find matrix of linear transformation in another basis

I can't get the workflow.

I have matrix:

$$A=\begin{pmatrix}-1&1&2\\0&3&1\\5&-1&1 \end{pmatrix}$$

I have to find matrix of this transformation in tha basis of $\langle f_1,f_2,f_3 \rangle$ if $$f_1 =e_2+e_3 \\f_2=e_1\\f_3=e_1-e_3$$

I know that there's formula $$A'=T^{-1}AT$$

where I believe that $A$ is given matrix, but do not know what are $T$ and $T^{-1}$

Let $\mathbf{x} = (x_{1}, x_{2}, x_{3})^T$ be the coordinates of a point in the $e$-basis and let $\mathbf{y} = (y_{1}, y_{2}, y_{3})^T$ be the coordinates of the same point in the $f$-basis.

It is the same point, so we require the following condition. $$x_{1} \mathbf{e}_{1} + x_{2} \mathbf{e}_{2} + x_{3} \mathbf{e}_{3} = y_{1} \mathbf{f}_{1} + y_{2} \mathbf{f}_{2} + y_{3} \mathbf{f}_{3}$$

The question gives the way of writing the $f$-basis vectors in terms of the $e$-basis vectors: \begin{aligned} \mathbf{f}_1 &= \mathbf{e}_1 + \mathbf{e}_2 \\ \mathbf{f}_2 &= \mathbf{e}_2 \\ \mathbf{f}_3 &= \mathbf{e}_1 - \mathbf{e}_3 \end{aligned}

We can substutite these formulas into the equation for the coordinates above

$$x_{1} \mathbf{e}_{1} + x_{2} \mathbf{e}_{2} + x_{3} \mathbf{e}_{3} = y_{1} (\mathbf{e}_{1} + \mathbf{e}_{2}) + y_{2} \mathbf{e}_{2} + y_{3} (\mathbf{e}_{1} - \mathbf{e}_{3})$$

$$x_{1} \mathbf{e}_{1} + x_{2} \mathbf{e}_{2} + x_{3} \mathbf{e}_{3} = y_{1} \mathbf{e}_{1} + y_{1} \mathbf{e}_{2} + y_{2} \mathbf{e}_{2} + y_{3} \mathbf{e}_{1} - y_{3} \mathbf{e}_{3}$$

$$x_{1} \mathbf{e}_{1} + x_{2} \mathbf{e}_{2} + x_{3} \mathbf{e}_{3} = (y_{1} + y_{3}) \mathbf{e}_{1} + (y_{1} + y_{2} ) \mathbf{e}_{2} - y_{3} \mathbf{e}_{3}$$

Now $\mathbf{e}_1$, $\mathbf{e}_2$ and $\mathbf{e}_3$ are three linearly independent vectors so that the components of each vector can be equated on both sides of the above. I.e. we can write:

\begin{aligned} x_1 &= y_1 + y_3 \\ x_2 & = y_1 + y_2 \\ x_3 &= -y_3 \end{aligned}

The following is exactly the same as the above with slightly different spacing \begin{aligned} x_1 &= y_1 & &+y_3 \\ x_2 & = y_1 &+ y_2 & \\ x_3 &= & &-y_3 \end{aligned}

Writing the equations that relate the coordinates in this way, we can see how the set can be written as a single matrix equation:

$$\begin{pmatrix} x_{1} \\ x_{2} \\ x_{3} \end{pmatrix} = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \end{pmatrix} \Rightarrow \mathbf{x} = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \mathbf{y}$$

This shows how we can use a matrix to convert coordinates in the $f$-basis to coordinates in the $e$-basis, i.e. $\mathbf{x} = T \mathbf{y}$, i.e. it represents the matrix $T$ in the formula

$$A' = T^{-1} A T$$

where $A$ is the transformation that is applied to coordinates in the $e$-basis. The above formula is applied

Having found $T$, we can find its inverse (by hand or with some software): $$T^{-1} = \begin{pmatrix} 1& -1& 0 \\ 0& 1& 0 \\ 1& -1& -1 \end{pmatrix}$$

and finally, we can calculate $A'$

$$A' = \begin{pmatrix} -3& -2& -2 \\ 3& 3& -1 \\ -7& -1& -6 \end{pmatrix}$$ which is the matrix of the transformation that is applied to coordinates in the $f$-basis.

This next part goes into how the formula relating the two transformation matrices in the different bases is derived.

If we write $\mathbf{u}$ for the result of applying $A$ to the $e$-basis vector $\mathbf{x}$ and if we write $\mathbf{v}$ for the result of applying $A'$ to the $f$-basis vector $\mathbf{y}$.

\begin{aligned} \mathbf{u} &= A \mathbf{x} \\ \mathbf{v} &= A' \mathbf{y} \\ \end{aligned}

The vectors $\mathbf{x}$ and $\mathbf{y}$ correspond to the same point in the two different bases and so do the pair of vectors $\mathbf{u}$ and $\mathbf{v}$. In other words they can be written: \begin{aligned} \mathbf{x} &= T \mathbf{y} \\ \mathbf{u} &= T \mathbf{v} \\ \end{aligned}

This means we can write the following \begin{aligned} \mathbf{u} &= T \mathbf{v} \\ A \mathbf{x} &= T \mathbf{v} \\ A \mathbf{x} &= T A' \mathbf{y} \\ A T\mathbf{y} &= T A' \mathbf{y} \\ T^{-1} A T\mathbf{y} &= A' \mathbf{y} \\ \end{aligned} As this works for all $\mathbf{y}$ we can conclude that $T^{-1} A T = A'$.

• How did you find right matrix ?:) Jun 29, 2017 at 6:47
• Apologies, it was a bit brief, I'll edit to expand the answer and clarify steps that were missing. Jun 29, 2017 at 9:22
• So, my $T$ is matrix $$\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$ which consists of $y$ сoefficients at right side? Jun 29, 2017 at 10:03
• Yes, on the right side of the equations that has been spaced out to show coefficients of $y_1$, $y_2$ and $y_3$ in different 'columns' - it is equivalent to a matrix multiplication. Jun 29, 2017 at 10:12
• aha, I just confused about matboy's answer then, will try other examples to make sure that I am right with understanding, thank you for your time! Jun 29, 2017 at 10:14

$T$ is the matrix whose columns are $f_1, f_2, f_3$. That is:

$$T=\begin{pmatrix}2&-1&-3\\6&0&-1\\-2&5&4 \end{pmatrix}$$

You just have to compute the inverse and apply your formula.

• see my edits, am I right about $f_1\ldots f_3$ computing? Jun 28, 2017 at 17:51
• and also the way of computing $T$? Jun 28, 2017 at 18:18
• otherwise how did you find $T$? Jun 28, 2017 at 19:49
• You got $f_2$ wrong. And I don't get that double matrix you wrote. Jun 28, 2017 at 20:17
• Why f 2 is wrong? (what exactly), and how did you get your T then? Jun 28, 2017 at 21:02