Matrix Exponential of a Singular Matrix?

Question

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?

Answer 1

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: $$ e^C=I+\sum_{k=1}^{\infty}\frac{C^k}{k!}=I+C=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$$

Answer 2

Note the answer given by @carmichael561 is the same as the one given by wolfram but using a $tC$ matrix to get

$$e^{tC} = I + tC = \begin{bmatrix}1+t&t\\-t&1-t\end{bmatrix}$$

where $I$ is the unit matrix

$$I := \begin{bmatrix}1&0\\0&1\end{bmatrix}$$

Answer 3

Alright, you called your matrix $C.$ Once we get the Jordan normal form, call it $J,$ using $$ A^{-1} C A = J, $$ so that $$ A J A^{-1} = C, $$ we get $$ e^C = e^{A J A^{-1}} = A e^J A^{-1}. $$ Note that AJA was a platinum selling album by Steely Dan.

In the material below, what matrix is $A$ and what is $J \; ?$ For that matter, what is $e^J,$ which is just a finite sum?

I cannot tell whether you have heard of Jordan Normal Form, here it is. Note how the first matrix is the inverse of the third.

$$ \left( \begin{array}{rr} 0 & -1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{rr} 1 & 1 \\ -1 & -1 \end{array} \right) \left( \begin{array}{rr} 1 & 1 \\ -1 & 0 \end{array} \right) = \left( \begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array} \right) $$

First and third: $$ \left( \begin{array}{rr} 0 & -1 \\ 1 & 1 \end{array} \right) \left( \begin{array}{rr} 1 & 1 \\ -1 & 0 \end{array} \right) = \left( \begin{array}{rr} 1 & 0 \\ 0 & 1 \end{array} \right) $$