Change matrix basis I believe I understand how to change the basis of a certain vector, but how do I change the basis of an entire matrix? I tried changing the basis of each column vector separately but that didn't seem to work.
The problem I have to solve is as follows:
$A=\left(\matrix{3&2&0\\-1&0&0\\1&1&2}\right)$
Write this matrix in the  $\{(1,0,1);(6,2, −1);(0,0,2)\}$ basis.
How do I solve this?
Any help is greatly appreciated!
Edit: Matrix A represents a linear transformation.
 A: Compute the change of basis matrix $P$ from the new basis to the standard basis; this matrix represents the identity linear transformation with respect to the new basis and the standard basis. Concretely, since we are working with the standard basis, the columns of $P$ are just the vectors of the new basis. Then the matrix of $A$ with respect to the new basis is $P^{-1}AP$.
To clarify: we want to obtain the representation of $A$ in the new basis in terms of the representation of $A$ in the standard basis. Writing $[T]_{\text{old}}^{\text{new}}$ for the representation of the linear map $T$ with respect to the bases $\text{old}$ and $\text{new}$ (that is, $\text{old}$ is a basis for the domain and $\text{new}$ is a basis for the codomain), we are using the following relation:
$$ [A]_{\text{new}}^{\text{new}} = [\mathit{Id}]_{\text{standard}}^{\text{new}} [A]_{\text{standard}}^{\text{standard}} [\mathit{Id}]_{\text{new}}^{\text{standard}}.$$
This works because, more generally, if you have linear maps between vector spaces
$$ U \xrightarrow{f} V \xrightarrow{g} W $$
and bases $\mathcal{U}$, $\mathcal{V}$ and $\mathcal{W}$ for those spaces, then
$$ [g \circ f]_\mathcal{U}^\mathcal{W} = [g]_\mathcal{V}^\mathcal{W}[f]_\mathcal{U}^\mathcal{V­}.$$
So in this particular case, I took $P = [\mathit{Id}]_{\text{new}}^{\text{standard}}$, which is just the matrix whose columns are the basis vectors in $\text{new}$. But it might be more natural to call $P = [\mathit{Id}]_{\text{standard}}^{\text{new}}$ the change of basis matrix (since it goes from $\text{standard}$ to $\text{new}$), in which case you'd want to compute $PAP^{-1}$ instead.
A: Write, for convenience
$$
v_1 = (1,0,1), \quad v_2 = (6,2, −1), \quad v_3 = (0,0,2)
$$
Note that, for example,
$$
A v_1 = (3,-1,3) = 
6v_1 -\frac 12 v_2 - \frac 74 v_3
$$
So, the first column of your matrix with respect to this basis will be
$$
\pmatrix{6\\-\frac 12 \\ -\frac 74}
$$
