proof is by induction on the dimension of $V$; as with all proofs by induction, that means that we need to explicitly show that the statement is true for some base case(s) (in this case, when the dimension of $V$ is $1$), and that if the statement is true up to some dimension $n$ then it remains true in dimension $n+1$.  
The approach is to take a vector space $V$ of dimension $n+1$ and breaking it up into two pieces, namely the subspace $U$ spanned by an eigenvector of $T$ and the subspace $U^\perp$ that is orthogonal to $U$.  If $\alpha$ is an orthonormal basis of $U$ and $\beta$ is an orthonormal basis for $U^\perp$, then $\alpha \cup \beta$ is an orthonormal basis for $V$, so all we need to do is find $\alpha$ and $\beta$, each consisting of eigenvectors of $T$.
Finding $\alpha$ is easy, because $U$ is 1-dimensional and spanned by an eigenvector of $T$; just take any vector in $U$ and scale it to have norm $1$.
To find $\beta$ we'd like to apply the induction hypothesis.  We do have $\dim(U^\perp) = n < \dim(V) = n+1$, which is good: If we have a self-adjoint operator from $U^\perp$ to $U^\perp$ then the induction hypothesis will give us the basis for $U^\perp$ that we're looking for.  The operator we'd like to use is $T$, but $T$ is an operator from $V$ to $V$, not from $U^\perp$ to $U^\perp$.  It would be nice, though, if we could think of $T$ as an operator from $U^\perp$ to $U^\perp$.  For that reason we define $S : U^\perp \to U^\perp$ to do the same thing as $T$: for all $v \in U^\perp$, $S(v) = T(v)$.  There's a little checking to do to make sure this makes sense (specifically, that if $v \in U^\perp$ then $T(v) \in U^\perp$ too), and that $S$ is self-adjoint.
Once those steps are done, we've now got a space ($U^\perp$) of dimension strictly less than the dimension of $V$, and a self-adjoint operator on that space.  By induction hypothesis there is an orthonormal basis, call it $\beta$, for $U^\perp$ consisting of eigenvectors of $S$.  But $S$ does the same thing as $T$, so the vectors in $\beta$ are also eigenvectors of $T$, which is what we wanted.
A: @Chris Eagle gave a very detailed explanation. I want to make a brief summary of his answer.
The logic of the proof in Linear Algebra Done Right is as follows:

*

*it's true for $dim V = 1$.

*let $dim V = n+1$. we assume it's true for real inner product spaces with dimensions up to $n$, in this proof, this real inner product space is $U^\perp$.

*$dim U = 1$, $dim U^\perp = n$, we can find an orthonormal basis of $V$.

