Characteristic polynomial of the symmetric matrix

Question

I tried to solve the following example:

Find all symmetric and real matrices A with the characteristic polynomial:$$ p_A(\lambda) = \lambda^2 – 4 \lambda + 4 $$

I have tried to approach this with diagonal matrices that have roots of the polynomial (or its n-th square root) on the diagonal. Yet this seems unsatisfactory. What do I miss?

Thank you in advance for any clue.

Answer 1

Here is easy beacuse the degree of the polynomial says you that the matrix has order $2$ and the symmetric property says you that it is similar to a diagonal matrix

$$\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$

Thus

$$\begin{cases} \lambda_1+\lambda_2=4\\ \lambda_1\lambda_2=4 \end{cases}$$

That tells you $\lambda_1=\lambda_2=2$.

Hence the symmetric matrices having that characteristic polynomial are $B(2I)B^{-1}=2I$. Thus there is only one, $2I$.