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Eigen value is given as $\lambda = 2,-3,5$..

$v_{1} = \begin{bmatrix} 1\\-3\\-2 \end{bmatrix}$ $v_{2} =\begin{bmatrix} -2\\7\\5 \end{bmatrix}$ $v_{2} =\begin{bmatrix} 0\\0\\1 \end{bmatrix}$

A) WRITE DOWN THE VALUES OF Sv1, Sv2 and Sv3 by definition this is the eigen values times the corresponding eigen vector (where eigen value 2 corresponds to --> $V_{1}$)

B) Let $w=7v_{1}+3v_{2}+3v_{3}$ calculate w and SW I basically used basic constant time vector and vector addition...

$$w = 7\begin{bmatrix} 1\\-3\\-2 \end{bmatrix} + 3\begin{bmatrix} -2\\7\\5 \end{bmatrix}+\begin{bmatrix} 0\\0\\1 \end{bmatrix} = \begin{bmatrix} 2\\-6\\4 \end{bmatrix} $$

By the definition of an eigenvalue multiplying the product of S and its Eigen with a constant multiple then the answer is in proportion to the normal eigen value and factor.

$$Sw = 2*7\begin{bmatrix} 1\\-3\\-2 \end{bmatrix} + -3*3\begin{bmatrix} -2\\7\\5 \end{bmatrix}+5\begin{bmatrix} 0\\0\\1 \end{bmatrix} = \begin{bmatrix} 32\\-105\\-78 \end{bmatrix} $$ Is this right?

c) This is where i get lost. It says "By referring to b or otherwise evaluate $S*\begin{bmatrix} 0\\1\\0 \end{bmatrix}$"

How do i apply the eigen vectors heere?

D) hence find the matrix S... What!!! How even?

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Hint

Write $(0,1,0)$ in terms of the eigenvectors, say $(0,1,0)=a_1v_1+a_2v_2+a_3v_3.$ Then

$$S(0,1,0)=a_1S(v_1)+a_2S(v_2)+a_3S(v_3)$$ where you know all terms in RHS.

In a similar way you can get $S(1,0,0)$ and $S(0,0,1).$ Finally, isn't $S(1,0,0)$ the first column of $S?$

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  • $\begingroup$ Hey how did i go for a and b? How did the eigenvectors come into play? Are a1 to a3 the eigen values given... Or are the (0,1,0}? $\endgroup$ – Ivan Mar 19 '15 at 19:31
  • $\begingroup$ Solving the linear system $a_1v_1+a_2v_2+a_3v_3=(0,1,0)$. You should get $a_1=2,a_2=1,a_3=-1.$ $\endgroup$ – mfl Mar 19 '15 at 19:32
  • $\begingroup$ Was A and B fine? How does this connect to the result of b? $\endgroup$ – Ivan Mar 19 '15 at 19:56
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    $\begingroup$ Your answer to A and B are correct. Just a typo. Line $10$ should read $Sw=\cdots$ What do you mean by "the result of b"? $\endgroup$ – mfl Mar 19 '15 at 20:03
  • $\begingroup$ I was given a hint that said "By referring to b or otherwise evaluate $S*\begin{bmatrix} 0\\1\\0 \end{bmatrix}$" $\endgroup$ – Ivan Mar 19 '15 at 20:20
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If I well understand you have a matrix $S$ that has eigenvalues $\lambda_1=2$, $\lambda_2=-3$ and $\lambda_3=5$, with corresponding eigevectors:

$v_{1} = \begin{bmatrix} 1\\-3\\-2 \end{bmatrix}$ $v_{2} =\begin{bmatrix} -2\\7\\5 \end{bmatrix}$ $v_{2} =\begin{bmatrix} 0\\0\\1 \end{bmatrix}$

So, your matrix diagonalizable and it is: $ S= PDP^{-1} $ with: $$ P= \begin{bmatrix} 1&-2&0\\ -3&7&0\\ -2&5&1 \end{bmatrix} $$ and

$$ D= \begin{bmatrix} 2&0&0\\ 0&-3&0\\ 0&0&5 \end{bmatrix} $$ So you have the answer to D) and you can find the answer to C) eavluating the product.

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  • $\begingroup$ Do you think the OP is familiar with this result? $\endgroup$ – mfl Mar 19 '15 at 19:49
  • $\begingroup$ I don't think we've covered that topic yet $\endgroup$ – Ivan Mar 19 '15 at 20:21

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