Questions about eigenvectors and symmetric matrices I have a few questions about eigenvectors and eigenvalues, especially in symmetric matrices. So, straight to the point:

Why, for any matrix $A_{m \times n}$, is $A^{T}A$ an symmetric matrix?

I tried to figure it out by myself, i couldn't, so i looked up in the internet and couldn't find anything. I really dont have a clue about this.
Let's talk about another problem:

Given $A$ an symmetric matrix, with, for example, all elements on its main diagonal = $k$. If $A$ is $n \times n$, then the matrix will always have $n$ independent eigenvectors? And what about when there is $n$ distinct values in its main diagonal?
  Is there any proof of this?

And the last one:

Given $A$ a symmetric matrix, then $dim(colA)^{\perp} = Nul A^{T}$. How to prove?

Where $colA$ is the subspace generated by $S\{ u_1,u_2,...,u_n \}$, where $u_n$ are the vectors in the columns of the matrix $A$. What i did in this last one was:
for $T: W \rightarrow V, T(x)=Ax$
$$ dim W + ker T = dim V$$
in matrix form:
$$ dim(colA) + dim(NulA) = n$$,
where n is the number of columns of the matrix.
applying this to $A^{T}$:
$dim(colA^{T}) + dim(nulA^{T}) = m$, where $m$ is the number of lines of $A$.
Now, we know that: $dim(colA) + dim(colA)^{\perp} = m$, substituting in the first equation, i got to:
$dim(nulA^T) = dim(ColA)^{\perp}$. This proves about the dimension, but not about the space itself. How can i prove it?
Thanks.
Notes:
$ColA$ = subspace generated by the columns of $A$.
$NulA$ = null space of A.
$dim V$ = dimension of $V$.
 A: 
Why, for any matrix $A_{m \times n}$, is $A^{T}A$ an symmetric matrix?

For any two matrices $A, B$ with compatible dimensions, $(AB)^T = B^TA^T$ If you apply that to $A^TA$, you see that it's equal to its own transpose.

Given $A$ an symmetric matrix, with, for example, all elements on its main diagonal = $k$. If $A$ is $n \times n$, then the matrix will always have $n$ independent eigenvectors? And what about when there is $n$ distinct values in its main diagonal?

A symmetric matrix will always be diagonalizable, which, in turn, means that it will have $n$ independent eigenvectors. It doesn't matter what values are in its diagonal.

Given $A$ a symmetric matrix, then $\dim(\operatorname{col} A)^{\perp} = \operatorname{Nul} A^{T}$. How to prove?

$\dim(\operatorname{col} A)$ is the same as the rank of $A$, and the dimension of the perpendicular space is then $n$ minus said rank. By the rank-nulity theorem, this is exactly the dimension of the kernel of $A$. This has nothing to do with $A$ being symmetric.
