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Consider the map $f:\mathbb{R}^3 \rightarrow \mathbb{R}^3 $ defined by $$f\begin{pmatrix} x_1\\ x_2\\ x_3 \end{pmatrix}=(x_1-x_2+x_3)b_1 + (x_1-x_2-x_3)b_2 + (x_1+x_2-x_3)b_3$$

where $b_1 = \begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix},b_2 = \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix},b_3 = \begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}$ is a basis for $\mathbb{R}^3 $ (this has been shown).

Problem

Determine the matrix $C$ that is represented by the given $f$ with respect to the basis $(b_1,b_2,b_3)$ for both domain and codomain. Furthermore, determine the coordinates of $f(b_1) \in \mathbb{R}^3$ with respect to the basis $(b_1,b_2,b_3)$.

Try

First off this is a pratice problem. I would think that $C=BP$ where $P$ would just be the matrix of the $j$'s basis vectors as columns and $B$ would be matrix representing the standard basis evaluated in the map $f$ and thus give us the $j$'s standard basis would give the $j$'s column in the matrix. Is this correct or is there an easier way since it is a endomorphism?

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Note that$$f(b_1)=2b_1,\ f(b_2)=-2b_2\text{, and }f(b_3)=2b_3.$$So,$$C=\begin{bmatrix}2&0&0\\0&-2&0\\0&0&2\end{bmatrix}.$$I think that this is the simplest way.

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  • $\begingroup$ I see. How would you then $f(b_1)$ expressed in therms of the basis $b_n$ $\endgroup$
    – user831952
    Jan 26 at 18:56
  • $\begingroup$ Since $f(b_1)=2b_1$, it is represented in the basis $B$ as $(2,0,0)$. $\endgroup$ Jan 26 at 18:57
  • $\begingroup$ Hmm I guess my last note is that I see evaluating the different standard basis vectors would give the constants in front of the vectors in the basis $2,-2,2$ but why is the matrix $C$ not just that result i.e first column would be $2b_1=(2,0,2)$ $\endgroup$
    – user831952
    Jan 26 at 19:07
  • $\begingroup$ Because $2b_1\ne2b_1+2b_3$. Asserting that $2b_1$ is represented by $(2,0,2)$ in the basis $B$ means that $2b_1=2b_1+2b_3$. $\endgroup$ Jan 26 at 19:10

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