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Find the eigenvalues and corresponding eigen vectors of the matrix $\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$

The eigenvalue $\lambda_1 = $____ corresponds to the eigevector$( \ ,\ , \ )$.

The eigenvalue $\lambda_2 = $____ corresponds to the eigevector$( \ ,\ , \ )$.

The eigenvalue $\lambda_3 = $____ corresponds to the eigevector$( \ ,\ , \ )$.

I'm kind of stuck after a certain point. Here is what I have so far

I do know that $(A - \lambda I)X = 0$

so $\begin{bmatrix}-3&6&-43\\0&-1&9\\0&0&2\end{bmatrix}$ $\implies \lambda_1 = -3, \lambda_2 = -1, \lambda_3 = 2$ so I have the eigenvalues but how can I find the corresponding eigenvectors?

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  • $\begingroup$ Do you know how to find eigenvalues? This is an upper triangular matrix. What about it's characteristic polynomial? $\endgroup$ – Kushal Bhuyan Jul 14 '16 at 1:40
  • $\begingroup$ In any good textbook, class, video lectures or lecture notes on linear algebra, solving linear systems should be covered before finding eigenvalues. In particular, the Gaussian elimination method first puts a linear system into "upper triangular" form, and then finds the unknowns starting from the last ones. $\endgroup$ – arctic tern Jul 14 '16 at 3:27
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You need to solve the equations $(A-\lambda I)v=0$ for $v$ for each of the three eigenvalues $\lambda$.

For instance when $\lambda=2$ we're solving

$$\begin{pmatrix} -5 & 6 & -43 \\ 0 & -3 & 9 \\ 0 & 0 & 0\end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$

The last equation is $0=0$ so it's superfluous. So we have two equations in three unknowns. If $z=0$ then the second equation implies $y=0$ and then the first equation implies $x=0$ which gives the zero vector, and that's not very interesting. Otherwise if $z\ne0$, we can scale the vector $(x,y,z)$ (scaling preserves the property of being a solution to the above system) in order to make $z=1$. In which case the second equation gives $y=3$ and then the first equation gives $x=-5$. So the eigenvector is $(-5,3,1)$ up to scaling.

Your turn. Do $\lambda=-3$ and $\lambda=-1$. (Sophia did $\lambda=-3$ for you, so now most of the work is done for you. If we do the problem for you, at least learn what it is we're doing!)

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  • $\begingroup$ For $\lambda_2 = -1$ I get $\begin{bmatrix}-2&6&-43\\0&0&9\\0&0&3\end{bmatrix}$ ~ $\begin{bmatrix}1&-3&\frac{43}{2}\\0&0&9\\0&0&0\end{bmatrix}$ so I thought it would be (-3,0,0) ? $\endgroup$ – Yusha Jul 14 '16 at 3:27
  • $\begingroup$ When I multiply that matrix by $(-3,0,0)^T$ I don't get the zero matrix. In that situation, the last equation is redundant. What does the second equation tell you about $z$? What does the first equation then reduce to saying about $x$ and $y$? (BTW I wouldn't bother with the fractions.) $\endgroup$ – arctic tern Jul 14 '16 at 3:30
  • $\begingroup$ Wow, I'm an idiot! I see now that $z = 0 \implies 6y = 2x \implies (3,1,0)$ $\endgroup$ – Yusha Jul 14 '16 at 3:31
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You start with the understanding of this formula: $(A-\lambda I)\vec x=0$, which is equivalent to $\det(A-\lambda I)=0$ $$\begin{vmatrix}-3-\lambda&6&-43\\0&-1-\lambda&9\\0&0&2-\lambda\end{vmatrix}=(-3-\lambda)(-1-\lambda)(2-\lambda)=0$$ Therefore, $\lambda_1=-3, \ \lambda_2=-1, \ \lambda_3=2$.
Let's do one example for eigenvectors:
Plug in the value of $\lambda$ into the augmented form of the matrix:
With $\lambda_1=-3$, $$\left[\begin{array}{ccc|c}-3-(-3)&6&-43&0\\0&-1-(-3)&9&0\\0&0&2-(-3)&0\end{array}\right]=\left[\begin{array}{ccc|c}0&6&-43&0\\0&2&9&0\\0&0&5&0\end{array}\right]$$ Solve this matrix and get $v_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$
Now you can use similar approach to find the eigenvectors of the next two eigenvalues.

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  • $\begingroup$ @Yusha, double check your calculations. At least for $\lambda=-1$, I did not get $(-3,0,0)$ $\endgroup$ – Ron Jul 14 '16 at 2:18
  • $\begingroup$ I'm lost, how are you getting (1,0,0). You have in your work (0,0,0). $\endgroup$ – Yusha Jul 14 '16 at 2:43
  • $\begingroup$ Both answers are correct, since one is a scalar multiple of the other. But it is customary to use numbers are small as possible, so $(1,0,0)$ would be prefered. $\endgroup$ – imranfat Jul 14 '16 at 2:47
  • $\begingroup$ Does (1,0,0) @imranfat come from the 2nd column after its been put in RREF? $\endgroup$ – Yusha Jul 14 '16 at 2:50
  • $\begingroup$ RREF has to give infinite solutions, I looked at Sophia's work, realizing that both vectors serve as the same eigenvector. Your RREF must have all zero's in bottom row $\endgroup$ – imranfat Jul 14 '16 at 2:55
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Hint: In an upper triangular matrix the characteristic polynomial is $(x_1-\lambda_1)^{\alpha_1}(x_2-\lambda_2)^{\alpha_2}\ldots(x_n-\lambda_n)^{\alpha_n}$, where $x_i$ are diagonal entries with $\alpha$'s as their multiplicities.

In your case the characteristic polynomial is $(3+\lambda)(1+\lambda)(2-\lambda)$

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    $\begingroup$ Here is one: sosmath.com/matrix/eigen2/eigen2.html $\endgroup$ – imranfat Jul 14 '16 at 2:27
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    $\begingroup$ Khan academy usually rocks: khanacademy.org/math/linear-algebra/alternate-bases/… $\endgroup$ – imranfat Jul 14 '16 at 2:28
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    $\begingroup$ OK, so I assume that you DO know how to find eigenvalues theoretically, but somehow your algebra is letting you down? It would be helpful to see your work in attempting finding the eigenvalue, because a lot of times it is a small algebraic mistake... $\endgroup$ – imranfat Jul 14 '16 at 2:33
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    $\begingroup$ Here is a worked out example of a 2 by 2 matrix: calvin.edu/~scofield/courses/m256/materials/eigenstuff.pdf In my view you should first be fluent with 2by2's before going for 3by3's... $\endgroup$ – imranfat Jul 14 '16 at 2:34
  • $\begingroup$ It sounds like your exchange converged. Anyway, this conversation has been moved to chat. If you need to revisit the full exchange of comments, go to that chatroom. I left the posts with links to resources also here. $\endgroup$ – Jyrki Lahtonen Jul 14 '16 at 6:26

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