Finding a transition matrix from basis B1 to the standard basis S.

Question

This is the info I have:

• B1 = {(3,4,0), (-2,-1,1), (1,0,3)}
• S is the standard basis of $R^3$

Is the matrix supposed to look like this then?

$$\begin{bmatrix}1 & 0 & 0 & 3 & 4 & 0\\0 & 1 & 0 & -2 & -1 & 1\\0 & 0 & 1 & 1 & 0 & 3 \end{bmatrix}$$

And if so, doesn't that mean the transition matrix is supposed to be this?

$$\begin{bmatrix}3 & 4 & 0\\-2 & -1 & 1\\1 & 0 & 3 \end{bmatrix}$$

Answer 1

A vector $v$ can be written as $a_1 \overrightarrow{e_1} + a_2 \overrightarrow{e_2} + a_3 \overrightarrow{e_3}$

You want a matrix that takes $v$ written in the $B_1$ basis, $v = c_1 \overrightarrow{b_1} + c_2 \overrightarrow{b_2} + c_3 \overrightarrow{b_3}$ and gives you the $a$'s. So the $c$'s are the inputs and the $a$'s are the outputs. $\overrightarrow{b_1} = 3 \overrightarrow{e_1} + 4 \overrightarrow{e_2} + 0 \overrightarrow{e_3}$. If you make a matrix $$\begin{bmatrix} 3 & -2 & 1\\4 & -1 & 0 \\0 & 1 & 3\end{bmatrix}$$ then when you multiply that by column vector $[c_1, c_2, c_3]^T$ the result should be $[a_1, a_2, a_3]^T$

Answer 2

In general, $Ae_i$ is equal to the $i$-th column of $A$, where $e_i$ is the $i$-th standard basis vector. Thus we merely need to put the non standard basis vectors into the columns of a matrix.

Note: the trick to this is that the non-standard basis $\beta$ in terms of itself is $S$.