How to Find a Finite-Difference Matrix I have read several websites trying to explain finite-differential equations, but I haven't been able to find one that explains how it's put into the matrix form.
$f(x) = -\frac{d^2u}{dx^2}$ where $u(0) = 0$ and $u(1) = 0$ becomes
$\begin{bmatrix}
2 & -1 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{bmatrix}$
multiplied by one column of values between $u_1$ to $u_5$ is equal to
$h^2$$\begin{bmatrix}
f(h) \\
f(2h) \\
f(3h) \\
f(4h) \\
f(5h)
\end{bmatrix}$
Looking at the equation that causes this matrix, it confuses me.  The difference equation is
$$-u_{j+1}+2u_j-u_{j-1}=h^2f(jh)$$
The initial term, when $j=1$, should make it so that the equation is:
$$u_2+2u_1-u_0$$
Would this not cause the first row to become $(-1, 2, 0, 0, 0)$ because $u_0$ has been defined as 0?  I know that my thinking is wrong, since the book tells me so, but I don't understand how the first and last row is determined.

On a related note, how does the matrix equation if the boundaries are changed?  For an example, if $u_0 = 1$ and $u_1 = 2$ on the original equation.  Will the answer become
$\begin{bmatrix}
2 & 0 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{bmatrix}$
 A: The first and last row are determined by boundary conditions. For instance, the matrix you've described arises if $u_0$ and $u_6$ are taken to be zero (though other BC might give rise to the same matrix).
As for your second question, we don't get to set the values of $u_1,\ldots,u_5$. Your matrix is multiplying the vector ${\bf u}$ to give the set of finite difference equations. However, $u_0=0$ gives a constraint
$$h^2 f(h)=-u_0+2u_1-u_2=2u_1-u_2$$
which is correctly reproduced by the first matrix in your post.
A: Just adding to the answer, summarizing Jonathan. 
\begin{eqnarray}
\begin{pmatrix}
2 & 0 & 0 & 0 & 0 \\
-1 & 2 & -1 & 0 & 0 \\
0 & -1 & 2 & -1 & 0 \\
0 & 0 & -1 & 2 & -1 \\
0 & 0 & 0 & -1 & 2
\end{pmatrix}
\begin{pmatrix}
u_1 \\
u_2 \\
u_3 \\
u_4 \\
u_5
\end{pmatrix}
=
\begin{pmatrix}
f(h) \\
f(2h) \\
f(3h) \\
f(4h) \\
f(5h) \\
\end{pmatrix}
\end{eqnarray}
Only $ u_1 $ to $u_5$ goes in the matrix due the Boundary Condition $u_0=u_6=0$, that explains why your logic about the first and last rows were wrong. Just remember the multiplication logic for matrix/array. 
