LA: Find a the same vector in a new basis

Question

We have the vector v = (1, 2 -1) in the identity matrix
$$
e_1 =
\left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}
\right)
$$
Now we want to know its coordinates in the new basis
$$
e_2 =
\left(
\begin{array}{ccc}
1 & 4 & 2 \\
1 & 3 & 1 \\
1 & 2 & 2
\end{array}
\right)
$$
I thought of this problem as following:
The identity matris gives us back the same vector
But if we do a matris multiplication as following should we not get the new coordinates with the same vector in the other basis?
$$
\left(
\begin{array}{ccc}
1 & 4 & 2 \\
1 & 3 & 1 \\
1 & 2 & 2
\end{array}
\right)
\left(
\begin{array}{ccc}
1 \\ 2 \\ -1
\end{array}
\right)
=
\left(
\begin{array}{ccc}
7 \\ 6 \\ 3
\end{array}
\right)
$$
Why does this not work? I saw a video on Youtube (3Blue1Brown) regarding Linear Transformations and he stated for example that if you take a vector fron the identity matris and do a transformation then we get the output vector by multiplying its coordinates from the identity matris to the transformation matris.

Answer 1

First you have to decide if you are working in row-space or column-space. Then you have to decide if you wish to express your given vector v=[1,2,-1] in terms of B={[1,4,2],[1,3,1],[1,2,2]} or in terms of B'={[1,1,1],[4,3,2],[2,1,2]} or the transpose of v in terms of the transpose of B or the transpose of B'. Any one of these four possible interpretations can be done easily by multiplying by the inverse of the given matrix or its transpose, but first you must decide exactly what you wish to do.