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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?

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    $\begingroup$ $10-\lambda=\pm6$ $\endgroup$ – tired Feb 8 '17 at 9:52
  • $\begingroup$ Its 16... Calculate the other zero of that polynomial... $\endgroup$ – Laray Feb 8 '17 at 9:52
  • $\begingroup$ Obviously, $\lambda = 16$ is also a root of the polynomial. $\endgroup$ – Taufi Feb 8 '17 at 9:52
  • $\begingroup$ 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? $\endgroup$ – Amin235 Feb 8 '17 at 10:32
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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 $

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