# A matrix is diagonalizable when algebraic and geometric multiplicities are equal

## Main Question or Discussion Point

A matrix is diagonalizable when algebraic and geometric multiplicities are equal.
I know this is true, and my professor proved it, but I did not understand him fully. Can someone please explain?

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HallsofIvy
Homework Helper
This area is for "Learning Materials", not questions. I am moving this thread to "Linear and Abstract Algebra".

HallsofIvy
If all the basis vectors $\{v_1, v_2, \cdot\cdot\cdot\, v_n\}$ are eigenvectors, that is, if $Lv_1= \lambda_1v_1$, $Lv_2= \lamba_2v_2$, $\cdot\cdot\cdot$, $Lv_n= \lambda_nv_n$, then each column consists of the eigenvalue, in the appropriate position, and "0"s:
$$\begin{bmatrix}\lambda_1 & 0 & \cdot & \cdot & \cdot & 0 \\ 0 & \lambda_2 & \cdot & \cdot\ & \cdot & 0 \\ \cdot & \cdot\ & \cdot\ & \cdot & \cdot & \cdot \\ 0 & 0 & \cdot & \cdot & \cdot & \lambda_n\end{bmatrix}$$