# Thread: Verifying the diagonalizability of a 2x2 matrix with rank 1

1. ## Verifying the diagonalizability of a 2x2 matrix with rank 1

Hello,

If A is a 2x2 matrix with rank 1, and one eigenvalue is known to be nonzero, then is A diagonalizable? I think this matrix is diagonalizable because if you take a generic 2x2 matrix with rank one:
[ a b ]
[ax bx]
and you find the characteristic polynomial, (a-t)(bx-t) - bax = 0, then one eigenvalue must be zero and thus A is diagonalizable. Is this reasoning correct? Thanks!

2. $\displaystyle \begin{bmatrix}
a & b\\
0 & 0
\end{bmatrix}$

$\displaystyle p(\lambda_A)=\lambda^2-tr(A)\lambda=\lambda(\lambda-tr(A))=0$

$\displaystyle \lambda_1=tr(A)=a \ \ \begin{bmatrix}
0 & b\\
0 & -a
\end{bmatrix}\rightarrow\begin{bmatrix}
1 & 0\\
0 & 0
\end{bmatrix}\rightarrow x_2\begin{bmatrix}
0 \\
1
\end{bmatrix}$

$\displaystyle \lambda_2=0 \ \begin{bmatrix}
a & b\\
0 & 0
\end{bmatrix}\rightarrow x_2\begin{bmatrix}
\frac{-b}{a}\\
1
\end{bmatrix}$

It is only diagonalizable if $b\neq 0$.

A matrix is only diagonalizable if it has n linearly independent eigenvectors. Also, just because a matrix has distinct eigenvalues it may not be diagonalizable if the eigenvectors aren't lin. independent, and conversely, a matrix with repeated eigenvalues isn't necessarily non-diagonalizable either.

3. If a 2 by 2 matrix has rank 1 then it has non-trivial kernel. That is, there exist non-zero vectors, v, such that Av= 0= 0v so that 0 is an eigenvalue. If it also has a non-zero eigenvalue, then it has two distinct eigenvalues, so two independent eigenvectors and is diagonalizable.

d.w. smith, eigenvectors corresponding to distinct eigenvalues are independent so two distinct eigenvalues imply two independent eigenvectors so, for a 2 by 2 matrix, diagonalizable.

As for your example, if b= 0, then $\begin{bmatrix}a & 0 \\ 0 & 0\end{bmatrix}$ already is diagonal! It makes no sense to say it is "only diagonalizable if $b\ne 0$"/