How to diagonalize an $N\times N$ symmetric matrix? ($N>3$, show an example of $N=5$) Example and basic theory will help. These things can be done for $3\times 3$ fruitfully, but problem arises after $N\ge 5$. Basic theory will be helpful (if posted),  as it will help to generalize to any ill condition situation (using higher level concepts of-course).
 A: For your matrix $A$, find all eigenvalues and corresponding eigenvectors, then let $P$ be the matrix whose columns are the eigenvectors of $A$, then $P^{-1}AP=D$, where $D$ is diagonal, and the diagonal entries of $D$ are the eigenvalues of $A$.
A: For example, let's try
$$ A = \left[ \begin {array}{ccccc} 3&-2&1&2&-2\\ -2&-1&0&0
&2\\ 1&0&1&-1&-3\\ 2&0&-1&1&-2
\\ -2&2&-3&-2&-1\end {array} \right] $$
Its characteristic polynomial is $P(\lambda) = \det( \lambda I - A) = \lambda^5 - 3 \lambda^4 - 33 \lambda^3 + 41 \lambda^2 + 96 \lambda - 6$, which happens to 
be irreducible.  The polynomial is not solvable using radicals.
Now an eigenvector for eigenvalue $r$ is a vector in the null space of $A - r I$.
I want to do my computations symbolically.  One trick is to just look at the top $4$ rows of $A - rI$ and row-reduce.  We find that a vector in the null
space is 
$$ \left[ \begin {array}{c} -2\,{r}^{3}-9\,{r}^{2}+20\,r+15
\\ 2\,{r}^{3}-6\,{r}^{2}+8\,r-12
\\ -3\,{r}^{3}+9\,{r}^{2}+19\,r-9
\\ -2\,{r}^{3}+5\,{r}^{2}-6\,r-21
\\ {r}^{4}-4\,{r}^{3}-8\,{r}^{2}+18\,r+9\end {array}
 \right] 
$$
Normalize by dividing by the square root of the sum of the squares, which
(after simplifying using $P(r)=0$) happens to be
$$ \sqrt{1266\,{r}^{4}-3300\,{r}^{3}-1140\,{r}^{2}+5772\,r+666}$$
...
