# Find the eigenvalues and corresponding eigen vectors of the matrix

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?

• Do you know how to find eigenvalues? This is an upper triangular matrix. What about it's characteristic polynomial? Commented Jul 14, 2016 at 1:40
• 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.
– anon
Commented Jul 14, 2016 at 3:27

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.

• @Yusha, double check your calculations. At least for $\lambda=-1$, I did not get $(-3,0,0)$
– Ron
Commented Jul 14, 2016 at 2:18
• I'm lost, how are you getting (1,0,0). You have in your work (0,0,0). Commented Jul 14, 2016 at 2:43
• 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. Commented Jul 14, 2016 at 2:47
• Does (1,0,0) @imranfat come from the 2nd column after its been put in RREF? Commented Jul 14, 2016 at 2:50
• 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 Commented Jul 14, 2016 at 2:55

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!)

• 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) ? Commented Jul 14, 2016 at 3:27
• 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.)
– anon
Commented Jul 14, 2016 at 3:30
• Wow, I'm an idiot! I see now that $z = 0 \implies 6y = 2x \implies (3,1,0)$ Commented Jul 14, 2016 at 3:31

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 $(3+\lambda)(1+\lambda)(2-\lambda)$

• Here is one: sosmath.com/matrix/eigen2/eigen2.html Commented Jul 14, 2016 at 2:27
• Khan academy usually rocks: khanacademy.org/math/linear-algebra/alternate-bases/… Commented Jul 14, 2016 at 2:28
• 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... Commented Jul 14, 2016 at 2:33
• 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... Commented Jul 14, 2016 at 2:34
• 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. Commented Jul 14, 2016 at 6:26