Operator's matrix in canonical form A stupid question
Let $F:\Bbb R^3\rightarrow\Bbb R^3$ the operator such that
$F((1,0,0)=(0,-1,1))\\ F((0,1,0)=(3,1,2))\\ F((0,0,1))=(2,1,1)) $
Determine the matrix $A$ of $F$ in canonical basis.
I think that tha matrix $A$ is
$$
    \begin{pmatrix}
    0 & -1 & 1 \\
    3 & 1 & 2 \\
    2 & 1 & 1 \\
    \end{pmatrix}
$$
But in the correction of test my professor said "no"! So, those vector should be used as columns of $A$? I don't understand why
 A: The problem is that $\Bbb R^3$ means two different things. To many (I'd say 
'most', just guessing), it means the set of column vectors containing 3 real numbers; to others, it means the set of row-vectors containing 3 numbers. 
From the way your problem was given, I'd have said "Ah, the author here is thinking of row vectors!", but apparently your prof was just being a bit lazy and didn't want to write little "transpose" signs above the arguments to $F$. 
Once you've decided that there might be two different meanings for $\Bbb R^3$, there are also two different meanings for "the matrix of $F$ in the canonical basis." In the columns version, it's the matrix $A$ wiht the property that 
$$
F(x) = Ax
$$
for every column vector $x$; in the rows version, it's the matrix $A$ with the property that 
$$
F(x) = xA
$$
for every row vector $x$. 
Your answer was the correct one for this second version, and would also be the answer I'd have given if I were asked this exact question. But I'd probably first have asked "Do you mean for the arguments to be column vectors really?" :(
A: What is the matrix of a linear map $\psi$ with repect to two bases $(e_1,\ldots,e_n)$ and $(f_1,\ldots,f_m)$? It's an array$$A=\begin{pmatrix}a_{11}&\ldots&a_{1n}\\\vdots&\ddots&\vdots\\a_{m1}&\ldots&a_{mn}\end{pmatrix}$$such that the entries of the $i$th column (that is, the numbers $a_{1i},a_{2i},\ldots,a_{mi}$) are such that$$f(e_i)=a_{1i}f_1+a_{2i}f_2+\cdots+a_{mi}f_m.$$So, yes, your professor is right and you should have used those vectors as columns.
A: If you want to write $$F(X)=AX$$ 
with $X=\begin{pmatrix}
    x_1 \\
    x_2 \\
    x_3\\
    \end{pmatrix}
$
you need:
$$F(X)=\begin{pmatrix}
    0 & -3 & 2\\
    -1 & 1 & 1 \\
    1 & 2 & 1 \\
    \end{pmatrix}
   \begin{pmatrix}
    x_1 \\
    x_2 \\
    x_3\\
    \end{pmatrix}
$$
A: Let's rewrite it as following $$F\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}0\\-1\\1\end{bmatrix}\\ F\begin{bmatrix}0\\1\\0\end{bmatrix}=\begin{bmatrix}3\\1\\2\end{bmatrix}\\F\begin{bmatrix}0\\0\\1\end{bmatrix}=\begin{bmatrix}2\\1\\1\end{bmatrix}$$since for any $v\in\Bbb R^3$ we have $v=a_1$ , $v=a_1\begin{bmatrix}1\\0\\0\end{bmatrix}+a_2\begin{bmatrix}0\\1\\0\end{bmatrix}+a_3\begin{bmatrix}0\\0\\1\end{bmatrix}$ we obtain$$Fv=a_1F\begin{bmatrix}1\\0\\0\end{bmatrix}+a_2F\begin{bmatrix}0\\1\\0\end{bmatrix}+a_3F\begin{bmatrix}0\\0\\1\end{bmatrix}=a_1\begin{bmatrix}0\\-1\\1\end{bmatrix}+a_2\begin{bmatrix}3\\1\\2\end{bmatrix}+a_3\begin{bmatrix}2\\1\\1\end{bmatrix}=\begin{bmatrix}0&3&2\\-1&1&1\\1&2&1\end{bmatrix}\begin{bmatrix}a_1\\a_2\\a_3\end{bmatrix}=Av$$where $$A=\begin{bmatrix}0&3&2\\-1&1&1\\1&2&1\end{bmatrix}$$
A: The matrix you are looking for is $\begin{pmatrix}
    0 & -1 & 1 \\
    3 & 1 & 2 \\
    2 & 1 & 1 \\
    \end{pmatrix}^{T}$.
A: Yes, the coordinates of these vectors should be used as the column vectors of the matrix. 
Indeed, let 
$$A=\begin{bmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3 \end{bmatrix}$$
be the required matrix. By definition
\begin{align}A\cdot\begin{bmatrix}1\\0\\0\end{bmatrix}=\begin{bmatrix}a_1 \\b_1 \\c_1 \end{bmatrix}\end{align}
and it has to be equal to $\;\begin{bmatrix}\phantom-0\\-1\\\phantom-1\end{bmatrix}$. Similarly for the other vectors in the canonical basis.
