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Problem: Let T: $\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a linear map given by

$$T\left[ \begin{matrix} x\\y\\ z\end{matrix} \right]= \left[ \begin{matrix} 3x-y\\z-x\\z-y\\\end{matrix} \right]$$

  1. Find the Matrix representation of T with respect to the canonical basis of $\mathbb{R}^3$, and call it A.

I am not sure how this works. So the cananical basis of $\mathbb{R}^3$ is ${(1,0,0),(0,1,0),(0,0,1)}$ But I am unsure how to get a matrix represenation from a linear operator. Any help is appreciated.

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Remember how a matrix is defined. The $j$th column tells you where the $j$th basis vector goes. So the first column should be the image of $(1,0,0)$ under $T$, which is $(3,-1,0)$. You can continue like this – Daniel Freedman May 14 '12 at 22:26
Ok great! Thank you guys so much it is so much clearer now. – Mathstudent May 14 '12 at 22:37
up vote 1 down vote accepted

I will write the linear map $T$ as $$T(x,y,z)=(3x-y,z-x,z-y).$$ Then we have $$T(1,0,0)=(3,-1,0),$$ $$T(0,1,0)=(-1,0,-1),$$ $$T(0,0,1)=(0,1,1).$$ Then the matrix representation of $T$ is given by $$\left[ \begin{array}{ccc} 3 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 1 \\ \end{array} \right].$$ The way to get it is: the vector $T(1,0,0)=(3,-1,0)$ becomes the first column, the vector $T(0,1,0)=(-1,0,-1)$ becomes the second column, and so forth. My linear algebra teacher always says, "Put it to be vertical, put it to be vertical,...".

The advantage of the matrix representation is that; for example if I want to find $T(1,-2,0)$, then I can do it by $$\left[ \begin{array}{ccc} 3 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 1 \\ \end{array} \right]\left[ \begin{array}{c} 1 \\ -2 \\ 0 \\ \end{array} \right]=\left[ \begin{array}{c} 5 \\ -1 \\ 2 \\ \end{array} \right],$$ that is, $T(1,-2,0)=(5,-1,2)$.

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wait shouldn't the matrix represenation be $$\left[ \begin{array}{ccc} 3 & -1 & 0 \\ -1 & 0 & -1 \\ 0 & 1 & 1 \\ \end{array} \right].$$ – Mathstudent May 14 '12 at 22:44

Let $e_1=(1,0,0)$, $e_2=(0,1,0)$, and $e_3=(0,0,1)$.

From the definition of $T$: $$ T(e_1) =\left[\matrix{3\cr -1\cr 0 }\right],\quad T(e_2) =\left[\matrix{-1\cr 0\cr -1 }\right],\quad T(e_3) =\left[\matrix{0\cr 1\cr 1 }\right]. $$ For an arbitrary vector $v=\left[\matrix{a\cr b\cr c }\right] =a e_1+b e_2+c e_3$, we have using the linearity of $T$ $$\eqalign{ T(v) &=T(a e_1+b e_2+c e_3)\cr &=a T(e_1)+b T(e_2)+c T(e_3)\cr &=a \left[\matrix{3\cr -1\cr 0 }\right]+ b\left[\matrix{-1\cr 0\cr -1 }\right]+ c\left[\matrix{0\cr 1\cr 1 }\right]\cr &= \left[\matrix{3a-1\cdot b+0\cdot c\cr -1\cdot a+0\cdot b+1\cdot c\cr 0\cdot a+1\cdot -b+1\cdot c }\right]\cr &= \left[\matrix{3&-1&0 \cr -1&0& 1\cr 0&-1& 1}\right] \left[\matrix{a\cr b\cr c}\right].\cr } $$ So the matrix representation of $T$ is $\left[\matrix{3&-1&0 \cr -1&0&1\cr 0&-1&1}\right]$.

Note that the $i^{\rm th}$ column of the matrix representing $T$ is the vector $T(e_i)$. So, we can get the desired!

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