Number of Eigenvectors in a Symmetric Matrix Supposing symmetric matrix $A_{n\times n}$, how do I know that there are $n$ eigenvectors of $A$?
By way of trying to communicate context, I've spent a rather unproductive 5 or so hours watching various YouTube videos on the spectral theorem, orthogonal diagonalization, algebraic-bounding-geometric multiplicity, and thumbing through Anton's Elementary Linear Algebra (4.6 Change of Basis, 5.1 Eigenvalues and Eigenvectors, 5.2 Diagonalization, 6.4 Gram-Schmidt Process, 7.1 Orthogonalization, 7.2 Orthogonal Diagonalization, and 9.4 Singular Value Decomposition).  (This video I found the most helpful, but didn't quite understand why $A$ defines an $(n - 1)\times (n - 1)$ matrix)
I'm sure the answer to my question is in there somewhere, but something isn't quite clicking.
I understand:


*

*that $A = A^T$, and alternatively-stated that $\langle A\vec{x},\vec{y} \rangle = \langle \vec{x},A\vec{y}\rangle$.

*that, if $\vec{x}$ and $\vec{y}$ are eigenvectors of $A$, and if $\lambda \neq \mu$, then $\lambda\langle \vec{x},\vec{y} \rangle = \langle \lambda\vec{x},\vec{y} \rangle = \langle A\vec{x},\vec{y} \rangle = \langle \vec{x},A\vec{y}\rangle = \langle \vec{x},\mu\vec{y}\rangle = \mu\langle \vec{x},\vec{y}\rangle = 0$


*

*I don't understand how I know there's a next eigenvector of $A$, or how I would know that $\lambda \neq \mu$.


*that, from the fundamental theorem of algebra, every square matrix must have at least one eigenvector (since $det(\lambda I - A) = 0$ must have at least one solution.

*that I can iteratively generate, through the Gram-Schmidt process, an orthonormal basis given a set $B$ that forms a basis.  That is, if I know that I have $n$ eigenvectors, I know through G-S that I can generate an orthonormal basis for the eigenspace wherever I have repeated roots.


I don't yet know what Hermitian, unitary, and conjugate transpose mean.
I also don't understand how I know that there exists a symmetric matrix of size $(n-1)\times (n-1)$, such that I would simply be able to presume that another eigenvector has to exist in or come out of that iteratively-smaller matrix.
 A: 
Supposing symmetric matrix $An×n$, how do I know that there are $A$
  eigenvectors of $A$?

Many people who are smarter than I am have already provided answers to this question.  But out of extreme hubris, I didn't see why they were right, so I attempted the proof myself:
Establishment of at least one Eigenvector of $A$


*

*Note that $A = A^\top$

*Note that, due to the fundamental theorem of algebra, $|\lambda I - A|=0$ must have at least 1 solution.

*Ergo, $\exists\lambda_1$, an eigenvalue of $A$.

*Since each distinct eigenvalue implies the existence of at least one eigenvector, a basis for that eigenspace, $\exists \vec{v_1}$, a unit eigenvector of $A$ (e.g. $||\vec{v_1}||=1$)


Establishment of a Basis


*

*Let $B = \{\vec{v_1}, \vec{b_2}, ..., \vec{b_n}\}$, an orthonormal basis for $\mathbb{R}^n$, which is constructed from $\vec{v_1}$ by solving the system $\begin{bmatrix}{\vec{v_1}^\top}\end{bmatrix}=\vec{0}$.

*Let $P = \begin{bmatrix}\vec{v_1} & \vec{b_2} & ... & \vec{b_n}\end{bmatrix}$ $\therefore$ $P^\top = \begin{bmatrix}\vec{v_1}^\top\\
 \vec{b_2}^\top\\
...\\
\vec{b_n}^\top\end{bmatrix}$

*Note that, as $B$ has been defined as an orthonormal basis, $P^\top P = I_n \therefore P^{-1} = P^\top \therefore P$ is symmetric.


Multiplication by $P$


*

*Note that $AP = \begin{bmatrix}A\vec{v_1} & A\vec{b_2} & ... & A\vec{b_n}\end{bmatrix}$

*Note that, since $\vec{v_1}$ is an eigenvector of $A$, $A\vec{v_1} = \lambda_1\vec{v_1} \therefore AP=\begin{bmatrix}\lambda_1\vec{v_1} & A\vec{b_2} & ... & A\vec{b_n}\end{bmatrix}$

*Note that $P^{\top}AP = \begin{bmatrix}\vec{v_1}^{\top}\lambda_1\vec{v_1} & \vec{v_1}^{\top}A\vec{b_2} & ... & \vec{v_1}^{\top}A\vec{b_n}\\
\vec{b_2}^{\top}\lambda_1\vec{v_1} & \vec{b_2}^{\top}A\vec{b_2} & ... & \vec{b_2}^{\top}A\vec{b_n}\\
\vdots & \vdots & & \vdots\\
\vec{b_n}^{\top}\lambda_1\vec{v_1} & \vec{b_n}^{\top}A\vec{b_2} & ... & \vec{b_n}^{\top}A\vec{b_n}
\end{bmatrix}$


