Given the following matrix:
$$B=\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & a^2 \\ 1 & 1 & 0 \end{bmatrix}$$
I tried to find for which values of $a$, the matrix $B$ is diagonalizable.
I found that the characteristic polynomial is: $P_B(x) = (1-x)(x+a)(x-a)$.
(Stop reading here and skip to the Edit section below)
Therefore I tried to find the eigenspace for each eigenvalue, but eventually concluded that:
- for $a=(-1)$, the eigenspaces are linearly dependent.
- for $a=1$, the trace of the diagonal form matrix (call it
D) isn't equal to the trace of the matrix that's composed of the eigenvectors (call itQ). - ('
a' must be 1 or (-1) according to the homogeneous equations with which I found the eigenspaces)
The diagonal form matrix that I have found (D):
$$D=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$$
The matrix that's composed of the eigenvectors that I have found (Q):
$$Q=\begin{bmatrix} 0 & -1 & 0 \\ 1 & 2 & -1 \\ 1 & 1 & 1 \end{bmatrix}$$
Edit:
I have found a mistake in the row reduction process of the matrices…
So now we have:
The diagonal form matrix (D):
$$D=\begin{bmatrix} 1 & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & -a \end{bmatrix}$$
The matrix that's composed of the eigenvectors (Q):
$$Q=\begin{bmatrix} 1-\frac{a^2+1}{2} & 0 & 0 \\ \frac{a^2+1}{2} & a & -a \\ 0 & 1 & 1 \end{bmatrix}$$
D and Q should be similar, thus by comparing their trace, I have found that:
$a_1=(1+\sqrt{2})$ and $a_2=(1-\sqrt{2})$. Does it make sense?
Edit2 – To conclude:
I was confused about the relations between the matrices $B,Q \text{ and }D$:
At first I thought that matrices $Q \text{ and }D$ must be similar, but that isn't necessarily true!
Only matrices $B \text{ and }D$ must be similar.
Edit 3 – Response to Marc:
I understood everything until the last sentence. Also I tried an example using Wolfram Alpha on this matrix. Does it have something to do with a nilpotent characteristic of the matrix? Indeed I can compute $(B-pI)(B-qI)$ and also could've seen with the example that it's true, but don't understand the rules (or characteristics) which allow this to be true.
Best Answer
The eigenvalues are $1, a, -a$ so if $a^2 \neq 1,$ then $A$ has three different eigenvalues so $A$ is diagonalizable.
Case $a^2 = 1$ : the rank of $$A - I = \pmatrix{0&0&0\\1&-1&1\\1&1&-1}$$ is $2$ because the first and second columns are linearly independent, so the dimension of the null space of $A-I$ is $1$. Therefore when $a^2=1, A$ is not diagonalizable.
added after the user1551's comment. Case $a = 0$:
the rank of $A$ is $2$ therefore the null space of $A$ has dimension $1$ so again $A$ is not diagonalizable.