Let and be a by and by matrix respectively. Show that

.

*Solution (By Zorn Leung).*

For convenience, given any matrix , denote (abuse of notation) to be the linear operator . Plus let and be the null space and range space of operator respectively.

Consider the restriction of on , denoted by .

Observe first since is clearly a subspace of . Also, is a subspace of . Indeed, if , then for some . Hence implying . Therefore .

By dimension theorem,

, which is the conclusion.

**Corollary.** , as above. Then .

*Proof.*

Dimension theorem implies

,

, and

.

Using the previous result,

. The result follows.

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Let me 馬後砲 again. I think this inequality can be understood as “a tower of kernel inclusions” (perhaps you have noticed from Zorn Leung’s proof already). In an extremely rough language, is just “summing” and (well… in general if , then must goes to after passing through up to some , , so eventually ). OK, I know seems nonsense, so if you think for a second (which takes me some minutes actually), you should guess that (now I use the notations for and ):

I think this is quite clear and everyone can prove it.

The next observation is “” is injective (and hence “preserves dimension”) on the complement of in ). More precisely, (notation abused) is 1-1 (and is well-defined). This is obvious: if for some , of course . [In so many words, it’s nothing but just Zorn’s proof rephrased. ] Therefore by the above, we conclude that

The analysis also shows that the equality holds if and only if , the deficit of the inequality (i.e. RHS-LHS) is measured by .

The same analysis can be applied to the general case for finitely many ‘s, too. i.e. for , we have the tower of kernel inclusion:

. So by the telescoping trick, we can get:

Zorn Leung’s Theorem, 2010The equality holds if and only if for

Since the last comment has become a bit long, I start a new one here.

The Corollary of Zorn Leung’s theorem can be regarded as the “dual version” of it. The general setting is the same, for and , we have the tower of range inclusion: , note that for example when , by abuse of notations, is well-defined and is surjective (and hence “rank decreasing”), so we have and similary .

By telescoping,

Rearranging, we get In general, we have the following theorem:

Zorn Leung’s Theorem, 2010Question:When does the equality hold?Answer:Ask Zorn Leung, who is the tutor of linear algebra.Merry Christmas!!