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Couple of years ago, I saw a proof in Horn's "A Second Course in Linear Algebra" of the fact that if a matrix (over a field) has a right inverse then it is automatically invertible, essentially using the fact that $M_n(K)$ is Artinian (the same works in any Artinian ring, I expect). At the time I did not know much mathematics and I was rather impressed by this proof, as the ones I'd seen previously were always "inelegant" rank-related arguments, and as a consequence it got stuck in my head.

As such, I would like to ask you to share what proofs of this kind you know, if any. Ideally, they should be either elementary using "advanced ideas" (as the one I posted might arguably be) or short one/two/three-liners directly applying more advanced results/theories. When I say "basic facts", I mean ones at the level of a ~2nd-3rd year undergrad who's had linear algebra, the basic algebra sequence, and topology (this does not refer to me necessarily, rather the hypothetical audience).

Sambo
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Hilbert Jr.
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    Not sure that it counts, but if you have Lagrange's theorem the proof of Euler's theorem ($x^{\varphi(n)} \equiv 1 \bmod n$) is less than a sentence. – Randall Oct 20 '22 at 18:26
  • @Randall that's a good one, if a little trivial. – Hilbert Jr. Oct 20 '22 at 19:25
  • I guess at a lower level, proving $\binom{pa}{pb} \equiv \binom{a}{b} \pmod{p}$ by using $(1+x)^{pa} \equiv (1+x^p)^a \pmod{p}$ and taking the coefficients of $x^{pb}$ on both sides could be a minor example of such a thing. – Daniel Schepler Oct 20 '22 at 21:34
  • Oh, and also, studying sums of squares becomes much easier when you cast it as a question about $\mathbb{Z}[i]$ and then you can use unique factorization in that ring. – Daniel Schepler Oct 20 '22 at 21:51
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    "the fact that if a matrix (over a field) has a right inverse then it is automatically invertible" --is false. The statement does hold for square matrices though. – user8675309 Oct 20 '22 at 22:45

6 Answers6

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The existence of Jordan normal form of a linear transformation is a corollary of the structure theorem for finitely generated modules over a PID.

Namely, suppose $k$ is an algebraically closed field, and we have a finite-dimensional vector space $V$ over $k$, along with a linear operator $T : V \to V$. Then this induces a structure of $k[t]$-module on $V$, where $t$ acts as $T$. Since $k[t]$ is a PID, then this module decomposes as $V \simeq k[t]^n \oplus \bigoplus_{k=1}^m k[t] / \langle p_i \rangle$, where each $p_i$ is a power of an irreducible of $k[t]$.

Now, the fact that $V$ is finite-dimensional over $k$ implies that $n=0$. Furthermore, since $k$ is algebraically complete, we see that each $p_i$ is of the form $(t - \lambda_i)^{d_i}$. However, the matrix representation of the action of $t$ on the module $k[t] / \langle (t-\lambda_i)^{d_i} \rangle$ with respect to the basis $\{ (t-\lambda_i)^{d_i-1}, \cdots, t-\lambda_i, 1 \}$ is exactly a block of the Jordan canonical form for eigenvalue $\lambda_i$ of size $d_i$.

Daniel Schepler
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    Incredible. I remember sweating over a two page proof for the JNF and being very confused with certain hand wavy aspects, two years ago… but unfortunately I don’t know enough abstract algebra yet to understand this fully. +1 – FShrike Oct 20 '22 at 19:07
  • I know this one, but I've never found a source that works out all the details of this or the rational canonical form in detail, they're usually very cursory with that. To clarify, I do know the structure theorem and its proof, it's just the applications I've never learned properly (e.g. why the (tA-I) is exactly the matrix from which you get the invariant factors, etc.). – Hilbert Jr. Oct 20 '22 at 19:19
  • Regarding the $A - tI$ matrix, it turns out that the cokernel of $A - tI$ as a morphism of $k[t]$-modules $k[t]^n \to k[t]^n$ is canonically isomorphic to $V$; and then a Smith normal form computation can give you the decomposition of $V$. However, that calculation is not necessary for this proof. – Daniel Schepler Oct 20 '22 at 19:27
  • My favourite proof of the existence of JNF is to use the primary decomposition theory (which is weaker than the structure theorem for modules over a PID, and implied by it) to reduce to the nilpotent case. For the nilpotent case I typed up one short proof here: http://www.ma.rhul.ac.uk/~uvah099/Maths/JNFfinal.pdf – Mark Wildon Oct 21 '22 at 05:09
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    @V.Ch. I once tried to understand this better myself in response to another question here and wrote this answer: https://math.stackexchange.com/a/4156949/11853 – Eike Schulte Oct 21 '22 at 05:52
  • Great answer. Although, it's helpful to point out the polynomial ring $k[t]$ and quotient rings $k[t]/\langle (t-\lambda_i)^{d_i}\rangle$ are themselves vector spaces over $k$. – user760 Feb 02 '23 at 16:18
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There is a proof for the basic fact that there are infinitely many prime numbers, using Euler products. Suppose that there are only finitely many primes. Then $$\frac{\pi^2}{6} = \zeta(2)=\prod_{p\in \Bbb P} \frac{1}{1-\frac{1}{p^2}}$$ is rational. Hence also $\pi^2$ is rational, which is a contradiction.

