How to obtain a certain transition matrix from a non-standard basis to another non-standard basis?

Question

The transition matrix, P, from the basis B=$\{(0, 1, 2)^t ,(1, 2, 0)^t ,(2, 0, 1)^t\}$ to the basis B'=$\{(9, 0, 0)^t ,(0, 9, 0)^t ,(0, 0, 9)^t \}$ of $\mathbb R^3$ ?

The solution I got was $$P=\begin{pmatrix} 0 & 1/9 & 2/9 \\ 1/9 & 2/9 & 0 \\ 2/9 & 0 & 1/9\end{pmatrix}$$

I got the above solution using the general augmented matrices method [B'|B] -> [I | P] (via Gaussian elimination) where P is the transition matrix.

However this is none of the answers in the problem sheet I am doing.

The matrices in the options are

$\begin{pmatrix} 0 & 3 & 3 \\ 3 & 3 & 0 \\ 3 & 0 & 3\end{pmatrix}$

$\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0\end{pmatrix}$

$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$

$\begin{pmatrix} -2 & 1 & 4 \\ 1 & 4 & -2 \\ 4 & -2 & 1\end{pmatrix}$

Answer 1

When a vector $v$ is written in the first basis like this

$$v=a_1e_1+a_2e_2+a_3e_3, \ \ \text{which can be written} \ \ v=\underbrace{[e_1|e_2|e_3]}_E\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$$

and in the second basis like this

$$v=a'_1e'_1+a'_2e'_2+a'_3e'_3, \ \ \text{which can be written} \ \ v=\underbrace{[e'_1|e'_2|e'_3]}_{E'}\begin{pmatrix}a'_1\\b'_1\\c'_1\end{pmatrix}$$

Equating the two expressions for $v$, we get:

$$E\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}=E'\begin{pmatrix}a'_1\\b'_1\\c'_1\end{pmatrix}$$

which is equivalent to :

$$\begin{pmatrix}a'_1\\b'_1\\c'_1\end{pmatrix}=\underbrace{(E')^{-1}E}_P\begin{pmatrix}a_1\\b_1\\c_1\end{pmatrix}$$

And in this way, I find back your transition matrix !

Answer 2

Let me explain this way.

You have, for the basis $B$ $$b_1=e_2+2e_3,$$ $$b_2=e_1+2e_2,$$ $$b_3=2e_1+e_3,$$ and for the basis $B'$ $$c_1=9e_1,$$ $$c_2=9e_2,$$ $$c_3=9e_3.$$ So solving for $e_i$ you have $$e_1=1/9c_1,$$ $$e_2=1/9c_2,$$ $$e_3=1/9c_3,$$ which subbed into the $b_i$, you get $$b_1=1/9c_2+2/9c_3,$$ $$b_2=1/9c_1+2/9c_2,$$ $$b_3=2/9c_1+1/9c_3,$$ with matrix specification $P$ as you calculate but $$P^{-1}= \left[\begin{array}{ccc} -2&1&4\\1&4&-2\\4&-2&1 \end{array}\right],$$ which means that $$c_1=-2b_1+b_2+4b_3,$$ $$c_2=b_1+4b_2-2b_3,$$ $$c_3=4b_1-2b_2+b_3.$$

Answer 3

The transition matrix, $P$, from the basis $B=\{(0, 1, 2)^t ,(1, 2, 0)^t ,(2, 0, 1)^t\}$ to the basis $B'=\{(9, 0, 0)^t ,(0, 9, 0)^t ,(0, 0, 9)^t \}$ on $\mathbb R^3$.

A transition matrix from $B$ to $B'$ can be defined as $[I]_{B\leftarrow B'}$ where $I$ is the identity mapping on $\mathbb R^3$.

Check out my answer here for clarification about notations, definitions and other related explanations about transition matrix and matrix of a linear transformation, in general.

Let $S$ be the standard ordered basis of $\mathbb R^3$.

$$\begin{align}[I]_{B\leftarrow B'} &=[I]_{B\leftarrow S}\cdot[I]_{S\leftarrow B'}\\ &= \left([I]_{S\leftarrow B}\right)^{-1}\cdot[I]_{S\leftarrow B'}\\ &= \begin{pmatrix} 0&1&2\\1&2&0\\2&0&1 \end{pmatrix}^{-1}\begin{pmatrix}9&0&0\\ 0&9&0\\ 0&0&9\end{pmatrix}\\ &= \begin{pmatrix}-2&1&4\\ 1&4&-2\\ 4&-2&1\end{pmatrix}\end{align}$$

This matches with your last option. Of course, Gaussian elimination can be helpful in finding the inverse.