I have the following matrix:
$$A=\left(\begin{array}{ccc}{2} & {-\alpha} & {1} \\ {3 \alpha} & {-2} & {3} \\ {0} & {0} & {\alpha}\end{array}\right)$$
And I have to find the value or values of $\alpha$ for which the matrix only has two eigenvalues. Then, I need to calculate the change of basis matrix P, so that D is the resulting diagonal matrix ($D = P^{-1} A P$). But I am not sure if I am getting this right. I will state my question first, and I will then show my work.
If the problem is set in such a way so that we only get two eigenvalues, and therefore we get two spaces with dimension 1, shouldn’t this $3\times 3$ matrix become non-diagonalizable? Why, then, am I being asked to find the matrices $P$ and $P^{-1}$?
The first thing I did was to find out the characteristic polynomial, $p(\lambda)$, which turned out to be: $$p(\lambda) = -\lambda^3 + \alpha \lambda^2 + (4-3\alpha^2)\lambda + 3\alpha^3 - 4\alpha $$
It is apparent that making the coefficient of $\lambda$ and the independent term zero, we will have an order 3 polynomial with no more than two solutions. Therefore, we set $(4-3\alpha^2) = 0$ and $3\alpha^3 - 4\alpha$, being the solution $\alpha = \pm \frac{2}{\sqrt{3}}$. We have to different values of $\alpha$ for which the matrix $A$ only has two eigenvalues. Those eigenvalues turn out to be: $$p(\lambda) = 0 \rightarrow -\lambda^3 +\frac{2}{\sqrt{3}} \lambda ^2 = 0 $$ $$\lambda_1 = 0, \lambda_2 = \frac{2}{\sqrt{3}}$$ With these eigenvalues we calculate the basis for each eigenspace: $$\lambda=0 \rightarrow B_0 = \{ (1,\sqrt{3},0)\}$$ $$\lambda=2/\sqrt{3} \rightarrow B_{2/\sqrt{3}} = \{ ( 1,-3, -\frac{2}{3}(3+2\sqrt{3}) ) \}$$
So, since $\dim(B_0) + \dim(B_{2/\sqrt{3}}) = 2 \neq 3$, isn't $A$ non-diagonalizable? Thank you.
There are other possibilities for $\alpha$ than the two that you examined.
To answer your first question, you’re given that the matrix has two distinct eigenvalues and is also diagonalizable, so one of the eigenspaces must be two-dimensional. In other words, the algebraic and geometric multiplicities of one of the two eigenvalues must be $2$.
We can see at a glance that $\alpha$ is one of the eigenvalues. The trace of the matrix is equal to $\alpha$, so its other eigenvalues must be of the form $\pm\mu$. To have exactly two distinct eigenvalues, then, we must have either $\mu=0$ or $\mu=\alpha$. In either case, $\alpha\ne0$.
The condition that the repeated eigenvalue $\lambda$ have geometric multiplicity $2$ is equivalent to saying that $A-\lambda I$ is a rank-one matrix. This can only occur if $\alpha-\lambda=0$, which means that $\alpha$ must be the repeated eigenvalue. That eliminates the case $\mu=0$, therefore the eigenvalues of the matrix are $\pm\alpha$.
The determinant of the matrix is equal to $\alpha(4-3\alpha)^2$. This must be equal to the product of the eigenvalues, which is $-\alpha^3$, from which $\alpha=\pm1$. Examining $A-\alpha I$ for these two cases, for $\alpha=1$ we have $$A-I = \begin{bmatrix}1&-1&1\\3&-3&3\\0&0&0\end{bmatrix},$$ which clearly has rank $1$, and for $\alpha=-1$, $$A+I = \begin{bmatrix}3&1&1\\-3&-1&3\\0&0&0\end{bmatrix},$$ which obviously does not. Therefore, $\alpha=1$ is the unique value that produces a matrix meeting the criteria of the problem. Computing eigenvectors for it is a straightforward matter from here.
You could work with the characteristic polynomial instead, of course. Expanding $\det(\lambda I-A)$ along the last row gives $$(\lambda-\alpha)\left((\lambda-2)(\lambda+2)+3\alpha^2\right) = (\lambda-\alpha)(\lambda^2+3\alpha^2-4).$$ Expanding further is counterproductive, I think, because the roots are easily found from this partially-factored form: they are $\alpha$ and $\pm\sqrt{4-3\alpha^2}$. For two distinct real roots, we must have either $4-3\alpha^2=0$, which is the case you examined, or $4-3\alpha^2=\alpha^2$, which you missed.