Proving that restriction of linear operator to invariant set has same eigenvectors as linear operator

I am hung-up on one part of the proof of the spectral theorem for real matrices. The proof by induction makes perfect sense, but when I try to write it using only matrices I'm struggling to show that the eigenvalues of $$B$$ are the same as $$A$$.

Quick proof of spectral theorem of real, linear operators on $$\mathbb{R}^n$$ by induction

Here's a quick rundown of the proof by induction that I am referring to:

Base case. Let let $$V$$ be an inner product space of $$\dim V = 1$$ and $$T$$ a self-adjoint linear operator $$T: V \to V$$. Then $$T$$ has a real eigenvalue $$\lambda$$ and corresponding real eigenvector $$\mathbf{v}\in \mathbb{R}$$. Now, $$\frac{\mathbf{v}}{\| \mathbf{v} \|}$$ spans $$\mathbb{R}$$.

Inductive step. Now let $$\dim V = n+1$$. $$T$$ again has a real eigenvalue $$\lambda$$ and corresponding real eigenvector $$\mathbf{v}$$. Let $$U = \text{span} \{ \mathbf{v} \}$$, so $$U^\perp = \{ \mathbf{u} \in V \; | \; \langle \mathbf{u}, \mathbf{v} \rangle = 0\}$$.

Note that $$\dim U^\perp = n$$ and that $$U^\perp$$ is invariant under $$T$$ as $$\langle T(\mathbf{u}), \mathbf{v} \rangle = \langle \mathbf{u}, T(\mathbf{v}) \rangle = \langle \mathbf{u}, \lambda \mathbf{v} \rangle = \lambda \langle \mathbf{u}, \mathbf{v} \rangle = 0 \; , \; \forall \mathbf{u} \in U^\perp$$ Define $$E : U^\perp \to U^\perp$$, where $$E = \left. T \right|_{U^\perp}$$, then $$E$$ is self-adjoint $$\langle E(\mathbf{u}_1) , \mathbf{u}_2 \rangle = \langle T(\mathbf{u}_1) , \mathbf{u}_2 \rangle = \langle \mathbf{u}_1 , T(\mathbf{u}_2) \rangle = \langle \mathbf{u}_1 , E(\mathbf{u}_2) \rangle$$ Thus, by the inductive hypothesis, $$E$$ has $$n$$ orthonormal eigenvectors that span $$U^\perp$$ and because $$E(\mathbf{x}) = \omega \mathbf{x} = T(\mathbf{x})$$ where $$(\omega, \mathbf{x})$$ is an eigenpair of $$E$$, then all eigenvectors of $$E$$ are also eigenvectors of $$T$$, so $$T$$ admits a spectral decomposition $$\, \blacksquare$$

Now I'd like to show this by using matrices.

Attempt to prove with matrix math

Base case. Let $$A = A^\top \in \mathbb{R}^{1 \times 1}$$. $$A$$ has a real eigenvalue $$\lambda \in \mathbb{R}$$ and eigenvector $$\mathbf{v} \in \mathbb{R}$$, and $$\frac{\mathbf{v}}{\| \mathbf{v} \|}$$ spans $$\mathbb{R}$$

Inductive step. Let $$A = A^\top \in \mathbb{R}^{(n + 1) \times (n + 1)}$$. Again $$\exists \lambda \in \mathbb{R} \exists \mathbf{v} \in \mathbb{R}^{n+1} \; | \; A \mathbf{v} = \lambda \mathbf{v}$$ because the fundamental theorem of algebra tells us that the characteristic polynomial of $$A$$ will have $$n+1$$ roots counted with algebraic multiplicity, and $$A$$'s symmetry tells us that the root(s) are real, and let $$(\lambda, \mathbf{v})$$ be an eigenpair of $$A$$. Let $$U = \text{span}\{ \mathbf{v} \}$$ and thus $$U^\perp = \{ \mathbf{u} \in \mathbb{R}^{n+1} \; | \; \langle \mathbf{u}, \mathbf{v} \rangle = \mathbf{u} \cdot \mathbf{v} = \mathbf{u}^\top \mathbf{v} = 0 \}$$ Since $$A$$ is symmetric, $$U^\perp$$ is invariant under $$A$$: $$A \mathbf{u} \cdot \mathbf{v} = (A \mathbf{u})^\top \mathbf{v} = \mathbf{u}^\top A^\top \mathbf{v} = \mathbf{u}^\top A \mathbf{v} = \mathbf{u}^\top (\lambda \mathbf{v}) = \lambda \mathbf{u}^\top \mathbf{v} = \lambda (\mathbf{u} \cdot \mathbf{v}) = 0$$ thus $$A \mathbf{u} \in U^\perp$$, for all $$\mathbf{u} \in U^\perp$$.

Now let $$\{\mathbf{y}_1 , \dots, \mathbf{y}_n \}$$ be a basis of $$U^\perp$$, and let $$Y = \left[\begin{smallmatrix}\mathbf{y}_1 & \dots & \mathbf{y}_n \end{smallmatrix}\right] \in \mathbb{R}^{(n+1) \times n}$$. Note that $$\dim U^\perp = n$$ since $$\dim U = 1$$. Let $$B$$ be the transformation described by $$A$$ in the basis $$\{\mathbf{y}_i \}$$, defined as $$B_{ij} = \langle A\mathbf{y}_j, \mathbf{y}_i \rangle = \sum_{k=1}^n AY_{jk}Y_{ki} = Y^\top A Y$$ that is, the $$ij$$-th entry of $$B \in \mathbb{R}^{n \times n}$$ is the $$i$$-th coordinate (w.r.t. $$\{ \mathbf{y}_i \}$$) of the transformed $$\mathbf{y}_j$$. The punchline is $$B^\top = (Y^\top AY)^\top = Y^\top A^\top Y$$ so $$B$$ is symmetric because $$A$$ is symmetric.

