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Given a linear transformation on $\mathbb{R^3}$ as follows: $$\phi(x)=(x,a)a$$ where $(x,a)$ stands for the dot product of the vectors $x$ and $a$ ,and $a=(1,2,3)$. Find the matrix of this transformation on the basis $e_1=(1,0,0),e_2=(0,1,0),e_3=(0,0,1)$, which all the vectors above are given on, and also find the matrix on the basis $b_1=(1,0,1),b_2=(2,0,-1),b_3=(1,1,0)$.

I have already found the matrix on the basis $e_1,e_2,e_3$: $$A=\begin{bmatrix} 1 &2 &3 \\ 2 &4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$$ Also, the matrix that maps the first basis to the second: $$C=\begin{bmatrix} 1 &2 &1 \\ 0 &0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$$ On the new basis $a=(\cfrac{5}{3},-\cfrac{4}{3},2)$. I have done some trials to find the matrix $B$ (the second matrix asked in the question) but none matches with the answer of the book, while $A$ seems to be correct.

The answer is $$B=\begin{bmatrix} 20/3 &-5/3 &5 \\ -16/3 &4/3 & -4 \\ 8 & -2 & 6 \end{bmatrix}$$ Any help is appreciated.

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You have$$f(b_1)=(4,8,12),\ f(b_2)=(-1,-2,-3)\text{ and }b_3=(3,6,9).$$But, with respect to the basis $B=\{b_1,b_2,b_3\}$, you have$$f(b_1)=\left(\frac{20}3,-\frac{16}3,8\right)_B,\, f(b_2)=\left(-\frac53,\frac43,-2\right)_B\text{ and }f(b_3)=(5,-4,6)_B.$$So,$$C=\begin{bmatrix}\frac{20}3&-\frac53&5\\-\frac{16}3&\frac43&-4\\8&-2&6\end{bmatrix}.$$

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  • $\begingroup$ "$f(b_1)=\left(\frac{20}3,-\frac{16}3,8\right)_B$" to have this, do I have to find the coordinates of $b_1 \text{ and } a$ on the basis $B$ and then calculate $\phi(b_1)=(b_1,a)a$? $\endgroup$
    – VIVID
    Apr 20 '20 at 8:58
  • $\begingroup$ No. You solve the system $f(b_1)=\alpha b_1+\beta b_2+\gamma b_3$. The only solution is $\alpha=\frac{20}3$, $\beta=-\frac{16}3$, and $\gamma=8$. $\endgroup$ Apr 20 '20 at 9:00

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