Dietrich Burde
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  • I’ve never been sure how valid this is, but you could also evaluate the Euler product at one. If there were finitely many primes then the harmonic series would converge, because it would “imply” that $\zeta(1)$ is finite, which is false – FShrike Oct 20 '22 at 18:51
  • @FShrike The Euler product only converges for $Re(s)>1$. – Dietrich Burde Oct 20 '22 at 18:51
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    Right. Because there are infinitely many primes… – FShrike Oct 20 '22 at 18:52
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    The argument for $\zeta(1)$, I think, should also prove that $\sum_{p\in \mathbb{P}} \frac{1}{p}$ diverges (and then the infinitude of primes is an easy consequence of that). – Daniel Schepler Oct 20 '22 at 19:18
  • I assume $\pi^2$ is irrational because $\pi$ is transcendental? – Hilbert Jr. Oct 20 '22 at 19:21
  • Yes, $\pi^2$ is transcendental, because $\pi$ is, see here. – Dietrich Burde Oct 20 '22 at 19:25
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    Is this a circular proof? Can we show that $\Pi_{p}\frac{1}{1-\frac{1}{p^2}} = \frac{\pi^2}{6}$, if we do not already know that there are an infinite amount of primes? – Hagamena Oct 20 '22 at 20:14
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    @Hagamena: yes. Expanding by a geometric series, the left-hand side is $\prod_p (1+1/p^2 + 1/p^4 + ...)$. By unique factorization this is $\sum_n 1/n^2$. Almost any proof I can think of that $\sum_n 1/n^2 = \pi^2/6$ doesn't use that there are infinitely many primes. – Mark Wildon Oct 21 '22 at 05:08
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To show that matrix multiplication is associative you can compute it directly or instead we can consider the matrices to be functions that take other matrices as input. This allows us to use the associativity of function composition so you only have to establish that the multiplication is well-defined.

CyclotomicField
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    Another way to reduce to associativity of function composition would be to interpret the matrices as linear transformations with respect to a chosen basis. – Mark Wildon Oct 21 '22 at 05:14
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Here is a proof for the basic fact that $\sqrt{2}$ is irrational. We use Fermat's Theorem:

Theorem (Fermat, 1640): The number $1$ is not congruent, i.e., there is no right triangle with rational sides whose area is equal to $1$.

Proof: If $\sqrt{2}$ were rational then $\sqrt{2},\sqrt{2}$,and $2$ would be the sides of a rational right triangle with area $1$. This is a contradiction to the fact that $1$ is not a congruent number.

Dietrich Burde
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  • I think , the Answer (as well as the Question) is circular , because we are using some (Higher) theorem Y to Prove some Statement X , where X was actually used (Directly or otherwise) to Prove Y ! – Prem Oct 20 '22 at 18:36
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    @Prem Why do you think that Fermat's proof used that $\sqrt{2}$ is irrational? He proved, that $x^4+y^4=z^2$ has no nontrivial integer solution by descent. This does not involve the irrationality of $\sqrt{2}$. – Dietrich Burde Oct 20 '22 at 18:42
  • Well , that was why I used "Directly or otherwise" , because if we go back early enough , this Proof by Descent will use "No Solution to $x^2=2y^2$" – Prem Oct 20 '22 at 18:50
  • Can you show us this in Fermat's proof? In the proof here I don't see it. – Dietrich Burde Oct 20 '22 at 18:51
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    The proof that $1$ is not congruent is given after Theorem $2.1$ here. I don't see that it is used that $\sqrt{2}$ is irrational. – Dietrich Burde Oct 20 '22 at 18:59
  • In the middle of Page 3 of the Document you gave , I see the question [ (why?) ] , & before that there is this claim : "$g^2|d^2$ hence $g|d$" , which I interpret like : "Why? Because (InDirectly) $g$ can not Divide the $2$ and get away with it , because no Square can be 2". – Prem Oct 20 '22 at 19:47
  • I had a typo in that comment : Correction : In the middle of Page 3 of the Document you gave , I see the question [ (why?) ] , & before that there is this claim : "$g^2|2d^2$ hence $g|d$" , which I interpret like : "Why? Because (InDirectly) g can not Divide the 2 and get away with it , because no Square can be 2". – Prem Oct 20 '22 at 19:58
  • This is also trivial, if $h\mid ab$ with coprime $a,b$ then either $h\mid 2$ or $h\mid b$ in integers. So we only have that $g^2=2$ in the integers has no solution. This is trivial. Take $g=1$, then $g=2$. – Dietrich Burde Oct 21 '22 at 08:04
  • (A) Using the same argument , we can show that $x^2=2y^2$ will make $(x,y)$ have some common factor , but here we are not explicitly stating that conclusion that there is no Integer Solution , rather we are moving on to other conclusions. Either Way , that conclusion , stated or unstated , is hiding there (B) I assume , you mean $(h,a,b)=(g^2,2,d^2)$ where $(a,b)$ are coprime. In that case , it may be that $d/g=\sqrt{2}$ hence $2d^2/g^2=2\sqrt{2}\sqrt{2}=2.2=4$ & Divisibility is satisfied. We know that is not Possible (that is the WHY which Paper asks) (C) This is my opinion , I may be wrong. – Prem Oct 21 '22 at 08:47
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Given any field, the fact that the set of diagonalizable matrices are dense over the field in the Zariski topology is a fascinating one line proof.

