This looks like a good opportunity to test if a post I wrote a few months ago is understandable. Let me know if something isn't clear. I've been meaning to turn this into a FAQ post, but I haven't gotten around to adding the additional material I think should be included.
I moved your post to linear & abstract algebra, questions about concepts, definitions and theorems do not belong in the homework forums. They are mainly for questions about textbook-style problems.
Let X and Y be finite-dimensional vector spaces. Let ##T:X\to Y## be a linear transformation. Let ##A=(e_1,\dots,e_n)## and ##B=(f_1,\dots,f_m)## be ordered bases for X and Y respectively. For each ##y\in Y##, there's a unique m-tuple of scalars ##y_1,\dots,y_m## such that ##y=\sum_{i=1}^m y_i f_i##. These scalars are called the components of y with respect to B.
Let ##i\in\{1,\dots,m\}## and ##j\in\{1,\dots,n\}## be arbitrary. ##Te_j## is clearly a member of Y. Since we use the notation ##y_i## for the ith component of an arbitrary ##y\in Y## with respect to B, it's natural to use the notation ##(Te_j)_i## for the ith component of ##Te_j## with respect to B. The mn (m times n) scalars ##(Te_j)_i## with ##i\in\{1,\dots,m\}## and ##j\in\{1,\dots,n\}## are called the components, or matrix elements, of T with respect to (A,B). The m×n matrix
$$\begin{pmatrix}(Te_1)_1 & \cdots & (Te_n)_1\\ \vdots & \ddots & \vdots\\ (Te_1)_m & \dots & (Te_n)_m\end{pmatrix}$$ is called the matrix representation of T, or just the matrix of T, with respect to (A,B). It is often denoted by the same symbol as the linear transformation, in this case T. In situations where you would prefer to use different notations for the linear transformation and its matrix representation, a notation like ##[T]## or ##[T]^{A,B}## can be used for the latter.
The standard notation for the scalar on row i, column j of a matrix ##T## is ##T_{ij}##. In this notation, we have ##T_{ij}=(Te_j)_i##. This is the formula you need to remember. An alternative notation for the scalar on row i, column j is ##T^i_j##. In this notation, we have ##T^i_j=(Te_j)^i##. If you commit this version of the formula to memory, there's no chance that you will forget the correct order of the indices.
The following observation provides some motivation for the definitions. Let ##x\in X## be arbitrary. Define ##y\in Y## by ##y=Tx##. Let ##x_1,\dots,x_n## be the components of x with respect to A, and let ##y_1,\dots,y_m## be the components of y with respect to B.
\begin{align}
y &=\sum_{i=1}^m y_i f_i\\
Tx &=T\left(\sum_{j=1}^n x_j e_j\right) =\sum_{j=1}^n x_jTe_j =\sum_{j=1}^n x_j \left(\sum_{i=1}^m(Te_j)_i f_i\right) =\sum_{j=1}^n \sum_{i=1}^m x_j (Te_j)_i f_i =\sum_{i=1}^m \left(\sum_{j=1}^n x_j (Te_j)_i\right) f_i
\end{align}
Since ##\{f_1,\dots,f_m\}## is linearly independent, these results and the equality y=Tx imply that
$$\sum_{j=1}^n x_j (Te_j)_i= y_i$$ for all ##i\in\{1,\dots,m\}##. If you recall that the definition of matrix multiplication is ##(AB)_{ij}=\sum_k A_{ik}B_{kj}##, you can easily recognize the above as the ith row of the matrix equation
$$\begin{pmatrix}(Te_1)_1 & \cdots & (Te_n)_1\\ \vdots & \ddots & \vdots\\ (Te_1)_m & \dots & (Te_n)_m\end{pmatrix} \begin{pmatrix}x_1\\ \vdots\\ x_n\end{pmatrix} =\begin{pmatrix}y_1\\ \vdots\\ y_m\end{pmatrix}.$$ The following is a simple example of how to find a matrix representation of a linear transformation. Define ##S:\mathbb R^3\to\mathbb R^2## by ##S(x,y,z)=(3z-x,2y)##. This S is clearly linear. Let ##C=(e_1,e_2,e_3)## and ##D=(f_1,f_2)## be the standard ordered bases for ##\mathbb R^3## and ##\mathbb R^2## respectively. We will denote the matrix of S with respect to (C,D) by .
\begin{align}Se_1 &=S(1,0,0)=(-1,0) =-1f_1+0f_2\\
Se_2 &=S(0,1,0) =(0,2)=0f_1+2f_2\\
Se_3 &=S(0,0,1) =(3,0)=3f_1+0f_2\\
&=\begin{pmatrix}(Se_1)_1 & (Se_2)_1 & (Se_3)_1\\ (Se_1)_2 & (Se_2)_2 & (Se_3)_2\end{pmatrix} =\begin{pmatrix}-1 & 0 & 3\\ 0 & 2 & 0 \end{pmatrix}.
\end{align} Note that for all ##x,y,z\in\mathbb R##,
$$\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}-1 & 0 & 3\\ 0 & 2 & 0 \end{pmatrix} \begin{pmatrix}x\\ y\\ z\end{pmatrix} =\begin{pmatrix}-x+3z\\ 2y\end{pmatrix}.$$ If Y is an inner product space and B is orthonormal, the easiest way to find the matrix elements is often to use the inner product. We will use the physicist's convention for inner products. This means that the term "inner product" is defined so that when we're dealing with a vector space over ℂ (i.e. when the set of scalars is ℂ rather than ℝ), the map ##v\mapsto\langle u,v\rangle## is linear and the map ##u\mapsto\langle u,v\rangle## is antilinear (i.e. conjugate linear). Let ##i\in\{1,\dots,m\}## and ##y\in Y## be arbitrary. Let ##y_1,\dots,y_m## be the components of y with respect to B.
$$\left\langle f_i,y\right\rangle =\left\langle f_i,\sum_{j=1}^m y_j f_j \right\rangle =\sum_{j=1}^m y_j \left\langle f_i,f_j\right\rangle =\sum_{j=1}^m y_j \delta_{ij} =y_i.$$ Since i and y are arbitrary, this implies that for all ##i\in\{1,\dots,m\}## and all ##j\in\{1,\dots n\}##,
$$(Te_j)_i =\left\langle f_i,Te_j \right\rangle.$$ If X=Y, it's convenient to choose B=A, and to speak of the matrix representation of T with respect to A instead of with respect to (A,A), or (A,B). The formula for ##T_{ij}## can now be written as
$$T_{ij}=\left\langle e_i,Te_j \right\rangle.$$ In bra-ket notation (which is used in quantum mechanics), we would usually write the ith basis vector as ##\left|i\right\rangle##. This turns the formula into
$$T_{ij} =\left\langle i\right|T\left|j\right\rangle.$$
