If $v_1 = \begin{bmatrix}5\\3\end{bmatrix}$ and $v_2 = \begin{bmatrix}3\\1\end{bmatrix}$ are eigenvectors of a matrix $A$ corresponding to the eigenvalues $\lambda_1 = -1$ and $\lambda_2 = 4$ respectively, then

$A(v_1 + v_2)= ?$

What about $A(3v_1) = ?$

I get what I'm doing wrong. I did $v_1 + v_2 = \begin{bmatrix}5\\3\end{bmatrix} + \begin{bmatrix}3\\1\end{bmatrix} = \begin{bmatrix}8\\4\end{bmatrix}$ then I did $A(\begin{bmatrix}8\\4\end{bmatrix})= \begin{bmatrix}-1&0\\0&4\end{bmatrix}*\begin{bmatrix}8\\4\end{bmatrix} = \begin{bmatrix}-8\\16\end{bmatrix}$

  • $\begingroup$ Hint. Do not add up the vectors first. You must first examine the multiplication $Av=\lambda{v}$ on both $v1$ and $v2$. And your eigenvalues are given.... $\endgroup$
    – imranfat
    Jul 19, 2016 at 2:30

1 Answer 1


$A$ is only diagonal with respect to a basis of eigenvectors, namely $v_1$ and $v_2$, not the standard basis, which is why your computation fails. Try using linearity and the definition of eigenvectors instead:

$A(v_1+v_2)=Av_1+Av_2=\lambda_1 v_1+\lambda_2 v_2$

Similarly for $A(3v_1)$.

  • $\begingroup$ I did that and got the same answer. $\endgroup$
    – Yusha
    Jul 19, 2016 at 2:38
  • 1
    $\begingroup$ You must have made an error. Did you compute $-v_1+4v_2$? $\endgroup$
    – solstafir
    Jul 19, 2016 at 2:39
  • $\begingroup$ You're correct. I failed to see the property you were showing. I see it now and got (7,1) for the first one and (-15,-9) for the second. Thanks $\endgroup$
    – Yusha
    Jul 19, 2016 at 2:41

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