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S and T are transformation mappings, what does [ST] and [TS] mean?

Does it mean transform via S and then apply T to the result and vice versa?

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Juxtaposition of linear transformations means composition. In details, $ST$ means $S \circ T$ and $TS$ means $T \circ S$ (whenever they are defined, surely). This notation is used thinking of the following: $$T: V_1 \to V_2 \qquad S:V_2 \to V_3 \qquad S\circ T: V_1 \to V_3$$ Let $B_1, B_2, B_3$ be basis for the respective spaces (in finite dimension). Then: $$[S \circ T]_{B_1, B_3} = [S]_{B_2, B_3}[T]_{B_1,B_2}$$ At the bottom of our hearts, we know the distinction, so, writing, we oftem "confuse" the matrix with the transformation, so $S \circ T$ becomes $ST$. Actually, the identity above is the motivation for the definition of matrix multiplication.

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  • $\begingroup$ So if T: V1 -> V2 and S: V2 -> V1 then what is ST? $\endgroup$ – user83039 Oct 13 '14 at 1:52
  • $\begingroup$ Did you read what I wrote? Think a bit. $\endgroup$ – Ivo Terek Oct 13 '14 at 1:58
  • $\begingroup$ Yes I've read it a number of times. I'm thinking ST: V1 -> V1 $\endgroup$ – user83039 Oct 13 '14 at 2:07
  • $\begingroup$ Yes, the composition goes from $V_1$ to $V_1$. To understand that notation better, I think it is important to take a look at matrices of linear transformations. $\endgroup$ – Ivo Terek Oct 13 '14 at 2:09
  • $\begingroup$ Great, and for the format of B1,B3, can the B3 also be written above B1 or is it B3 above B1? $\endgroup$ – user83039 Oct 13 '14 at 2:23

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