# Matrix of a linear transformation on basis

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.

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,$$C=\begin{bmatrix}\frac{20}3&-\frac53&5\\-\frac{16}3&\frac43&-4\\8&-2&6\end{bmatrix}.$$
• "$f(b_1)=\left(\frac{20}3,-\frac{16}3,8\right)_B$" to have this, do I have to find the coordinates of $b_1 \text{ and } a$ on the basis $B$ and then calculate $\phi(b_1)=(b_1,a)a$? Apr 20 '20 at 8:58
• 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 $\gamma=8$. Apr 20 '20 at 9:00