There's a portion of a problem assigned to me of which I do not understand the answer. The problem is as follows: "Find a basis $B$ for $\mathbb{R}^3$ such that the matrix $A$ relative to $B$ is diagonal."
Now beforehand we found that the diagonal vector of A was $D=\begin{bmatrix} 2 & 0 & 0\\ 0 & -2 & 0\\ 0 & 0 & 3 \end{bmatrix}$ where the diagonal entries were the eigenvalues of A. My professor claimed the the basis B is the diagonal D. I can't seem to understand why that is. Can anyone explain?
Edit: Included more information This was the standard matrix $A=\begin{bmatrix} 1 & -1 & -1 \\ 1 & 3 & 1 \\ -3 & 1 & -1 \end{bmatrix}$
I think it would do you well to review diagonalization, so let's walk through this example. I'm assuming you have a basic knowledge of matrix manipulations and basic facts from linear algebra.
We would like to diagonalize the matrix $$ A = \begin{bmatrix} 1 & -1 & -1\\ 1 & 3 & 1 \\ -3 & 1 & -1 \end{bmatrix} $$ To do so, we need to compute the eigenvalues of $A$ and find bases for the eigenspaces of $A$. The eigenvalues are precisely the roots of the characteristic polynomial of $A$. In this case the characteristic polynomial is $$ \DeclareMathOperator{char}{char}\char_A(\lambda) =\det(A-\lambda\cdot I)=-(\lambda+2)(\lambda-2)(\lambda-3) $$ The eigenvalues of $A$ are thus \begin{align*} \lambda_1 &= -2 & \lambda_2 &= 2 & \lambda_3 &=3 \end{align*} and the algebraic multiplicity of each eigenvalue is one.
Now, for each eigenvalue $\lambda_k$, let $E_{\lambda_k}$ be the associated eigenspace $$ E_{\lambda_k}=\{\mathsf v\in\Bbb R^3:A\mathsf v=\lambda_k \mathsf v\}=\DeclareMathOperator{Null}{Null}\Null(A-\lambda_k\cdot I) $$ In our case, we have \begin{align*} E_{\lambda_1} &= \Null\begin{bmatrix}3&-1&-1\\ 1&5&1\\ -3&1&1\end{bmatrix} & E_{\lambda_2} &= \Null\begin{bmatrix}-1&-1&-1\\ 1&1&1\\ -3&1&-3\end{bmatrix} & E_{\lambda_3} &= \Null\begin{bmatrix}-2&-1&-1\\ 1&0&1\\ -3&1&-4\end{bmatrix} \\ &\DeclareMathOperator{Span}{Span}=\Span\left\{\begin{bmatrix}1\\-1\\4\end{bmatrix}\right\} && =\Span\left\{\begin{bmatrix}-1\\0\\1\end{bmatrix}\right\} && =\Span\left\{\begin{bmatrix}-1\\1\\1\end{bmatrix}\right\} \end{align*} This gives bases for each eigenspace (I'm skipping details here. If you don't understand what I have written, I strongly encourage you to review how to find bases for nullspaces).
Finally, since $A$ is a $3\times 3$ matrix and we have $3$ linearly independent eigenvalues of $A$, we know that $A$ is diagonalizable. It's diagonalization is $$ A=PDP^{-1} $$ where \begin{align*} D &= \begin{bmatrix}-2&0&-\\ 0& 2&0\\ 0&0&3\end{bmatrix} & P &= \begin{bmatrix}1&-1&-1\\ -1&0&1\\ 4&1&1\end{bmatrix} \end{align*} Note that the diagonal entries of $D$ are exactly the eigenvalues of $A$ and that the columns of $P$ are exactly the associated eigenvectors of $A$.
In short, the basis $B$ for $\Bbb R^3$ such that the matrix of $A$ relative to $B$ is diagonal is $$ B=\left\{\begin{bmatrix}1\\-1\\4\end{bmatrix},\begin{bmatrix}-1\\0\\1\end{bmatrix},\begin{bmatrix}-1\\1\\1\end{bmatrix}\right\} $$