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**Silverflow** Hello, I've been looking at question which states to diagonalize the matrix $\displaystyle A = \left(\begin{array}{ccc}1&t&0\\0&t&t\\0&0&0\end{ar ray}\right)$

I've found the eigenvalues to be 1, 0 and t. I know that if t = 1, the matrix is not diagonalizable, and if t = 0, then A is already diagonalized, and thus is diagonalizable. Now, I have to look at what happens when t is not equal to 1 or 0.

Looking at each eigenvalue, I'm having trouble finding the eigenvectors.

For $\displaystyle \lambda = 1$, they say, "Obviously $\displaystyle Ae_{1}=e_{1}$, therefore $\displaystyle e_{1}$ is a eigenvector." I don't follow that. It's not obvious to me.