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Very basic linear algebra question 
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#1
Dec513, 01:57 PM

P: 1,290

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
Dec513, 02:16 PM

P: 229

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]. 


#3
Dec513, 02:17 PM

Mentor
P: 21,215

A linear transformation T_{A}: R^{n} → R^{m} takes vectors from R^{n} and maps them to vectors in R^{m}. A matrix for T_{A} will by m X n. Think about how T_{B} would have to be defined (in terms of its domain and codomain) so that the composition T_{A} ° T_{B} would make sense. It might be helpful to use constants for the dimensions. 


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