# Does every (complex) $n\times n$ matrix have n linearly independent eigenvectors?

Every $$n\times n$$ matrix will produce an $$n$$-th order characteristic equation. This polynomial will have $$n$$ distinct roots or less than $$n$$ distinct roots (with some repeated).

In the former case, each eigenvalue corresponds to a distinct eigenvector, so you can form an eigenbasis of $$n$$ linearly independent eigenvectors.

In the latter case, I can see two possibilities:

1. There are still $$n$$ linearly independent eigenvectors, but some share the same eigenvalue, so a $$k$$-fold root of the characteristic equation (an eigenvalue repeated $$k$$ times) will correspond to $$k$$ linearly independent eigenvectors.

2. There are less than $$n$$ linearly independent eigenvectors, so a $$k$$-fold root may correspond to less than $$k$$ linearly independent eigenvectors.

Is one of these two possibilities always true, for all complex $$n\times n$$ matrices, or could either possibility be true depending on the matrix in question?

Is there a proof of this, or perhaps a counter example?

• $\begin{bmatrix}1&1\\0&1\end{bmatrix}$ – David C. Ullrich Dec 24 '18 at 17:23

Both possibilities apply. Case (1) means the matrix is diagonalisable over $$\mathbb{C}$$, of which there clearly are examples. However matrices like $$\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$ are ones that are not diagonalisable over $$\mathbb{C}$$; this one would fit your case (2).
The reason why we would consider eigenvalues over $$\mathbb{C}$$ rather than over $$\mathbb{R}$$ is precisely because the characteristic polynomial will split as a products of its roots, which does not always happen over $$\mathbb{R}$$. Hence we always find (counting multiplicities) $$n$$ eigenvalues, but it turns out this is not sufficient for diagonalisability.
• Thanks. I've also actually just managed to find $\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix}$ as an example of a matrix with one distinct eigenvalue and one degenerate eigenvalue, but three linearly independent eigenvectors. – Pancake_Senpai Dec 24 '18 at 17:38
Standard example: $$\pmatrix{0 & 1\cr 0 & 0\cr}$$ with only one linearly independent eigenvector.