Proving that a matrix is symmetric if it can be expressed as a spectral decomposition 
If $\{u_1, \cdots, u_n\}$ is an orthonormal basis for $\mathbb{R}^n$, and if $A$ can be expressed as
  $$A = c_1u_1u_1^T + \cdots + c_nu_nu_n^T$$
  then $A$ is symmetric and has eigenvalues $c_1, \cdots, c_n$.

I'm trying to prove this. Here's what I have so far.

I figure I need to show:


*

*$A$ is symmetric. I can achieve this by showing that $A$ has an orthonormal set of $n$ eigenvectors (or equivalently, that $A$ is orthogonally diagonalizable). If $P$ orthogonally diagonalizes $A$ then $D = P^TAP \equiv A = PDP^T$. $(PDP^T) = (PDP^T)^T$ by trivial manipulations, knowing that $D$ is diagonal and thus $D^T = D$.

*$c_1, \cdots, c_n$ are the eigenvalues of $A$.


I think both of these would be satisfied if I could show that $c_1u_1u_1^T + \cdots + c_nu_nu_n^T$ was equivalent to $PDP^T$ for the orthogonal matrix $P$ and a $D$ such that $D_{ij} = \begin{cases}0 & i \neq j\\c_i & i = j\end{cases}$.
If $P$ was an orthogonal matrix such that $P = \begin{bmatrix} u_1 & \cdots & u_n\end{bmatrix}$ where $u_j$ was an eigenvector of $A$ then it would also be a basis for $\mathbb{R}^n$, since we'd have $n$ linearly independent vectors. If I had this then I believe you can do the tedious matrix multiplication and get $PDP^T$ given the $D$ defined above and receive $A = c_1u_1u_1^T + \cdots + c_nu_nu_n^T$. Then I'd be done.
But to me the question implies that any orthonormal basis for $\mathbb{R}^n$ would satisfy this. Perhaps I need to show that if $A$ can be expressed with those basis vectors then those basis vectors must be the eigenvectors of $A$. I'm kind of stuck on this part though!
Edit: To be clear: I have outlined here my approach to the proof and what I know to be true. I'm ultimately stuck on how to prove the quoted question. I am asking how one can prove this.

This is exercise 7.2.26 of Anton and Rorres' Elementary Linear Algebra, 11th ed.
 A: I'm not quite sure what you're asking in your question, but if its helpful, here's how I would write this proof.
1) If $$A=\sum_{i=1}^n c_iu_iu_i^T,$$then observe that
$$A^T=\left(\sum_{i=1}^nc_iu_iu_i^T\right)^T=\sum_{i=1}^n c_i(u_i^T)^Tu_i^T=\sum_{i=1}^n c_iu_iu_i^T=A,$$
where the second equality follows since taking transposes reverses the order of multiplication for matrices, and we can always pull constants out front.
2) If $A$ has the form above, then to show $c_j$ is an eigenvalue, consider the following product:
$$Au_j= \sum_{i=1}^nc_iu_iu_i^Tu_j=\sum_{i=1}^nc_iu_i\delta_{ij}=c_ju_j.$$
The second equality follows from the fact that the $u_i$ form an orthonormal basis so $u_i^Tu_j=\delta_{ij}$ (by definition of orthonormal).
A: By inspection from the hypotesis we have that
$$A^T= (c_1u_1u_1^T + \cdots + c_nu_nu_n^T)^T=A$$
and
$$A\cdot u_i=c_iu_i$$
therefore the thesis follows.
A: Note that
$(u_i u_i^T)^T = (u_i^T)^Tu_i^T = u_i u_i^T; \tag 1$
thus each matrix $u_i u_i^T$ is symmetric; hence every $c_i u_i u_i^T$ and hence their sum.  This shows that
$A^T = A. \tag 2$
We further note that, since the $u_i$ are orthnormal,
$u_i^T u_j = \delta_{ij}, \tag 3$
whence
$A u_j = \displaystyle  \left ( \sum_{i = 1}^n c_i u_i u_i^T \right ) u_j = \sum_{i = 1}^n c_i u_i u_i^Tu_j = \sum_{i = 1}^n c_iu_i \delta_{ij} = c_j u_j, \tag 4$
which shows that $c_j$ is an eigenvalue of $A$ with associated eigenvector $u_j$, $1 \le j \le n$.
