Symmetric matrices - eigenvalues I am trying , without success, to find Symmetric matrix 2x2 (Not Diagonal) with eigenvalues 2 and 3.
I tried to start from here :
Av = 3v  ==> (A-3I)v = 0
Aw = 2w  ==> (A-2I)w = 0

It didn't help me.
Your help is appricaited. 
 A: Start with the general symmetric matrix:
$$\begin{pmatrix}
a & b \\
b & c \\
\end{pmatrix}$$
Calculate its characteristic polynomial and compare it with the characteristic polynomial you want:
$$(\lambda-2)(\lambda-3)$$
Then try to choose a, b, c in order to make both polynomials equal.
A: $$\begin{vmatrix}
a-x & c \\
c & b-x \\
\end{vmatrix}=0$$
The characteristic polynomial will be:
$$(a-x)(b-x)-c^2=0$$
$2$ and $3$ are the roots, so:
$$(a-2)(b-2)-c^2=0 \quad (1)$$
$$(a-3)(b-3)-c^2=0 \quad (2)$$
Then $(1)-(2)$:
$$a+b=5$$
Now choose $a$ and $b$ that fit the above equation and then find $c$.
Can you finish?
A: Use the spectral theorem with an orthogonal basis different from the standard one:
$$
A=
2
\begin{bmatrix}\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}\end{bmatrix}
\begin{bmatrix}\frac{1}{\sqrt2}&\frac{1}{\sqrt2}\end{bmatrix}
+3
\begin{bmatrix}\frac{1}{\sqrt2}\\-\frac{1}{\sqrt2}\end{bmatrix}
\begin{bmatrix}\frac{1}{\sqrt2}&-\frac{1}{\sqrt2}\end{bmatrix}
$$
A: The companion matrix for $(x-2)(x-3)=x^2-5x+6$ is 
$$ \left[
        \begin{matrix}
        0 & 1 \\
        -6 & 5 \\
        \end{matrix} \right]
$$
See this.  I used the transpose of the companion matrix here. It isn't symmetric, but it isn't the identity.
