# Linear operators and matrices

I'm working through a proof that every linear operator, $$A$$, can be represented by a matrix, $$A_{ij}$$. So far I've got

$$Let \textbf{p}=\sum_{i}p_{i}\widehat{\textbf{e}}_{i}$$
$$A(\textbf{p}) = \sum_{i}p_{i}A(\textbf{e}_{i})$$

which is fine. Then it says that $$A(\textbf{e}_{i})$$ is a vector, given by:

$$A(e_{i}) = \sum_{j}A_{j}(p_{i})e_{j} = \sum_{j}A_{ji}e_{j}$$.

The fact that its a vector is fine with me, but I can't get my head around the equation for it. why does the operator acting on one of the base vectors depend on $$p_{i}$$? Surely the base vectors are independent of $$p_{i}$$ and so should be any operation acting on them.

CompuChip
Homework Helper
Indeed, they don't.
I would write it like
$$A(\hat e_i) = \sum_j (A_i)_j \hat e_j$$
where Ai is some vector of coefficients.

Fredrik
Staff Emeritus
Gold Member
This is how I do this thing: Suppose $A:U\rightarrow V$ is linear, and that $\{u_j\}$ is a basis for U, and $\{v_i\}$ is a basis for V. Consider the equation y=Ax, and expand in basis vectors.

$$y=y_i v_i$$

$$Ax=A(x_j u_j)=x_j Au_j= x_j (Au_j)_i v_i$$

I'm using the Einstein summation convention: Since we're always supposed to do a sum over the indices that appear exactly twice, we can remember that without writing any summation sigmas (and since the operator is linear, it wouldn't matter if we put the summation sigma to the left or right of the operator). Now define $A_{ij}=(Au_j)_i$. The above implies that

$$y_i=x_j(Au_j)_i=A_{ij}x_j$$

Note that this can be interpreted as a matrix equation in component form. $y_i$ is the ith component of y in the basis $\{v_i\}$. $x_j$ is the jth component of x in the basis $\{u_j\}$. $A_{ij}$ is row i, column j, of the matrix of A in the pair of bases $\{u_j\}$, $\{v_i\}$.

$$A(e_{i}) = \sum_{j}A_{j}(p_{i})e_{j} = \sum_{j}A_{ji}e_{j}$$
This one should be

$$Ae_{i} = \sum_{j}(Ae_{i})_j e_{j} = \sum_{j}A_{ji}e_{j}$$

Note that the first step is just to express the vector $Ae_i$ as a linear combination of basis vectors, and that $(Ae_i)_j$ is just what I call the jth component.

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