# Is a matrix $A$ diagonal with respect to its eigenbasis?

I guess I am trying to understand the ideas of diagonalizability and eigenobjects. I get the basics of what an eigenvalue and eigenvector are, but I am not sure how these relate to diagonalizaility...

Does the existence of eigenvalues (period) imply diagonalizability? Or do the eigenvalues have to be distinct? Also, let's say I have a matrix $A$ and this matrix has some eigenvalues. To these eigenvalues are corresponding eigenvectors which form an eigenbasis for the matrix. Will $A$ always be diagonal with respect to this eigenbasis?

• It needs to be the case that the eigenvectors of the matrix span $\mathbb{R}^n$, given that the matrix is $n \times n$. Commented Apr 1, 2017 at 17:50
• I see so in other words they must be distinct. Commented Apr 1, 2017 at 17:53
• @SocraticaFan if they're distinct then you can directly conclude that it's diagonalizable but if they're not distinct the endomorphism can still be diagonalizable if the multiplicity of each eigenvalue is equal to the dimension of its associated eigenspace. Commented Apr 1, 2017 at 17:57
• By endomorphism you just mean the linear transformation that A represents, right? We have not discussed this term in class yet. Commented Apr 1, 2017 at 18:03
• yes an endomorphism $f$ of a vector space $V$ is a linear transformation $f: V → V$ Commented Apr 1, 2017 at 18:08

A matrix $A_{n \times n}$ is diagonalizable precisely when there are exactly $n$ linearly independent eigenvectors. A sufficient condition is having $n$ distinct eigenvalues, as each eigenvalue is guaranteed at least one eigenvector (by definition, or else it wouldn't be an eigenvalue) and eigenvectors from different eigenvalues are automatically linearly independent. When we have repeated eigenvalues, we need to slow down and double check.
$$A= \begin{bmatrix} 3& 0 \\ 0 & 3 \end{bmatrix} \qquad B=\begin{bmatrix} 3 & 1 \\ 0 & 3 \end{bmatrix}.$$ Both matrices have $\lambda= 3$ as a repeated eigenvalue (Algebraic Multiplicity $2$). However,
• $A$ is (trivially) diagonalizable, as $\dim \mathcal{N}(A-3I) =2$, i.e., we can find two linearly independent eigenvectors for the eigenvalue of $\lambda=3$. We could also say that "the geometric multiplicity of $\lambda=3$ is two, which equals the algebraic multiplicity." Thus $A$ is diagonalizable: it is a $2 \times 2$ matrix with exactly $2$ linearly independent eigenvectors across all of its eigenvalues.
• $B$, on the other hand, is not diagonalizable! $\dim \mathcal{N}(B-3I) = 1$, i.e., we can only find $1$ linearly independent eigenvector for the double eigenvalue of $3$---"the geometric multiplicity is smaller than the algebraic multiplicity." As such, $B$ is not diagonalizable as it is a $2 \times 2$ matrix with only $1$ linearly independent eigenvector.