Explain why the vectors you determined together form a basis for $\mathbb{R}^3$. Let $A = \begin{bmatrix}
1 & -1 & 1 \\
1 & 0 & 2 \\
-1 & 2 & 0
\end{bmatrix}$ and $B = \begin{bmatrix}
1 & 0 & 2 \\
0 & 1 & 1 \\
0 & 0 & 0
\end{bmatrix}$.


*

*Explain why $0$ is an eigenvalue of the matrix $A$.
$0$ is an eigenvalue of $A$ because this forces the determinant of $A$ to be $0$ (as the product of eigenvalues of an $n\times n$ matrix is the determinant) which means that $A$ is singular (not invertible). This is how we find eigenvalues of $A$.  

*Determine the other eigenvalues of the matrix $A$.
The characteristic polynomial of $A$ is given by
\begin{equation*}
\begin{split}
c(\lambda) = \text{det}(A-\lambda I) = \text{det}\begin{bmatrix}
1-\lambda & -1 & 1 \\
1 & -\lambda & 2 \\
-1 & 2 & -\lambda
\end{bmatrix} &= (1-\lambda)\begin{vmatrix}
-\lambda & 2 \\
2 & -\lambda 
\end{vmatrix} + \begin{vmatrix}
1 & 2 \\
-1 & -\lambda 
\end{vmatrix} + \begin{vmatrix}
1 & -\lambda \\
-1 & 2 
\end{vmatrix} \\
&= -\lambda^3+\lambda^2+2\lambda \\
&= -\lambda(\lambda^2-\lambda-2) \\
&= -\lambda(\lambda+1)(\lambda-2),
\end{split}
\end{equation*}
using cofactor expansion along the first row. The eigenvalues of $B$ are the roots of $c(\lambda)$, which are $\lambda = 0$, $\lambda = -1$ and $\lambda = 2$.  

*Explain why the vectors you determined in 2. together form a basis for $\mathbb{R}^3$.
This is the question I'm having trouble with. Is it because since there are 3 distinct eigenvalues so there are 3 distinct eigenvectors which means the set $\{v_1,v_2,v_3\}$ are linearly independent. Hence form a basis for $\mathbb{R}^3$? Thanks.

 A: 1. First Solution
You just find the eigenvectors of matrix A. The eigenvectors corresponding to the eigenvalues $0,2,-1$ are $[-2,-1,1]^T$, $[0,1,1]^T$ and $[-1,-1,1]^T$ respectively.
You can see easily that the there three elements of $\mathbb{R}^3$ are linearly independent, computing the determinant of the matrix $P$ $$P=\left[\begin{matrix}-2 & 0 & -1\\-1 & 1 & -1\\1 & 1 & 1\end{matrix}\right]$$
they define, which is $\text{det}(P)=-2\neq0$.
Then, using the fact that you correctly stated above, which is that $n$ linearly independent vectors of $\mathbb{R}^n$ form a basis for $\mathbb{R}^n$, you prove that the eigenvectors of $A$  form a basis for $\mathbb{R}^3$.
Εdit: In fact, now that I think of it again, we don't even need to compute the eigenvalues of the matrix, because, as you stated above, we know that eigenvectors corresponing to different eigenvalues are linearly independent. 
2. Second Solution
The matrix $A$  has $3$ distinct eigenvalues, hence is diagonalizable: there exists a basis $\hat{a}=\{a_1,a_2,a_3\}$ of $\mathbb{R}^3$ and a diagonal matrix $D$ (with the eigenvalues of $Α$ in its diagonal)  such that $(f: \hat{a})= D$, where $f$ is the linear map that corresponds to the matrix $Α$. As a result, $f(a_i)=\lambda_ia_i$, which shows that the elements of the basis of $\mathbb{R}^3$ are just the eigenvectors of $A$. 
