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?
 A: 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.
A: 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 $(3+\lambda)(1+\lambda)(2-\lambda)$

A: 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!)
