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?

  • $\begingroup$ 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 '20 at 2:07

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 $\;T\;$ :

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


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

  • $\begingroup$ So i should do the same process(creating the linear combination) but for B1? $\endgroup$
    – SWAT
    Mar 26 '20 at 1:50

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 $$T\begin{pmatrix}2\\ 2\\ 0\end{pmatrix} = \color{blue}{2}\begin{pmatrix}0\\ 1\\ 1\end{pmatrix} = \begin{pmatrix}0\\ 2\\ 2\end{pmatrix}$$

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