# Matrix Exponential for defective eigenvalues

I am trying to compute the matrix exponential ($$e^{At}$$) of the following matrix:

$$\begin{bmatrix}1 &0&0&0\\1&0&0&0\\1&0&0&0\\1&0&0&0\\\end{bmatrix}$$

The eigenvalues were 0 and 1, but they were defective so I was having difficulty finding the diagonal matrix.

I also tried to check if this was a nilpotent matrix, but it was not. I do know that

$$A^n=A \\n=1,2,....$$

I could use this to expand $$e^{At}=I+At+{A^2t^2\over 2!}+...$$ but I am not sure where to go from there.

Since all positive powers of A are the same as $$A$$
$$e^{At}=I+At+{A^2t^2\over 2!}+...= I+ (e^t-1)A$$
Actually, $$A^2=A$$ (just multiply it out) and therefore $$A^n=A^{n-1}=A^{n-2}=\cdots=A$$ for all $$n$$, meaning that $$e^{At}=I+At+A\frac{t^2}{2!}+\cdots=I+A(e^t-1).$$