Matrix functions of a non-diagonalizable matrix Let $A$ be the following $3 \times 3$ matrix:
$$
A =
\begin{pmatrix}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1 \\
\end{pmatrix}
$$
I'm supposed to calculate $A^n$, where $n \in \Bbb R$, $\exp(tA)$ and $\sin(\pi A)$. Obviously $A$ is not diagonalizable. Since we haven't had anything about Jordan decomposition in the lecture, I'm not sure how to solve this.
The eigenvalues $\lambda_1 = -1 , \lambda_{2,3} = 1$ can be read off. I tried to expand the two eigenvectors into a orthonormal basis, i.e.:
$$
\mathbf{x}_{\lambda_1} =
\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix} \qquad
\mathbf{x}_{\lambda_2} =
\begin{pmatrix}
1 \\
0 \\
0 \\
\end{pmatrix} \qquad
\mathbf{x}_3 =
\begin{pmatrix}
0 \\
1 \\
0 \\
\end{pmatrix}
$$
But I'm rather unsure how to continue. I suspect that
$$
A^n =
\begin{pmatrix}
1 & n & 0 \\
0 & 1 & 0 \\
0 & 0 & (-1)^n \\
\end{pmatrix} \qquad \text{for} \qquad n \in \Bbb N_0,
$$
But how to expand this to $n \in \Bbb R$? In general, how can I solve such a problem of matrix functions, if I've not heard anything about Jordan decomposition?
EDIT: Thanks for your help. I could show that the above mentioned matrix for $A^n$ is correct even for $n \in \Bbb Z$. The two other functions are straightforward then. If someone has an idea or hint about $A^n$ for $n \notin \Bbb Z$, i would appreciate it.
 A: Like you said, 
$$
A^n =
\begin{bmatrix}
1 & n & 0 \\
0 & 1 & 0 \\
0 & 0 & (-1)^n \\
\end{bmatrix} \qquad \text{for} \qquad n \in \Bbb N.$$
Then
$$
\exp(tA)=\sum_{n=0}^\infty\frac{t^nA^n}{n!}
=\begin{bmatrix}
\sum_{n=0}^\infty \frac{t^n}{n!}
&\sum_{n=0}^\infty n\frac{t^n}{n!}&0\\0&\sum_{n=0}^\infty \frac{t^n}{n!}&0\\
0&0&\sum_{n=0}^\infty \frac{(-1)^nt^n}{n!}
\end{bmatrix}
=\begin{bmatrix}
e^{t}&te^t&0\\
0&e^t&0\\
0&0&e^{-t}
\end{bmatrix}.
$$
You can play the same game for the sine. 
About the powers, you could define
$$
A^t=\begin{bmatrix}1&t&0\\0&1&0\\0&0&e^{\pi i t}\end{bmatrix}.
$$
This agrees with the integer powers of $A$ and satisfies the exponential property $A^{t+s}=A^{t}A^{s}$. It is important to notice that for non-integer $t$ this choice is rather arbitrary and not the result of a calculation. 
A: Here's another strategy, take it or leave it.  You can write $ tA = tD+tB$ where $tD$ is a diagonal matrix and $tB$ is an upper triangular "nilpotent" matrix.  Nilpotent matrices have the special property that $B^k = 0$ for some finite $k$ (in your case $k=2$).  Then, because any matrix commutes with a diagonal matrix, the following formula holds:
$$
\exp(t(D+B)) = \exp(tD)\exp(tB)
$$
You can compute $\exp(tD)$ easily, and $\exp(tB)$ will just be a polynomial of $tB$ since the series for $\exp()$ will terminate after finitely many terms.  In fact, because $B^2=0$ for your matrix, $\exp(tB) = I+tB$.  Then you just multiply $\exp(tD)$ with $I+tB$ to get the result.
Note: the exponential formula above only holds because the matrices commute.  $\exp(A+B)\neq \exp(A)\exp(B)$ in general!
