Find missing Eigenvalues and check invertability

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?
• Isn't it $v_1=(p,q,r)$? – Mathlover Jan 7 '17 at 17:54
• Yes it is, will edit. Thanks! – Stijn Hoste Jan 7 '17 at 17:55
• Do you know $Av=\lambda v$ so $\lambda_1=0$? – Mathlover Jan 7 '17 at 17:56
• And distinct eigenvalues implies diagonalizability? – Mathlover Jan 7 '17 at 17:57
• So that would imply that $\lambda_3 = \frac{-1}{3}$? – Stijn Hoste Jan 7 '17 at 18:06

As @mathlover said: because of $Av = \lambda v \to \lambda_1=0$
Since $\lambda_1+\lambda_2+\lambda_3=\frac{2}{3} \to \lambda_3 = -\frac{1}{3}$
$\to$ A is diagonalisable.
$\to$ Since $\lambda_1 = 0$, A is not invertable.