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Say that $T : \mathbb{R}^3 \to \mathbb{R}^4$ is linear and satisfies $T(1,1,1) = (3,2,0,1)$, $T(1,1,0) = (2,1,3, −1)$ and $T(1,0,0) =(5, −2,1,0)$. Find $T(x, y, z)$ and then find the matrix representation for $T$.

I am not sure how to proceed. I thought of two approaches, one the matrix $A$ from $Ax=b$ could be expressed as

$$A=\begin{pmatrix} 3 & 2 & 5\\ 2 & 1 & -2\\ 0 & 3 & 1\\ 1 & -1 & 0 \end{pmatrix}$$

Then possibly solve for $Ax=0$?

Or since $T(1,0,0)=v_3$ etc, we can also write $T(x,y,z)=x\cdot v_1+y \cdot v_2+z \cdot v_3$, but I am not sure how to find $v_1$, $v_2$ or $v_3$.

Appreciate your help.

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First question: What is the dimension of the matrix of this transformation? –  imranfat Mar 19 at 19:18
    
It should be 4x3. Because we are going from R3->R4 –  user3424493 Mar 19 at 19:20
    
Good, now set up this matrix, using $a,b,c,...$ letters as entries of the matrix. You can set up 3 matrix multiplications like $A*(1,1,1) = (3,2,0,1)$ etc. and this gives you a whole bunch of systems of equations which you can solve for all your entries of matrix A. A Ti calculator would be handy (use RREF) –  imranfat Mar 19 at 19:26
    
Did you get it? –  imranfat Mar 19 at 19:34

1 Answer 1

up vote 1 down vote accepted

$T(0,0,1) = T(1,1,1) - T(1,1,0) = (1, 1, -3, 2)$, and $T(0,1,0) = T(1,1,0) - T(1,0,0) = (-3, 3, 2, -1)$, and $T(1,0,0) = (5, -2, 1, 0)$. So \begin{align*}T(x,y,z) &= xT(1,0,0) + yT(0,1,0) + zT(0,0,1) \\ &= x(5,-2,1,0) + y(-3,3,2,-1) + z(1,1,-3,2) \\ &= (5x-3y+z, -2x+3y+z, x+2y-3z, -y+2z).\end{align*} And from this the matrix $A$ is:

\begin{pmatrix} 5 & -3 & 1 \\ -2 & 3 & 1 \\ 1 & 2 & -3 \\ 0 & -1 & 2 \\ \end{pmatrix}

You can continue.

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2  
273 answers and still didn't learn basic latex? I think it's time. But that is just my opinion. –  Cortizol Mar 19 at 20:12
    
Thank you very much! –  user3424493 Mar 19 at 21:42
    
@Cortizol: yes yes, soon on mathjax –  OC-Sansoo Mar 19 at 21:43

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