Change of Basis In Matrix Form

Question

Suppose $S_1, S_2,$ and $S_3$ are three bases for $ℝ^3$ and base transition matricies $_{S1}T_{S2}$ and $_{S1}T_{S3}$ are (respectively)

$$\begin{bmatrix} 2&&3&&-1\\ 1&&0&&-2\\ 0&&-3&&5 \end{bmatrix} $$ $$\begin{bmatrix} 0&&-3&&1\\ 4&&6&&0\\ 8&&0&&-3 \end{bmatrix} $$

How would one compute $_{S2}T_{S3}$?

Answer 1

For any vector space $E$ endowed with bases $S_1, S_2, S_3$, $_{S_i}T_{S_j}$ is the matrix of the identity map from $(E,S_j)$ to $(E, S_i)$ and its inverse is the matrix of the identity map the other way. Now if you consider the commutative diagram: \begin{array}{rcl} (E, S_2)&\xrightarrow[_{S_2}T_{S_3}]{=} &(E, S_3) \\ =\searrow&\enspace\scriptstyle\llap{{}_{S_2}T_{S_1}}\quad_\rlap{{}_{S_1}{\scriptstyle T} _{S_3}}, &\nearrow=\\[] &(E, S_1) \end{array} you obtain instantly $$_{S_2}T_{S_3}={}_{S_1}T_{S_3}{\,}_{S_2}T_{S_1}={}_{S_1}T_{S_3}{\,}_{S_1}T_{S_2}^{-1}$$

Answer 2

We have that

• $v_1=_{S1}T_{S2}\cdot v_2 \implies v_2=(_{S1}T_{S2})^{-1}\cdot v_1$
• $v_1=_{S1}T_{S3}\cdot v_3$

then

$$v_2=(_{S1}T_{S2})^{-1}\cdot_{S1}T_{S3}\cdot v_3$$

that is

$$_{S2}T_{S3}=(_{S1}T_{S2})^{-1}\cdot_{S1}T_{S3}$$

In the derivation I've assumed that:

• $_{S1}T_{S2}$ is the transition from $S_2$ to $S_1$

• $_{S1}T_{S3}$ is the transition from $S_3$ to $S_1$