Let $v_1 \begin{bmatrix}0\\-2\\2\end{bmatrix}, v_2 = \begin{bmatrix}1\\2\\0\end{bmatrix}$ and $v_3 = \begin{bmatrix}2\\0\\-1\end{bmatrix}$ be eigenvectors of the matrix $A$ which correspond to the eigenvalues $\lambda_1 = -1, \lambda_2 = 1$ and $\lambda_3 = 4$ respectively and let $v = \begin{bmatrix}0\\0\\5\end{bmatrix}$ Express $v$ as a linear combination of $v_1,v_2,$ and $v_3$, and find $Av$
$$v =\ \ …v_1 +\ \ …v_2+\ \ …v_3$$
$Av = $ (Should be a $3\times 1$ Matrix)
I know $A(v_1 + v_2 + v_3) = Av_1 + Av_2 + Av_3 = \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3$ though I'm not sure how to get the linear combination? Are they wanting me to put this matrix in RREF form and solve a system of equations? Or should I let $v = \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3$? I'm sort of lost
Best Answer
Assuming $av_1+bv_2+cv_3=v$, we have $$\begin{cases} b+2c=0\\ -2a+2b=0\\ 2a-c=5 \end{cases}$$ and by solving the sysytem, you have got $a=2$, $b=2$, and $c=-1$.
To find $Av$, as you mentioned, we have $$Av=A(2v_1+2v_2-v_3)=-2v_1+2v_2-4v_3= \begin{bmatrix}-6\\8\\0\end{bmatrix}$$