Thus, by the inductive hypothesis, $$B$$ admits an orthonormal eigenbasis of $$\mathbb{R}^n$$.

The point where I'm stuck

At this point, the proofs I've read usually say something along the lines of:

Since $$B$$ is the transformation $$A$$ in the basis $$\{ \mathbf{y}_i \}$$, then all eigenvectors of $$B$$ are also eigenvectors of $$A$$, and further, these eigenvectors are orthogonal to $$\mathbf{v}$$, thus $$A$$ admits an orthonormal eigenbasis.

Which makes perfect sense; however, since we know this is true we should be able to show the fact that the eigenvectors of $$B$$ are also eigenvectors of $$A$$ using just matrices (i.e. the math should work out).

Attempt

My attempt was to let $$(\mu, \mathbf{x})$$ be an eigenpair of $$B$$. Then $$B \mathbf{x} = \mu \mathbf{x} = Y^\top A Y \mathbf{x}$$ So $$Y \mathbf{x}$$ is $$\mathbf{x}$$ in the standard basis of $$\mathbb{R}^{n+1}$$, so I'll nickname it $$\mathbf{m}$$, so this gives us $$Y B \mathbf{x} = Y \mu \mathbf{x} = \mu \mathbf{m} = Y Y^\top A \mathbf{m}$$ at this point, I wanted to say something about $$Y Y^\top$$, as in if $$Y$$ is an orthogonal matrix then we have the desired result, but it isn't since $$Y \in \mathbb{R}^{(n+1) \times n}$$, so it can't be.

My question/TLDR

My question written out explicitly is:

If $$A \in \mathbb{R}^{(n+1)\times (n+1)}$$ and $$B \in \mathbb{R}^{n \times n}$$ are both symmetric, and $$B = Y^\top A Y$$, prove that all eigenvectors of $$B$$ are also eigenvectors of $$A$$, i.e. show that if
$$B \mathbf{x} = \mu \mathbf{x}$$ and $$Y \mathbf{x} = \mathbf{m}$$, then $$A \mathbf{m} = \mu \mathbf{m}$$

• What does it mean for a non-square matrix to be symmetric? Jun 7 at 23:10

The piece of data you are missing is that, because $$y_1,\ldots,y_n$$ is an orthonormal basis for $$U^\perp$$, then $$YY^\top$$ is the orthogonal projection onto $$U^\perp$$. Thus from $$B=Y^\top AY$$ and $$Bx=\mu x$$ you get, since $$Yx\in U^\perp$$, $$A(Yx)=YY^\top AYx=YBx=Y(\mu x)=\mu(Yx).$$
Another requirement/something that is worth emphasizing: $$\{ \mathbf{y}_i \}$$ is an orthonormal basis of $$U^\perp$$, so $$\begin{gather} Y^\top Y = I \end{gather}$$ since $$\begin{gather} (Y^\top Y)_{ij} = \sum_k Y^\top_{ik} Y_{kj} = \sum_k Y_{ki} Y_{kj} = \langle \mathbf{y}_i, \mathbf{y}_j \rangle = \mathbf{y}_i \cdot \mathbf{y}_j = \delta_{ij} \end{gather}$$ Then, since $$Y \mathbf{x} = \mathbf{m}$$ $$\begin{gather} Y^\top \mathbf{m} = \mathbf{x} \end{gather}$$ then $$\begin{gather} B \mathbf{x} = \mu \mathbf{x} \xrightarrow[B = Y^\top A Y]{} Y^\top A Y \mathbf{x} = \mu \mathbf{x} \end{gather}$$ Now replace $$\mathbf{x}$$ with $$Y^\top \mathbf{m}$$, and replace $$Y \mathbf{x}$$ with $$\mathbf{m}$$ $$\begin{gather} Y^\top A \mathbf{m} = \mu Y^\top \mathbf{m} \to Y^\top (A\mathbf{m} - \mu \mathbf{m}) = \mathbf{0} \end{gather}$$ Which means that $$A\mathbf{m} - \mu \mathbf{m}$$ is either orthogonal to $$U^\perp$$, or $$A\mathbf{m} - \mu \mathbf{m}= \mathbf{0}$$. Since $$\text{im } Y = U^\perp$$, $$\mathbf{m} \in U^\top$$, and since $$U^\top$$ is invariant under $$A$$, then $$A\mathbf{m} - \mu \mathbf{m} \in U^\perp$$ and thus $$A\mathbf{m} - \mu \mathbf{m} = \mathbf{0} \quad \blacksquare$$
So we can take $$B$$'s orthonormal eigenbasis, $$\{ \mathbf{x}_i \}$$, and let $$X = \begin{bmatrix} \mathbf{x}_1 & \dots & \mathbf{x}_n \end{bmatrix}$$ then $$YX = \begin{bmatrix} Y\mathbf{x}_1 & \dots & Y\mathbf{x}_n \end{bmatrix}$$ are all eigenvectors of $$A$$, and the eigenvectors are all orthogonal since the eigenvalues are real. So if we take all of those eigenvectors of $$A$$ with the original eigenvector $$\mathbf{v}$$ (and normalize), we get the orthonormal eigenbasis of $$A$$: $$\{ Y\mathbf{x}_i \} \cup \{ \mathbf{v} \}$$