Eigenvalues & Eigenvectors: Repeated vs Distinct

[LikeMath]

Hi there!

Let A be a square matrix of order n.
It is well known that if we have n distinct eigenvalues then we surely have n distinct eigenvectors. But if there are repeated eigenvalues then the tow possibilities may happen.
My question is: How can I know that do the eigenvectors are distinct or not?

Thank you very much.

 

[HallsofIvy]

The only way to be certain is by trying to find the eigenvectors!

For example both
\begin{bmatrix}a & 0 \\ 0 & a\end{bmatrix}
and
\begin{bmatrix} a & 1 \\ 0 & a\end{bmatrix}
have a as a double eigenvalue.
To find the eigenvectors, we need to solve
\begin{bmatrix}a & 0 \\ 0 & a\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}ax \\ ay\end{bmatrix}
and
\begin{bmatrix}a & 1 \\ 0 & a\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}ax \\ ay\end{bmatrix}

The first gives the two equations ax= ax and ay= ay. Clearly, those are true for all x and y- any vector in R² and, in particular <1, 0> and <0 1>, which are independent, are eigenvectors.

The second gives the two equations ax+ y= ax and ay= ay. The second equation is true for any y but the first equation reduces to y= 0. Given that a can be anything but we have that all eigenvectors are of the form <a, 0>, a one dimensional space so we have only one "independent" eigenvector.