One notes that the set of diagonalizable matrices contain the matrices with distinct eigenvalues which is equal to the set of all those matrices such that the discriminant of the characteristic polynomial of the matrix does not vanish. Since the discriminant is a polynomial in the matrix entries, this is the complement of a zero-set of some polynomials and hence is closed in the Zariski topology, and hence is dense. Thus the set of diagonalizable matrices is Zariski dense over the given field.

This is very helpful to reduce many questions of linear algebra over any field to just the diagonalizable cases, because then by the Zariski topology, it would hold for all cases!

HackR
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  • Don't you need the field to be infinite for the "open => dense" proposition to hold? E.g. take any finite field with $q$ elements, then the zero set $V(x)$ is both closed and open (as the complement of the zero set $V(x^{q-1}-1)$), so it is its own closure and not dense. What you said definitely holds for infinite fields, though. – Hilbert Jr. Jun 20 '23 at 12:04
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Here is a proof(paraphrased) I found of the division algorithm for polynomials in my Linear Algebra textbook that I thought was really nice.

Proposition: Let $p,s\in P(F)$ with $s\ne 0$. Then there exist unique polynomials $q,r\in P(F)$ such that $p=sq+r$ and $\deg r<\deg s$.

Proof: Let $\deg p =n$ and $\deg s=m$. If $n<m$, take $q=0$ and $r=p$. So we can assume that $n \ge m$. Define a linear map $T : P_{n-m}(F)\times P_{m-1}(F)\to P_n(F)$ by $$T(q, r)=sq+r$$

To show uniqueness it suffices to show that null $T = \{0\}$. If $(q, r)\in$ null $T$, then $sq+r=0$ implies that $q=0$ and $r=0$ because otherwise if $q\ne 0$, then $\deg sq \ge m$ and $sq\ne -r$. This shows that null $T=\{0\}$ and $T$ is injective and therefore $q$ and $r$ are unique.

To show existence, by the fundamental theorem of linear maps we know that $$\dim \text{range } T=\dim V -\dim \text{null }T $$ Because $\dim \text{null } T=0$, we have that $$\dim \text{range }T = \dim(P_{n-m}(F)\times P_{m-1}(F))=(n-m+1)+(m-1+1)=n+1$$ Which equals $\dim P_n(F)$. Thus, range $T=P_n(F)$, $T$ is surjective and such $q$ and $r$ always exist.

Seeker
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  • That's actually really neat! – Hilbert Jr. Oct 21 '22 at 00:13
  • Since we don’t assign degree to zero polynomial, we have to be careful and consider cases. For instance: $(1)$ if $p=0$ or $n\lt m$, take $q=0$ and $r=p$. $(2)$ if $\deg (s)=m=0$, then $P_{m-1}(F)$ don’t make any sense. If $m=0$, then $s$ is scalar polynomial, let say $c$. Take $q=c^{-1}p$ and $r=0$. Summary: there are $3$ cases, $(1)$ $p=0$ or $n\lt m$, $(2)$ $n\geq m=0$, $(3)$ $n\geq m\geq 1$. Note cases $(2)$ & $(3)$ implicitly assume $p\neq 0$. – user264745 Dec 04 '22 at 14:13
  • Following is my proof of $sq+r=0$$\implies$$ q=0$ and $r=0$. ATC, $q\neq 0$. Case$(1)$: $r=0$. Then $sq=0$. Since $s\neq 0$, we have $q=0$. Which contradicts our initial assumption of $q\neq 0$. Case$(2)$: $r\neq 0$. Then $\deg (r)=\deg (-r)$. Since $sq=-r$, we have $\deg (sq)=\deg (s)+\deg (q)=\deg (-r)\leq m-1$. So $m+\deg (q)\leq m-1$. Which implies $\deg (q)\leq -1$. Thus we reach contradiction. Hence $q=0$ and $r=0$. – user264745 Dec 04 '22 at 14:20