I ran into a difficult question today as I was trying to find the matrix exponential for a matrix that has a determinant of $0$. Here is the matrix:
$$C = \begin{bmatrix}
1 & 1 \\
-1 & -1 \\
\end{bmatrix} $$
I got only one eigenvalue from the characteristic polynomial, that eigenvalue being $\lambda = 0.$ From there, I got one eigenvector and it was $$\begin{bmatrix}
-1 \\
1 \\
\end{bmatrix}$$
I was trying to begin the process of how I usually find the constants to later write the solution vector, and all I had was:
$$\begin{bmatrix}
1 \\
0 \\
\end{bmatrix} = c_1\begin{bmatrix}
-1 \\
1 \\
\end{bmatrix} $$
I am really confused after this step, and I feel like I cannot calculate the the matrix exponential of $C$ because it has no inverse.
What should I do? Is it possible to calculate?
I tried to go to wolfram alpha, and it should be a solution, but I was not sure how they obtained it. Could someone please help me?
Best Answer
Note that $C$ is nilpotent: $$ C^2=\begin{bmatrix}1&1\\-1&-1\end{bmatrix}\begin{bmatrix}1&1\\-1&-1\end{bmatrix}=0$$ Therefore $e^C$ can be computed directly from the definition: $$ e^C=I+\sum_{k=1}^{\infty}\frac{C^k}{k!}=I+C=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$$