\begin{bmatrix}-13&40&-48\\-8&23&-24\\0&0&3\end{bmatrix}
Consider the matrix above. This corresponds to
\begin{align*}
x_{1}'&=-13x_{1}+40x_{2}-48x_{3}\\
x_{2}'&=-8x_{1}+23x_{2}-24x_{3}\\
x_{3}'&=3x_{3}
\end{align*}
What I want to do is use eigenvectors to find the general solution. First I computed $\det(A-\lambda I)=0$. From this I got my eigenvalues to be $\lambda = 7$ and $\lambda = 3$ (this one is multiplicity 2).
Next I would find $(A-\lambda I)x=0$ where $x$ is the eigenvector I am looking for. For $\lambda=7$, my eigenvector came out to be,
\begin{bmatrix} 2\\1\\0 \end{bmatrix}
So no problems here. On the other hand, I had some difficulties for when $\lambda = 3$. The matrix I would be solving for is
\begin{bmatrix} -16&40&-48&0 \\ -8&20&-24&0 \\ 0&0&0&0 \end{bmatrix}
When I do this I obtain,
\begin{bmatrix} 1&-2.5&3&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{bmatrix}
so $$x_{1} -2.5x_{2}+3x_{3} = 0.$$
From here I am not sure how to get the two eigenvectors. The computer told me they are,
\begin{bmatrix} -3 \\ 0 \\ 1 \end{bmatrix} and \begin{bmatrix} 5 \\ 2 \\ 0 \end{bmatrix}
but I have no idea how to get these from what I have remaining. Also how would I write the general solution?
Best Answer
Solving the System of linear Equations
So it seems you want to solve the system of equations $$ Ax = \begin{pmatrix}-13&40&-48\\-8&23&-24\\0&0&3\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x'_1 \\ x'_2 \\ x'_3 \end{pmatrix} = x' $$ Where the $x'$ is given and the $x_j$ are looked for.
You would normally solve this by calculating the inverse of $A$, since then $$ x = A^{-1} x' $$ But let's say you really want to solve this by using the eigenvalues and eigenvectors. You have already determined both of these, providing the change of base matrix $S$ which contains the eigenvectors $$ S = \begin{pmatrix} 2 & 5 & -3 \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ which accomplishes $$ S^{-1} A S = D = \begin{pmatrix} 7 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{pmatrix} $$ We can now solve for $x$. Using the above, \begin{align*} x &= A^{-1} x' \\&= (SDS^{-1})^{-1} x' \\&= SD^{-1}S^{-1} x' \\&= \begin{pmatrix} 2 & 5 & -3 \\ 1 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{7} & 0 & 0 \\ 0 & \frac{1}{3} & 0 \\ 0 & 0 & \frac{1}{7} \end{pmatrix} S^{-1} x' \end{align*} As you can see, we do not yet have the solution, since we don't know what $S^{-1}$ looks like. So what's left to do is: calculate the inverse of $S$, multiply everything and be done.
(So whatever method you use, you have to calculate the inverse of some matrix)
Getting the Eigenvectors
To determine the last two eigenvectors, you got the equation $$ x_1 - 2.5 x_2 + 3 x_3 = 0 $$ whose solutions $(x_1,x_2,x_3)$ you want to find. What the equation tells us is, that if we are given two of the variables, the third one (for example $x_1$) will be determined. $$ x_1 = 2.5 x_2 - 3 x_3 $$ Thus the solutions can be written as $$ ( 2.5 x_2 - 3 x_3 , x_2 , x_3 ) \equiv \begin{pmatrix} 2.5 x_2 - 3 x_3 \\ x_2 \\ x_3 \end{pmatrix} = x_2 \begin{pmatrix} 2.5 \\ 1 \\ 0 \end{pmatrix} + x_3 \begin{pmatrix} - 3 \\ 0 \\ 1 \end{pmatrix} ~~~~~~~ \text{with} ~~ x_2 , x_3 \in \mathbb{R} $$ Any solution of this form is an eigenvector. Since you need two eigenvectors, you take two different solutions. For example the solutions with $$ (x_2,x_3) = (2,0) ~~~\text{or}~~~ (0,1) $$ Which yield the eigenvectors that the computer gave you.