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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.

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    Why not diagonal? – 2017-01-08

4 Answers 4

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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.

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    Thank you for the fast reply. – 2017-01-08
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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.

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    Thanks for the reply , do you mean (a-λ)(c-λ) - b^2 = (λ−2)(λ−3) ? – 2017-01-08
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    Yes, compare the coefficients. – 2017-01-08
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$$\begin{vmatrix} a-x & c \\ c & b-x \\ \end{vmatrix}=0$$

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?

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    Awesome! , thank you very much. – 2017-01-08
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    You are very welcome! – 2017-01-08
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Use the spectral theorem with an orthogonal basis different from the standard one: $$ A= 2 \begin{bmatrix}\frac{1}{\sqrt2}\\\frac{1}{\sqrt2}\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt2}&\frac{1}{\sqrt2}\end{bmatrix} +3 \begin{bmatrix}\frac{1}{\sqrt2}\\-\frac{1}{\sqrt2}\end{bmatrix} \begin{bmatrix}\frac{1}{\sqrt2}&-\frac{1}{\sqrt2}\end{bmatrix} $$