# Why is $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ not diagonalizable

I have a number of sufficient conditions as to when a matrix $A$ is diagonalizable, namely:

1. When $A$ is symmetric
2. When $A$ has distinct eigenvalues

Given $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

$A$ has nondistinct eigenvalues $\lambda = 0$ with algebraic multiplicity $2$, is there some conditions that says this is when the matrix fails to be diagonalizable?

• The matrix is already in reduced row echelon form. From here you can see the dimension of the eigenspace corresponding to $\lambda = 0$ is 1 due to the second row of zeros. Oct 17, 2016 at 21:41
• Algebraic and geometric multiplicity of every eigenvalue should be equal for a matrix to be diagonalizable Oct 17, 2016 at 21:41
• Also its determinant is 0. Oct 17, 2016 at 21:44
• @Astyx that means nothing, just consider null matrix which is diagonal, but has determinant zero Oct 17, 2016 at 21:46
• Do you agree that zero is the only eigenvalue of $A$? In this case just solve $Ax=0x$ and see what happens. Oct 17, 2016 at 21:47

Assume it is diagonalisable the eigenvalue being $0$ with multiplicity $2$. This means that there exists an invertible matrix $P$ such that

$$\begin{bmatrix}0 & 1\\0 &0\end{bmatrix}=P\cdot\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\cdot P^{-1}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$

You can check that $A^2 = 0$. If $A$ were diagonalizable, what would then be the options for values on the diagonal?

A general condition may be described as follows: For every eigenvalue $\lambda$ we assume that you are able to calculate $$Z_1(\lambda) = \ker (A - \lambda) \ \ \ \mbox{and} \ \ \ Z_2(\lambda)=\ker(A-\lambda)^2$$ The matrix is diagonalizable iff for every eigenvalue $Z_1(\lambda)=Z_2(\lambda)$ or equivalently $\dim Z_1(\lambda)=\dim Z_2(\lambda)$. In the given example, $\lambda=0$ and we have $Z_1(A) = {\rm Span} \{e_1\}$ but $Z_2(A) = {\rm Span} \{ e_1,e_2\}= {\Bbb R}^2$

Let $A$ be diagonalizable $n\times n$, so $A=SDS^{-1}$ for some diagonal matrix $D$. This can be rewritten as $AS=SD$ or, if we consider $S=[v_1\;v_2\;\dots\;v_n]$ (where $v_i$ represents a column) and $D=\operatorname{diag}(d_1,d_2,\dots,d_n)$ (I hope the notation is clear), $$Av_i=d_iv_i$$ Since $S$ is invertible, its columns form a linearly independent set. Moreover, we see that each column is an eigenvector for $A$. If an eigenvalue $\lambda$ appears $k$ times in the diagonal of $D$, we see that there are at least $k$ linearly independent eigenvectors relative to $\lambda$. On the other hand, the characteristic polynomial of $D$ is the same as the characteristic polynomial of $A$, so each eigenvalue must appear in $D$ exactly as many times as its algebraic multiplicity.

The eigenspace of $A$ relative to $\lambda$ is the subspace formed by the vectors $v$ such that $Av=\lambda v$. So we have proved that the dimension of the eigenspace (usually called the geometric multiplicity of the eigenvalue) is at least equal to the algebraic multiplicity.

Since the sum of the algebraic multiplicities of all eigenvalues is the size $n$ of the matrix $A$, we see that the geometric multiplicity of each eigenvalue must equal the algebraic multiplicity.

Thus this is a necessary condition for a matrix to be diagonalizable. It is also sufficient, because if it holds we can make a basis of $\mathbb{R}^n$ consisting of eigenvectors and we can use it for diagonalizing $A$.

In the case of your matrix, the geometric multiplicity of the eigenvalue $0$ is $1$, whereas the algebraic multiplicity is $2$.