Finding the eigenvalues and the eigenvectors 
The matrix:
  $$
        A=\begin{pmatrix}
        1 & -1 & 1 \\
        -1 & 1 & -1 \\
        -1 & 1 & -1 \\
        \end{pmatrix}
$$
  has two real eigenvalues, one of multiplicity $1$ and one of multiplicity $2$. Find the eigenvalues and a basis of each eigenspace. 

$\lambda_1$ has multiplicity $1$, with basis $\begin{bmatrix} ? \\ ? \\ ? \end{bmatrix}$, and $\lambda_2$ has multiplicity $2$, with basis $\begin{bmatrix} ? \\ ? \\ ? \end{bmatrix}$, $\begin{bmatrix} ? \\ ? \\ ? \end{bmatrix}$.
The polynomial I get upon doing $(\lambda I-A)$ is: $$ (\lambda^3-\lambda^2) = (\lambda-1)\lambda^2.$$
I'm not entirely sure what to do after this point... Any input?
 A: The multiplicity of a root $r$ of a polynomial $p(x)$ is the number of times that $x-r$ appears as a factor when $p(x)$ is fully factored into linear factors. You have the equation $\lambda^2(\lambda-1)=0$, which is fully factored into the linear factors $\lambda$, $\lambda$, and $\lambda-1$. Thus, $0$ is the root of multiplicity $2$, and $1$ is the root of multiplicity $1$.
Now you want to find the eigenvectors. For a given eigenvalue $\lambda$, these are the vectors $v$ such that $Av=\lambda v$ or, equivalently, $(A-\lambda I)v=\vec 0$. For $\lambda=0$ this equation becomes
$$\begin{bmatrix}1&-1&1\\-1&1&-1\\-1&1&-1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\;,$$
a matrix equation that you solve by your favorite method. I’d just row-reduce the augmented matrix to
$$\left[\begin{array}{rrr|r}1&-1&1&0\\0&0&0&0\\0&0&0&0\end{array}\right]$$
and choose a basis for the two-dimensional solution space.
Then do the same thing for $\lambda=1$; your starting point should be the equation
$$\begin{bmatrix}0&-1&1\\-1&0&-1\\-1&1&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\;.$$
This will actually be a little easier, since the solution space will be one-dimensional.
