http://homepage.divms.uiowa.edu/~frohman/pyth2.pdf
[In page 4]
why is $l_{ij}$ square? I can't see why the dimensions of domain and image must be equal..
Explanation of a step in General Pythagoras
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$\begingroup$
linear-algebra
proof-explanation
1 Answers
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WLOG, we can take $V=W$ and assume $V$ is finite-dimensional ; let $n:=dim V$.
By definition of $(\ker L)^{\perp}$, we have a direct sum:
$$(\ker L)\oplus (\ker L)^{\perp}=V,$$
from which we can deduce
$$dim(\ker L)+dim(\ker L)^{\perp}=n,$$
i.e.,
$$\tag{1}dim(\ker L)=n-dim(\ker L)^{\perp}.$$
Besides, the rank-nullity theorem (https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem) gives:
$$\tag{2}dim(\ker L)=n-dim(im L).$$
Comparing (1) and (2) gives
$$dim(\ker L)^{\perp}=dim(im L).$$