Simplification


*

*Note that the first column of $P^{\top}AP$ can be rewritten as $\begin{bmatrix}\lambda_1(\vec{v_1}^{\top}\vec{v_1}) & \vec{v_1}^{\top}A\vec{b_2} & ... & \vec{v_1}^{\top}A\vec{b_n}\\
\lambda_1(\vec{b_2}^{\top}\vec{v_1}) & \vec{b_2}^{\top}A\vec{b_2} & ... & \vec{b_2}^{\top}A\vec{b_n}\\
\vdots & \vdots & & \vdots\\
\lambda_1(\vec{b_n}^{\top}\vec{v_1}) & \vec{b_n}^{\top}A\vec{b_2} & ... & \vec{b_n}^{\top}A\vec{b_n}
\end{bmatrix}$ 

*Note that, since we have the dot product of orthonormal vectors in the first column, $P^{\top}AP$ simplifies to $\begin{bmatrix}1 & \vec{v_1}^{\top}A\vec{b_2} & ... & \vec{v_1}^{\top}A\vec{b_n}\\
0 & \vec{b_2}^{\top}A\vec{b_2} & ... & \vec{b_2}^{\top}A\vec{b_n}\\
\vdots & \vdots & & \vdots\\
0 & \vec{b_n}^{\top}A\vec{b_2} & ... & \vec{b_n}^{\top}A\vec{b_n}
\end{bmatrix}$

*Note that, since we have the dot product of orthonormal vectors in the first column, $P^{\top}AP$ simplifies to $\begin{bmatrix}1 & \vec{v_1}^{\top}A\vec{b_2} & ... & \vec{v_1}^{\top}A\vec{b_n}\\
0 & \vec{b_2}^{\top}A\vec{b_2} & ... & \vec{b_2}^{\top}A\vec{b_n}\\
\vdots & \vdots & & \vdots\\
0 & \vec{b_n}^{\top}A\vec{b_2} & ... & \vec{b_n}^{\top}A\vec{b_n}
\end{bmatrix}$

*Note that $(P^{\top}AP)^\top = P^{\top}A^{\top}(P^{\top})^\top = P^{\top}AP \therefore P^{\top}AP$ is symmetric. 

*Since $P^{\top}AP$ is symmetric, it must be that $P^{\top}AP = \begin{bmatrix}1 & 0 & ... & 0\\
0 & \vec{b_2}^{\top}A\vec{b_2} & ... & \vec{b_2}^{\top}A\vec{b_n}\\
\vdots & \vdots & & \vdots\\
0 & \vec{b_n}^{\top}A\vec{b_2} & ... & \vec{b_n}^{\top}A\vec{b_n}
\end{bmatrix}$


Recursion


*

*Since $P^{\top}AP$ is symmetric, it must be that the block matrix formed by removing the first row and first column of $P^{\top}AP$ is also symmetric; that is, $\begin{bmatrix}\vec{b_2}^{\top}A\vec{b_2} & ... & \vec{b_2}^{\top}A\vec{b_n}\\
\vdots & & \vdots\\
\vec{b_n}^{\top}A\vec{b_2} & ... & \vec{b_n}^{\top}A\vec{b_n}
\end{bmatrix}$ is symmetric.

*Now, we're back to the very beginning, namely supposing a symmetric matrix, although now of $B_{n-1\times n-1}$.


Base Case


*

*Note that, for any $A_{k\times k}$ where $k < n$, we'll still have leftover symmetric matrix to pull eigenvectors from, and we don't need to consider a base case.

*However, when $k = n$, that is when we're actually considering a symmetric matrix with the same number of columns and rows as the dimension of the space, eventually (i.e., for our very last eigenvector) we'll get down to a symmetric submatrix denoted $M_{1\times 1}$, that is $M = \begin{bmatrix}k\end{bmatrix}$ where $k \in \mathbb{R}$.

*For this 1-by-1 matrix, note that $|\lambda I - M| = 0 = \lambda - k$, $\therefore \lambda_0 = k$, $\therefore$ the eigenvector $\begin{bmatrix}1\end{bmatrix}$ forms a basis for the eigenspace of $M$, $\therefore P = \begin{bmatrix}1\end{bmatrix}$.

*Note that, since $P = \begin{bmatrix}1\end{bmatrix}$, $P^{\top}MP = \begin{bmatrix}1\end{bmatrix}\begin{bmatrix}k\end{bmatrix}\begin{bmatrix}1\end{bmatrix} = \begin{bmatrix}k\end{bmatrix} = M$, $\therefore P$ orthogonally diagonalizes $M$, which is actually already diagonalized, being a 1-by-one matrix.


Some additional links to helpful proofs follow.
http://www.swarthmore.edu/NatSci/dmcclen1/math028S/m028f2011realspectral.pdf
https://math.berkeley.edu/~ogus/old/Math_54-05/webfoils/symmetric.pdf
https://www.math.ucdavis.edu/~linear/old/notes22.pdf

