# Find $A(v_1+v_2)$ and $A(3v_1)$ given eigenvectors and eigenvalues

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.

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

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: $$A(3v_1)=3A(v_1)=-3(v_1)=[3;-15].$$