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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 : /

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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$$A_{BB}^f=T^B_{B_0^3}A_{B_0^3B_0^3}^fT^{B_0^3}_B=\begin{pmatrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3\end{pmatrix}.$$

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  • $\begingroup$ 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
    Commented May 1, 2018 at 13:17
  • $\begingroup$ @cnmesr Thanks a lot. I don't know how I made that mistake. $\endgroup$ Commented May 1, 2018 at 13:25

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