1
$\begingroup$

Given is linear mapping $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3$. Matrix of $f$ in terms of the ordered standard basis $B_0^3 = \left\{\vec{e_1}, \vec{e_2}, \vec{e_3}\right\}$ of $\mathbb{R}^3$ is $\,\,\,\, A_{B_0^3 B_0^3}^{f} = \begin{pmatrix} 4 & 0 & -2\\ 1 & 3 & -2\\ 1 & 2 & -1 \end{pmatrix} \in \mathbb{R}^{3 \times 3}$, moreover $B=\left\{\vec{b_1}, \vec{b_2}, \vec{b_3}\right\} = \left\{\begin{pmatrix} 2\\ 2\\ 3 \end{pmatrix}; \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}; \begin{pmatrix} 2\\ 1\\ 1 \end{pmatrix}\right\}$ is an ordered basis of $\mathbb{R}^3$.

Determine the transformation matrix $A_{BB}^f$ and the change of basis $T_{B_0^3}^{B}$.

I try to calculate the transformation matrix $A_{BB}^f$ first. So I think because we have $A_{B_0^3 B_0^3}^{f} = \begin{pmatrix} 4 & 0 & -2\\ 1 & 3 & -2\\ 1 & 2 & -1 \end{pmatrix}$, where $B_0^3$ is standard basis, we have that $A^f = \begin{pmatrix} 4 & 0 & -2\\ 1 & 3 & -2\\ 1 & 2 & -1 \end{pmatrix}$ because standard basis shouldn't have any changes on it.

But for calculating $A^f_{BB}$, I need to know the mapping, but I just know $\mathbb{R}^3 \rightarrow \mathbb{R}^3$.. I don't know the precise mapping from matrix to another matrix. I need that so I can insert the given basis $B$ and determine the transformation matrix like that. But how is this supposed to work here? :S

And about the change of basis I really have no idea because it requires the previous thing I mentioned : /

$\endgroup$
5
$\begingroup$

The change of basis matrix $T_B^{B_0^3}$ is simply$$\begin{pmatrix}2&1&2\\2&1&1\\3&1&1\end{pmatrix}$$and $T_{B_0^3}^B$ is$$\begin{pmatrix}2&1&2\\2&1&1\\3&1&1\end{pmatrix}^{-1}=\begin{pmatrix}0 & -1 & 1 \\ -1 & 4 & -2 \\ 1 & -1 & 0\end{pmatrix}.$$So$$A_{BB}^f=T^B_{B_0^3}A_{B_0^3B_0^3}^fT^{B_0^3}_B=\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}.$$

$\endgroup$
  • $\begingroup$ I think there is a little error in the last line (multiplication)? $$A^{f}_{BB} = T^{B}_{B^3_0}A^{f}_{B^3_0 B^3_0} T^{B^3_0}_{B} = \begin{pmatrix} 0 & -1 & 1\\ -1 & 4 & -2\\ 1 & -1 & 0 \end{pmatrix} \begin{pmatrix} 4 & 0 & -2\\ 1 & 3 & -2\\ 1 & 2 & -1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 2\\ 2 & 1 & 1\\ 3 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3 \end{pmatrix}$$ $\endgroup$ – cnmesr May 1 '18 at 13:17
  • $\begingroup$ @cnmesr Thanks a lot. I don't know how I made that mistake. $\endgroup$ – José Carlos Santos May 1 '18 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.