0
$\begingroup$

Currently im on Lineal Algebra class, seing linear transformations with matrices. My question is what are the steps to resolve this question?

$\text{Let }T: \mathbb{R}^3 \longrightarrow \mathbb{R}^3 \text{ a lineal transformation. Consider the following basis of }\mathbb{R}^3: \\ B_1= (1,1,1),(1,1,0),(1,0,0)\\ B_2= {(1,0,1),(0,1,1),(0,1,0)}\\$

$\text{If we know that:} [T]_{B_1}^{B_2}=\begin{pmatrix}3&0&2\\ 0&1&1\\ 1&0&2\end{pmatrix} $

Find $T(2,2,0)$.

How should i resolve this question?

$\endgroup$
  • $\begingroup$ Seems to me that there should’ve been at least one example of this sort of thing in the material you’re meant to have studied before attempting the exercise. What have you learned about representing a linear transformation as a matrix and how to change basis? $\endgroup$ – amd Mar 26 at 2:07
1
$\begingroup$

It all depends on your notation. For me, $\;[T]_{B_1}^{B_2}\;$ means the matrix of basis change from $\;B_1\;$ to $\;B_2\;$ , which means:

$$\begin{cases}T(1,1,1)=3(1,0,1)+0(0,1,1)+1(0,1,0)=(3,1,3)\\{}\\ T(1,1,0)=0(1,0,1)+1(0,1,1)+0(0,1,0)=(0,1,1)\\{}\\ T(1,0,0)=2(1,0,1)+1(0,1,1)+2(0,1,0)=(2,3,3)\end{cases}$$

Since $\;(2,2,0)=0(1,1,1)+2(1,1,0)+(-2)(1,0,0)\;$, we get that by linearity of $\;T\;$ :

$$T(2,2,0)=0\cdot T(1,1,1)+2T(1,1,0)+(-2)T(1,0,0)=0(3,1,3)+2(0,1,1)-2(2,3,3)=$$

$$=(-4,-4,-4)$$

The last vector is expressed in the coordinates of $\;B_2\;$ . Try now to end the argument.

$\endgroup$
  • $\begingroup$ So i should do the same process(creating the linear combination) but for B1? $\endgroup$ – SWAT Mar 26 at 1:50
1
$\begingroup$

I suggest the following interpretation:

  • Vector $\begin{pmatrix}2\\ 2\\ 0\end{pmatrix}=\color{blue}{2}\begin{pmatrix}1\\ 1\\ 0\end{pmatrix}$ has wrt. basis $B_1$ the coordinates: $$\begin{pmatrix}0\\ \color{blue}{2}\\ 0\end{pmatrix}_{B_1}$$
  • We know the coordinates of $T\begin{pmatrix}2\\ 2\\ 0\end{pmatrix}$ wrt. basis $B_2$:

$$ [T]_{B_1}^{B_2}\begin{pmatrix}0\\ 2\\ 0\end{pmatrix}_{B_1}=\begin{pmatrix}3&0&2\\ 0&1&1\\ 1&0&2\end{pmatrix}\begin{pmatrix}0\\ 2\\ 0\end{pmatrix}_{B_1} = \begin{pmatrix}0\\ 2\\ 0\end{pmatrix}_{B_2}=\color{blue}{2}\begin{pmatrix}0\\ 1\\ 0\end{pmatrix}_{B_2}$$

  • Since the second basis vector in $B_2$ is $\begin{pmatrix}0\\ 1\\ 1\end{pmatrix}$, you get $$T\begin{pmatrix}2\\ 2\\ 0\end{pmatrix} = \color{blue}{2}\begin{pmatrix}0\\ 1\\ 1\end{pmatrix} = \begin{pmatrix}0\\ 2\\ 2\end{pmatrix}$$
$\endgroup$

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.