Let $$A = \begin{pmatrix}-1 & 0 & 0 & 0 \\\ a & -1 & 0 & 0 \\\ b & d & 1 & 0 \\\ c & e & f & 1\end{pmatrix}$$
Its eigenvalues are $1$ and $-1$. Both are double roots. Discuss the conditions $a,b,c,d,e,f$ must satisfy to make matrix $A$ diagonalizable.
Well, I hesitate in this exercise because when I try to get its eigenvectors, particularly the ones related to $\lambda = 1$ and I arrive to something like this
$$\begin{pmatrix}-2 & 0 & 0 & 0 \\\ a & -2 & 0 & 0 \\\ b & d & 0 & 0 \\\ c & e & f & 0\end{pmatrix}$$
which multiplied by the column vector $(x;y;z;t)$ must result $(0;0;0;0)$. The solutions I obtain are $x = 0, y = 0, z = 0$ and $t = \mu $(parameter), so the dimension of the subspace $V_{1}$ is $1$. As the multiplicity of 1 was 2, and this is not the dimension of the subspace, the matrix can't be diagonalized.
Is this correct?
Thanks in advance.
Let $A\in M_n(K)$ where $K$ is a field. We will assume that the characteristic polynomial of $A$ splits into linear factors over $K$ , which is an obvious necessary condition for diagonalizability.
Let $\lambda_1,\cdots,\lambda_r\in K$ the distinct eigenvalues of $A$.
Then it is well-known that $A$ is diagonalizable if and only if $K^n=\bigoplus_i \ker(A-\lambda_i I_n)$. It is not difficult to deduce the following:
Theorem. Assume that the characteristic polynomial of $A$ splits into linear factors over $K$ . Let $\lambda_1,\cdots,\lambda_r$ the distinct eigenvalues of $A$. Then $A$ is diagonalizable if and only if $\prod_i(X-\lambda_i)$ annihilates $A$.
This gives you a criterion which avoid explicit computations of eigenspaces.
In your case, the eigenvalues of $A$ are $1$ and $-1$, so $A$ will be diagonalizable if and only if $A^2=I_4$.
Now, $A^2=\begin{pmatrix} 1 & 0 & 0 & 0 \cr -2a & 1 & 0 & 0 \cr ad & 0 & 1 & 0 \cr ae+bf & df & 2f & 1\end{pmatrix}.$
We easily deduce that $A$ is diagonalizable if and only if $a=0$ and $f=0.$