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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

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    $\begingroup$ To begin with forget about the fact that $v_1, v_2, v_3$ are eigenvectors. You should know by now how to find numbers $x,y,z$ such that $$xv_1 + yv_2 + zv_3 = v$$ Hint: can you represent this equation in the form $Ax=b$? $\endgroup$
    – user137731
    Commented Jul 19, 2016 at 3:04
  • $\begingroup$ I did $\begin{bmatrix} 0&1&2\\-2&2&0\\2&0&-1\end{bmatrix}$ and got the REF of $\begin{bmatrix}2&0&-1\\0&2&-1\\0&0&-5\end{bmatrix}$ so then I got $2x - z =0$, $2y-z = 0$ and $-5z = 5 \implies z = -1$ so then I get $y = -1/2$ which is different than what Majid is doing $\endgroup$
    – Yusha
    Commented Jul 19, 2016 at 3:39

2 Answers 2

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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 $$Av=A(2v_1+2v_2-v_3)=-2v_1+2v_2-4v_3= \begin{bmatrix}-6\\8\\0\end{bmatrix}$$

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  • $\begingroup$ I did $\begin{bmatrix} 0&1&2\\-2&2&0\\2&0&-1\end{bmatrix}$ and got the REF of $\begin{bmatrix}2&0&-1\\0&2&-1\\0&0&-5\end{bmatrix}$ so then I got $2x - z =0$, $2y-z = 0$ and $-5z = 5 \implies z = -1$ so then I get $y = -1/2$ why is mine different $\endgroup$
    – Yusha
    Commented Jul 19, 2016 at 3:41
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    $\begingroup$ Actually, in your way, you updated the matrix without updating the vector $v$. In fact, you should put the vector $v$ by side of the matrix and do the same to this vector as you do to the matrix, and at the end, you have a system according to updated version of vector $v$. $\endgroup$
    – Majid
    Commented Jul 19, 2016 at 3:47
  • $\begingroup$ That makse sense! $\endgroup$
    – Yusha
    Commented Jul 19, 2016 at 3:48
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Hint: Try this. Assume arbitrary scalars $\alpha,\beta,\gamma$ and take $$v=\alpha v_1+\beta v_2+\gamma v_3.$$ Now solve for $\alpha,\beta,\gamma$.

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