I have this question here…
Let $V$ be the span of $v_{1}=(0,1,2)$, $v_{2}=(-1,0,1)$ and $v_{3}=(-1,1,3)$.
$(a)$Construct an orthonormal basis $B'$ for $V$ (usual dot product).
$(b)$ Express each of $v_1$,$v_2$,$v_3$ in terms of the basis $B'$ found in $(a)$.
$(c)$ For any $x$ in $\mathbb{R^3}$, let $T(x)=(x \cdot v_1)v_1+(x \cdot v_2)v_2$ and let $B=<i,j,k>$ be the standard basis for $\mathbb{R^3}$. Construct $[T]_{B'B}$, the matrix representation of $T$ with respect to the two basis $B,B'$.
$(d)$ If $x=(1,1,1)$, find the coordinates of $T(x)$ with respect to the basis $B'$.
I already figured out parts $(a)$ and $(b)$.
In part $(a)$, I just need to use the gram schmidt procedure. Since $v_{3}$ is linearly dependent, that cannot be part of the orthonormal basis $B'$ and hence I just need to use the gram schmidt procedure on $v_1$ and $v_2$.
Thus, using the gram schmidt prodedure, I obtain:
$U_{1}=V_{1}=(0,1,2)$
Orthonormalizing, we obtain $v_{1}'=(0,\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})$
Then $U_2=V_2-\frac{V_2 \cdot U_1}{||U_1||^2}U_1$
Doing so gives: $U_2=(-1,\frac{-2}{5},\frac{1}{5})$
Orthonormalizing, we get $v_{2}'=(\frac{-5}{\sqrt{30}},\frac{-2}{\sqrt{30}},\frac{1}{\sqrt{30}})$
$B'=\left\{\begin{array}{@{}c@{}}\left(\begin{matrix}0 \\ \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{matrix}\right),\left(\begin{matrix} \frac{-5}{\sqrt{30}} \\ \frac{-2}{\sqrt{30}} \\ \frac{1}{\sqrt{30}} \end{matrix}\right)\end{array}
\right\}$
Where $v_1'=\left(\begin{matrix}0 \\ \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{matrix}\right)$ and $v_2'= \left(\begin{matrix} \frac{-5}{\sqrt{30}} \\ \frac{-2}{\sqrt{30}} \\ \frac{1}{\sqrt{30}} \end{matrix}\right)$
In part $(b)$, I just need to express $v_1$ , $v_2$ and $v_3$ as a linear combination of $B'$, so in essence:
$v_1=av_1'+bv_2'$
$v_2=av_1'+bv_2'$
$v_3=av_1'+bv_2'$
If I solve with row reduction, I obtain that:
$v_1=\sqrt{5}v_1'+0v_2'$
$v_2=\frac{2}{\sqrt{5}}v_1'+\frac{6}{\sqrt{30}}v_2'$
$v_3=\frac{7}{\sqrt{5}}v_1'+\frac{6}{\sqrt{30}}v_2'$
Can anyone help on parts $(c)$ and $(d)$ though?
In part (c) I think it's asking for the compostition of the transformation matrix, namely $[T]_{b} [T]_{b'}$ but I am not even sure how to compute that.
I'm pretty sure I have to use a change of basis matrix in part $(d)$ but I am not sure.
Any help would be much appreciated! This isn't for marks or anything by the way. I'm just reviewing linear algebra after 6 years because I didn't understand it that well so I wanted to make sure I have a good grasp on this now.
Best Answer
Since the notation is sometimes different: In the following I assume $T_{B',B}$ takes a vector in its representation w.r.t. $B$ and evaluates $T$ and gives back a vector in representation w.r.t. $B'$. (Sometimes the order of $B$ and $B'$ is different.)
The idea in (c) is to change basis from $B$ to $B'$ and evaluate T after that. Try to calculate the matrix $T_{B',B'}$ first. After that you can calculate the base change matrix $M_{B',B}$, changing the representation of $x$ w.r.t. $B$ into the representation of $x$ w.r.t. $B'$. Then the answer is the product of these matrices (Which one do you have to multiply with $x$ first? So which order is the right one?).
For (d) you just need to evaluate T on (1,1,1), i. e. multiply the matrix in (c) from right with (1,1,1).