I have this basis $B = ((1,0,1),(0,1,-1),(1,-1,0))$
That is represented by:
$$[T]_B = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{pmatrix}$$
I want to convert this representing matrix into standard basis, but I somehow misunderstood it since I have mistakes.
I know I can always do: $M_E^B[T]_BM_B^E$ to get $[T]_E$ but I'm looking for more of a manual way of doing this in order to fully understand it.
Can someone please elaborate on this process?
Thanks.
First you find $ [i]_{E^B} $ and $ [i]_{B^E} $, where $i$ is the identity map. Now I assume you are calling $ E $ as the standard basis vectors. To find $ [i]_{B^E} $, the $j^{th} $ column is given by the linear combination of $E_j$ in terms of the vectors from $B$. This is easy because this is already how $B$ is written. So $ [i]_{B^E} = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{bmatrix} $. Now to find $ [i]_{E^B} $, you must solve for the standard basis vectors in terms of the vectors from $B$. Note that $$ E_1 = (1,0,0) = \frac{1}{2}(1,0,1) + \frac{1}{2}(0,1,-1) + \frac{1}{2}(1,-1,0) $$ $$ E_2 = (0,1,0) = \frac{1}{2}(1,0,1) + \frac{1}{2}(0,1,-1) - \frac{1}{2}(1,-1,0) $$ $$ E_3 = (0,0,1) = \frac{1}{2}(1,0,1) - \frac{1}{2}(0,1,-1) - \frac{1}{2}(1,-1,0) $$ This means $$ [i]_{E^B} = \frac{1}{2} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1& -1 & -1\end{bmatrix}$$ Now multiplying these guys all together will give you just what you want: $$ [T]_{E} = [i]_{B^E}[T]_{B}[i]_{E^B} = \frac{1}{2} \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1& -1 & -1\end{bmatrix} \begin{bmatrix} 1 & 0& 1 \\ 1 & 1 & 2 \\ 1 & 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 0 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 8 & -3 & 1 \\ 2 & -1 & 1 \\ -4 & 1 & 1\end{bmatrix}$$