Diagonalize symmetric matrix $\small \begin{bmatrix} 4 & 3 \\ 3 & 1 \end{bmatrix}$

Question

please help me to find out the diagonalization of this matrix:

$$\begin{bmatrix} 4 & 3 \\ 3 & 1 \end{bmatrix}$$

I am stuck in finding out the Eigenvectors. My eigenvalues are $\frac{5+3\sqrt{5}}{2} , \frac{5-3\sqrt{5}}{2}$

Answer 1

You have found the two eigenvalues. What is an eigenvector that corresponds to the first eigenvalue?

$A\mathbf v = \lambda \mathbf v\\ (A - \lambda I)\mathbf v = \mathbf 0$

Find $\mathbf v$ such that: $(A - \lambda I)\mathbf v = \mathbf 0$

$\lambda = \frac {5 + 3 \sqrt 5}{2}\\ A - \lambda I = \begin{bmatrix}\frac{3 - 3\sqrt 5}{2}&3\\3&\frac{-3 - 3\sqrt 5}{2}\end{bmatrix}\\ \begin{bmatrix}\frac{3 - 3\sqrt 5}{2}&3\\3&\frac{-3 - 3\sqrt 5}{2}\end{bmatrix}\begin{bmatrix}2\\-1+\sqrt5\end{bmatrix}=\mathbf 0$

Can you follow this example to find the remaining eigenvector?