Matrix Exponential of a Singular Matrix? 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?
 A: 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:
$$ e^C=I+\sum_{k=1}^{\infty}\frac{C^k}{k!}=I+C=\begin{bmatrix}2&1\\-1&0\end{bmatrix}$$
A: 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:
$$
\left(
\begin{array}{rr}
0 & -1 \\
1 & 1
\end{array}
\right)
\left(
\begin{array}{rr}
1 & 1 \\
-1 & 0
\end{array}
\right) =
\left(
\begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}
\right)
$$
A: 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
$$I := \begin{bmatrix}1&0\\0&1\end{bmatrix}$$
