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Very basic linear algebra question

  1. Dec 5, 2013 #1
    So as I'm preparing for finals, I'm wondering:

    The multiplication of two matrices is only defined under special circumstances regarding the dimensions of the matrices.

    Doesn't that require that compositions of linear transformations are only defined in the same circumstances? I can't imagine not being able to not define a composition of linear transformations, can someone demonstrate this?
  2. jcsd
  3. Dec 5, 2013 #2
    If [itex]f:V\to W[/itex] and [itex]g:X\to Y[/itex] are linear transformations, it only makes sense to talk about the composition [itex]g\circ f[/itex] if [itex]W=X[/itex]. In particular, if [itex]f:\mathbb R^K \to \mathbb R^L[/itex] and [itex]g:\mathbb R^J \to \mathbb R^I[/itex] are linear transformations, it only makes sense to talk about the composition [itex]g\circ f[/itex] if [itex]\mathbb R^L=\mathbb R^J[/itex], i.e. if [itex]L=J.[/itex]

    Phrasing the last point a different way now: If [itex]F[/itex] is an [itex]L\times K[/itex] matrix and [itex]G[/itex] is an [itex]I\times J[/itex] matrix, it only makes sense to talk about the matrix product [itex]GF[/itex] if [itex]L=J[/itex].

    So the matrix dimension rule you learned is really there exactly because only certain functions can be composed. The expression [itex]g\circ f[/itex] only has meaning if the outputs of [itex]f[/itex] are valid inputs for [itex]g[/itex].
  4. Dec 5, 2013 #3


    Staff: Mentor

    If A is an m X n matrix, and B is an n X p matrix, then the product AB is defined, and will be an m X p matrix.

    A linear transformation TA: Rn → Rm takes vectors from Rn and maps them to vectors in Rm. A matrix for TA will by m X n. Think about how TB would have to be defined (in terms of its domain and codomain) so that the composition TA ° TB would make sense. It might be helpful to use constants for the dimensions.
  5. Dec 5, 2013 #4
    Crystal clear, thanks you two.
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