\( \newcommand{\matr}[1] {\mathbf{#1}} \newcommand{\vertbar} {\rule[-1ex]{0.5pt}{2.5ex}} \newcommand{\horzbar} {\rule[.5ex]{2.5ex}{0.5pt}} \newcommand{\E} {\mathrm{E}} \)
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Math and science::Algebra

Matrix vs. linear operator

Every linear operator can be expressed as a matrix. Why then do we bother with the term linear operator?

Consider the Sobel operator, which is a smoothing followed by derivative applied to an image. The operator satisfies \( T(A + B) = T(A) + T(B) \) and \( T(\alpha A) = \alpha T(A) \) so it must have a matrix form. However, this matrix is far more unweildy than the standard 3x3 matrix associated with the Sobel operator that is convolved with an image. The linear operator's matrix would have shape \( nm \times nm \) for an \( n \times m \) image. So while the Sobel operator has an associated matrix, it's more semantically convenient and general to talk about the Sobel operator rather than the transformation matrices the operator implies.