# How do I determine if matrix A is diagonalizable?

I am trying to figure out how to determine the diagonalizability of the following matrix:

$A=\begin{pmatrix} 1 &0 &0 &0 \\ 2&1 & -3 & -2\\ 3& 0 & 0 &-9 \\ -1& 0& -1& 0 \end{pmatrix}$

There are two distinct eigenvalues: $λ_1=λ_4=1$,$λ_2=3$ and $λ_3=-3$.

Can someone help to define geometric multiplicity? And I found that matrix is diagonalizable geometric multiplicity is equal to the algebraic multiplicity.

• For single eigenvalues, the algebraic and geometric multiplicities are always the same. It is double, triple and higher order eigenvalues that are the problem. What does that tell you in this case? – Cameron Williams May 3 '16 at 21:11
• @CameronWilliams Well nothing – jebiovo May 3 '16 at 21:11
• It tells you a lot. You only need to check the eigenvalue $1$ since $3$ and $-3$ are single eigenvalues. – Cameron Williams May 3 '16 at 21:12
• @CameronWilliams Done that I get 3 vectors. – jebiovo May 3 '16 at 21:13
• And it is "matrix", not "matrice". – Dietrich Burde May 3 '16 at 21:14

The characteristic polynomial is $$\det \begin{pmatrix} 1-X &0 &0 &0 \\ 2&1-X & -3 & -2\\ 3& 0 & 0-X &-9 \\ -1& 0& -1& 0-X \end{pmatrix} =(1-X)^2(3-X)(-3-X)$$ You want to determine the geometric multiplicity of the eigenvalue $1$, that is the dimension of the eigenspace, which is the null space of $A-I$. Let's do an elimination: \begin{align} A-I=\begin{pmatrix} 0 &0 &0 &0 \\ 2&0 & -3 & -2\\ 3& 0 & -1 &-9 \\ -1& 0& -1& -1 \end{pmatrix} &\to \begin{pmatrix} -1& 0& -1& -1 \\ 2&0 & -3 & -2\\ 3& 0 & -1 &-9 \\ 0 &0 &0 &0 \end{pmatrix} \\[6px]&\to \begin{pmatrix} 1& 0& 1& 1 \\ 0& 0 & -5 & -4\\ 0& 0 & -4 &-12 \\ 0 &0 &0 &0 \end{pmatrix} \\[6px]&\to \begin{pmatrix} 1& 0& 1& 1 \\ 0& 0 & 1 & 4/5\\ 0& 0 & 0 &-44/5 \\ 0 &0 &0 &0 \end{pmatrix} \\[6px]&\to \begin{pmatrix} 1& 0& 1& 1 \\ 0& 0 & 1 & 4/5\\ 0& 0 & 0 &1 \\ 0 &0 &0 &0 \end{pmatrix} \end{align} So the matrix $A-I$ has rank $3$ and its null space has dimension $1$.