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If $v_1=[-1;5]$ and $v_2=[-3;5]$ are eigenvectors of a matrix $A$

corresponding to the eigenvalues $\lambda_1=-1$ and $\lambda_2=1$, find $A(v_1+v_2)$ and $A(3v_1).$

I managed to find $A,$ which I believe is $[[2,\frac 35];[-5,-2]]$, but I'm unsure of how to continue.

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    $\begingroup$ Please use MathJax. By linearity, $A(v_1+v_2)=A(v_1)+A(v_2)$ $\endgroup$ – J. W. Tanner Mar 19 '19 at 19:05
  • $\begingroup$ How do you get two As? I only get one A. $\endgroup$ – Amy Kulp Mar 19 '19 at 19:10
  • $\begingroup$ My equation with two As holds because $A$ is a linear map $\endgroup$ – J. W. Tanner Mar 19 '19 at 19:13
  • $\begingroup$ You don’t need to construct $A$ explicitly to solve this. Use the definition of an eigenvector and linearity. $\endgroup$ – amd Mar 19 '19 at 20:59
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You could say $v_1+v_2=[-1;5]+[-3;5]=[-4;10]$,

and when you multiply that by the matrix you found, the result is $[-2;0].$

Alternatively, by linearity, $A(v_1+v_2)=A(v_1)+A(v_2)=-1v_1+1v_2=[-2;0].$

To find $A(3v_1)$, you could say $3v_1=[-3;15],$

and when you multiply that by the matrix you found, the result is $[3;-15],$

but I find it again easier to use linearity: $A(3v_1)=3A(v_1)=-3(v_1)=[3;-15].$

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