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 $\frac{5+3\sqrt{5}}{2} , \frac{5-3\sqrt{5}}{2}$
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try youtube – user66081 Aug 01 '16 at 17:05
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1You don't need to know the eigenvectors to diagonalize it. As long as it has $2$ distinct eigenvalues $\lambda_1,\lambda_2$ it can be diagonalized to the diagonal matrix with entries $\lambda_1,\lambda_2$ – Asinomás Aug 01 '16 at 17:07
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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?
Doug M
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@Ankit In this case, there’s a shortcut to finding the second eigenvector. Since the matrix is symmetric, its eigenvectors will be orthogonal, so once you have one eigenvector, the other can be written down without further ado. – amd Aug 01 '16 at 17:59