Find the spectral decomposition of $A$ $$
A= \begin{pmatrix} -3 & 4\\ 4 & 3 
\end{pmatrix}
$$ 
So i am assuming that i must find the evalues and evectors of this matrix first, and that is exactly what i did.
The evalues are $5$ and $-5$, and the evectors are $(2,1)^T$ and $(1,-2)^T$
Now the spectral decomposition of $A$ is equal to $(Q^{-1})^\ast$ (diagonal matrix with corresponding eigenvalues) * Q
$Q$ is given by [evector1/||evector1|| , evector2/||evector2||]
and for Q i got the matrix 
$$
Q= \begin{pmatrix} 2/\sqrt{5} &1/\sqrt{5} \\ 1/\sqrt{5} & -2/\sqrt{5} 
\end{pmatrix}
$$ 
the inverse of Q is the matrix...
$$
 \begin{pmatrix} 2 \sqrt{5}/5 & \sqrt{5}/5 \\ \sqrt{5}/5 & -2 \sqrt{5}/5 
\end{pmatrix}
$$ 
and the diagonal matrix with corresponding evalues is
$$
A= \begin{pmatrix} 5 & 0\\ 0 & -5 
\end{pmatrix}
$$ 
so now i found the spectral decomposition of $A$, but i really need someone to check my work.
Did i take the proper steps to get the right answer, did i make a mistake somewhere?

\begin{align}
\begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin{bmatrix} 2 \\ 1\end{bmatrix}= \begin{bmatrix} -2 \\ 11\end{bmatrix}
\end{align}
The eigenvector is not correct. The correct eigenvecor should be $\begin{bmatrix} 1 & 2\end{bmatrix}^T$ since 
\begin{align}
\begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin{bmatrix} 1 \\ 2\end{bmatrix}= 5 \begin{bmatrix} 1 \\ 2\end{bmatrix}
\end{align}
$\begin{bmatrix} 1 & -2\end{bmatrix}^T$ is not an eigenvector too.
\begin{align}
\begin{bmatrix} -3 & 4 \\ 4 & 3\end{bmatrix}\begin{bmatrix} -2 \\ 1\end{bmatrix}= -5 \begin{bmatrix} -2 \\ 1\end{bmatrix}
\end{align}
Also, at the end of the working, $A$ remains $A$, it doesn't become a diagonal matrix. 
You should write $A$ as $QDQ^T$ if $Q$ is orthogonal.
