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If $A$ is a $3\times 3$ matrix with eigenvector $\begin{bmatrix} 3\\ 0\\ -2\end{bmatrix}$ corresponding to an eigenvalue of $5$ and $\begin{bmatrix} -1\\ 2\\ 7\end{bmatrix}$ corresponding to an eigenvalue of $2$, and $\textbf{v}=\begin{bmatrix}10 \\ 4 \\10\end{bmatrix}$ find $A\textbf{v}$.

My attempt at solving this problem was to begin by stating that $Ax=\lambda x$, Where $\lambda$ is the eigenvalue and $x$ is the eigenvector and then saying that $$A=5*(3,0,-2)+2*(-1,2,7)=(13,4,4)=A.$$ Then I followed by saying that $Av=(13,4,4)*(10,4,10)=186$ but I'm sure If I'm moving in the right direction..

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  • $\begingroup$ We are not here to do your homework for you. You must tell us what you have tried and where you are stuck to receive help. $\endgroup$ – AnotherPerson May 7 '16 at 1:12
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    $\begingroup$ My attempt at the solution is clearly stated above thank you. $\endgroup$ – Peter B. May 7 '16 at 1:14
  • $\begingroup$ Why did you write $A$ as a 1x3 matrix when it says that it is a 3x3? $\endgroup$ – AnotherPerson May 7 '16 at 1:20
  • $\begingroup$ What is the vector v ? What does it represent ? Tell me and I'll give you the full answer below because it seems you have misunderstood things about that. $\endgroup$ – Rebellos May 7 '16 at 1:20
  • $\begingroup$ v represents the eigenvector of A ? $\endgroup$ – Peter B. May 7 '16 at 1:24
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With the eigenvectors and eigenvalues $$ v_1 = (3, 0, -2)^t \quad \lambda_1 = 5 \\ v_2 = (-1, 2, 7)^t \quad \lambda_2 = 2 $$ the systems $A v_i = \lambda_i$ give only $6$ equations for the $9$ unknown matrix elements $a_{ij}$.

It turns out that $v = (10,4,10) \not\in \langle v_1, v_2 \rangle$: We have to choose $$ v = 2 v_2 + c v_1 = (3c-2, 4, -2c + 14) $$ because of the middle component. To get the first component right, we need $c = 4$, to get the third component right, we need $c = 2$. Otherwise we could have used $$ A v = A (c_1 v_1 + c_2 v_2) = c_1 (A v_1) + c_2 (A v_2) = 5 c_1 v_1 + 2 c_2 v_2 $$

So this problem seems not to have a unique solution.

One possible solution is $$ A = \begin{pmatrix} 5 & 3/2 & 0 \\ 0 & 2 & 0 \\ -10/3 & 16/3 & 0 \end{pmatrix} $$ For this matrix we have $A v = (56, 8, -12)^t$.

This matrix $A$ was derived from the system $$ [M|b] = \left[ \begin{array}{rrrrrrrrr|r} 3 & 0 & -2& 0 & 0 & 0 & 0 & 0 & 0 & 15 \\ 0 & 0 & 0 & 3 & 0 & -2 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 3 & 0 & -2 & -10\\ -1 & 2 & 7 & 0 & 0 & 0 & 0 & 0 & 0 & -2\\ 0 & 0 & 0 & -1 & 2 & 7 & 0 & 0 & 0 & 4\\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 2 & 7 & 14\\ \end{array} \right] $$ and its row echelon form $$ \left[ \begin{array}{rrrrrrrrr|r} 1 & 0 &-2/3 & 0 & 0 & 0 & 0 &0 & 0 & 5 \\ 0 & 1 & 19/6 & 0 & 0 & 0 & 0 &0 & 0 & 3/2 \\ 0 & 0 & 0 & 1 & 0 & -2/3 & 0 &0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 19/6 & 0 &0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 &0 & -2/3 & -10/3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 &1 & 19/6 & 16/3 \\ \end{array} \right] $$ which can be extended into $$ \left[ \begin{array}{rrrrrrrrr|r} 1 & 0 &-2/3 & 0 & 0 & 0 & 0 &0 & 0 & 5 \\ 0 & 1 & 19/6 & 0 & 0 & 0 & 0 &0 & 0 & 3/2 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 &0 & 0 & a_{13} \\ 0 & 0 & 0 & 1 & 0 & -2/3 & 0 &0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 19/6 & 0 &0 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 &0 & 0 & a_{23} \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 &0 & -2/3 & -10/3 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 &1 & 19/6 & 16/3 \\ 0 & 0 & 0 & 0 &0 & 0 & 0 & 0 & 1 & a_{33} \\ \end{array} \right] $$ and where the unknowns are the matrix elements $x = (a_{11}, a_{12}, a_{13}, a_{21}, \dotsc, a_{33})^t$.

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$Ax = λx \Leftrightarrow (Α-λ)x = 0$

Let :

$ A = \left( \begin{array}{ccc} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{array} \right) $

Then, $ \left( \begin{array}{ccc} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{array} \right) \left( \begin{array}{ccc} 3 \\ 0 \\ -2 \end{array} \right) = 5 \left( \begin{array}{ccc} 3 \\ 0 \\ -2 \end{array} \right) $

and

$ \left( \begin{array}{ccc} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{array} \right) \left( \begin{array}{ccc} -1 \\ 2 \\ 7 \end{array} \right) = 2 \left( \begin{array}{ccc} -1 \\ 2 \\ 7 \end{array} \right) $

Proceed first of all by solving the system that is created by the matrix equations above and you should be able to find the matrix $A$. After finding $A$, do the multiplication $Av$, where $v =[10,4,10]^T$ and find the solution to your exercise.

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  • $\begingroup$ There are 9 unknowns , I tried to assemble the matrices regarding these ax=b systems and wasn't able to figure out any feasible way of determining matrix A $\endgroup$ – Peter B. May 7 '16 at 2:43
  • $\begingroup$ Maybe you are missing a part of the exercise because indeed the problem does not have a unique solution .Make sure that the exercise is exactly as you posted. $\endgroup$ – Rebellos May 7 '16 at 2:45
  • $\begingroup$ No the question is not missing anything i made sure to include everything the question asks for $\endgroup$ – Peter B. May 7 '16 at 2:50
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This isn't a real answer, just a sketch for the answer.

May A be 3 x 3 matrix

$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$

With

$Av = \lambda_1 v_1 \ , \ Av = \lambda_2 v_2$

I would solve those first and then proceed like you did before.

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