# Eigenvalue of symmetric matrix with same diagonal elements

How to find the eigenvalues of the following matrix $$M = \begin{bmatrix} 10 & 6 \\ 6 & 10 \end{bmatrix}$$ I have used the following method $$\begin{vmatrix} (10- \lambda) & 6 \\ 6 & (10-\lambda) \end{vmatrix} = 0$$ Solving this I'm getting $$(10 - \lambda)^2 = 36$$ which gives me $\lambda = 4$. But that's the only eigenvalue I'm getting from this, how to calculate the other eigenvalue?

• $10-\lambda=\pm6$ – tired Feb 8 '17 at 9:52
• Its 16... Calculate the other zero of that polynomial... – Laray Feb 8 '17 at 9:52
• Obviously, $\lambda = 16$ is also a root of the polynomial. – Taufi Feb 8 '17 at 9:52
• For $2\times2$ matrix $M$, eigenvalues are two numbers like $\lambda_1$ and $\lambda_2$ such that $$\lambda_1+\lambda_1=tra(M)=20$$ and $$\lambda_1\times \lambda_2=det(M)=64$$. Now what are these two numbers? – Amin235 Feb 8 '17 at 10:32

## 1 Answer

Motivated from the comments of @tired
Your matrix $$M = \begin{bmatrix} 10 & 6 \\ 6 & 10 \end{bmatrix}$$
Calculating eigen values: $$\begin{vmatrix} (10- \lambda) & 6 \\ 6 & (10-\lambda) \end{vmatrix} = 0$$

The eigenvalues on solving this,

$$(10 - \lambda)^2 = 36$$
$100+\lambda^2-20 \lambda = 36$
$\lambda^2-20 \lambda + 64= 0$
$\lambda = 16,4$