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?
 A: The needed computation is
$$\mathsf{A} = \mathsf{Q\Lambda}\mathsf{Q}^{-1}$$
Where $\Lambda$ is the eigenvalues matrix. And your eigenvalues are correct.
Hence you have to compute
$$\mathsf{AQ} = \mathsf{Q\Lambda}$$
Which gives you the solutions
$$a = 2c ~~~~~~~~~~~ d = -\frac{b}{2}$$
You can then choose easy values like $c = b = 1$ to get
$$Q = \begin{pmatrix} 2 & 1  \\  1 & -\frac{1}{2} \end{pmatrix}$$
And easily
$$\mathsf{Q}^{-1} = \frac{1}{\text{det}\ \mathsf{Q}} \begin{pmatrix} -\frac{1}{2} & -1 \\ -1 & 2 \end{pmatrix}$$
Which you can compute alone.
A: \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.
A: I think of the spectral decomposition as writing $A$ as the sum of two matrices, each having rank 1.  Let $A$ be given.  Then compute the eigenvalues and eigenvectors of $A$.  Then 
$$
A = \lambda_1P_1 + \lambda_2P_2
$$
where $P_i$ is an orthogonal projection onto the space spanned by the $i-th$ eigenvector $v_i$.  
In your case, I get $v_1=[1,2]^T$ and $v_2=[-2, 1]$ from Matlab.  Then we have:
$$
\left[ \begin{array}{cc}
-3 & 4 \\
4 & 3\\
\end{array} \right] =
5\left[ \begin{array}{cc}
1/5 & 2/5 \\
2/5 & 4/5\\
\end{array} \right] -
5\left[ \begin{array}{cc}
4/5 & -2/5 \\
-2/5 & 1/5\\
\end{array} \right] 
$$
Each $P_i$ is calculated from $v_iv_i^T$.
