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I've been struggling with this problem for a couple of days now. I'm familiar with matrices and how to find eigenvalues and eigenvectors to write them as $QDQ^{-1}$. But I don't seem to able to crack this one. Can somebody maybe help out?

Given:

$\lambda_1+\lambda_2+\lambda_3=\frac{2}{3}$

$a_{1,1}+a_{2,2}+a_{3,3}=\frac{2}{3}$

$A * \begin{pmatrix}p\\q\\r\\\end{pmatrix} = 0$

$\lambda_1 =$ ? with eigenvector $v_1 = (p, q, r)$

$\lambda_2 = 1$ with eigenvector $v_2 = (1,0,0)$

$\lambda_3 =$ ? with eigenvector $v_3 = (-15, 4, 0)$

Find:

  • Find the missing eigenvalues $\lambda_1$ and $\lambda_2$.
  • Is matrix A diagonalizable?
  • Is matrix A invertable?
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  • $\begingroup$ Isn't it $v_1=(p,q,r)$? $\endgroup$ – Mathlover Jan 7 '17 at 17:54
  • $\begingroup$ Yes it is, will edit. Thanks! $\endgroup$ – Stijn Hoste Jan 7 '17 at 17:55
  • $\begingroup$ Do you know $Av=\lambda v$ so $\lambda_1=0$? $\endgroup$ – Mathlover Jan 7 '17 at 17:56
  • $\begingroup$ And distinct eigenvalues implies diagonalizability? $\endgroup$ – Mathlover Jan 7 '17 at 17:57
  • $\begingroup$ So that would imply that $\lambda_3 = \frac{-1}{3}$? $\endgroup$ – Stijn Hoste Jan 7 '17 at 18:06
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As @mathlover said: because of $Av = \lambda v \to \lambda_1=0$

Since $\lambda_1+\lambda_2+\lambda_3=\frac{2}{3} \to \lambda_3 = -\frac{1}{3}$

An n by n matrix A is diagonalisable if and only if a A has n lineair undependent eigenvectors.

$\to$ A is diagonalisable.

A n by n matrix A is invertable if and only if 0 is not a Eigenvector of A

$\to$ Since $\lambda_1 = 0$, A is not invertable.

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