Let $\beta=\{(1,0,0),(0,1,0),(0,0,1\}$be a basis of $\mathbb{R^3}$ and $g: \mathbb{R^3} \to \mathbb {R^3}$ a linear transformation, which matrix is:

$$G=\begin{bmatrix}1 & 0 &-1 \\ 6 & 4 & -2\\ 0 & 1 & -1 \end{bmatrix}$$ Now let $\lambda=\{(0,0,1),(0,1,0),(1,0,0)\}$ be a basis also.

What is the matrix of $M(g;\lambda, \lambda)$?

I started to write each vector of $\beta$ into $\lambda$ coordinates.




So, the change of basis's matrix, from $\beta$ to $\lambda$ is:

$$T=\begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0\\ 1 & 0 & 0 \end{bmatrix}$$

This last matrix transforms any vector on $\beta$ coordinates into a vector on $\lambda$ coordinates. Then I wrote this scheme:

$$[v]_{\lambda} \to C \to [v]_{\beta} \to G \to [u]_{\beta} \to T \to [u]_{\lambda}$$

Where C is the change of basis's matrix from $\lambda$ to the canonic basis.

My difficult is how to merge theese matrix in one. Can You help me? thanks.

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    $\begingroup$ en.wikipedia.org/wiki/Change_of_basis#Change_of_basis_2 $\endgroup$ May 24 '13 at 22:34
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    $\begingroup$ Try writing out the scheme as the multiplication of matrices rather than with the arrows. E.g. instead of $[\vec{v}]_\lambda\to C \to [\vec{v}]_\beta$, write $[\vec{v}]_\beta=C[\vec{v}_\lambda]$, and go from there. $\endgroup$ May 24 '13 at 23:01

What you need is nothing else but the matrix product $TGT^{-1}$.

Let's call the standard basis vectors $e_1,\ e_2,\ e_3$. Then we have $\beta=(e_1,e_2,e_3)$ and $\lambda=(e_3,e_2,e_1)$.

Now you have luckily $T^{-1}=T$, as $T$ simply exchanges the basis vectors $e_1$ and $e_3$.

But, for this specific exercise, we can find the answer more directly:

We need $G'=M(g;\lambda,\lambda)$. Because $G\cdot e_i$ gives the $i$th column of the matrix $G$, we have $$g(e_1)=G\cdot e_1 = e_1+6e_2$$ This means that the third column of $G'$ will be $\pmatrix{0\\6\\1}$, as these are the $\lambda$-coordinates of the $g$-image of the third basis vector of $\lambda$.

Similarly, you can see what happens to the rest of the matrix.


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