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The determinant is a homomorphism from the multiplicative monoid of matrices to the multiplicative monoid of a field (right?). I find this to be the most intuitive way to interpret some of the determinant's properties (notably the invertibility condition: obviously a matrix is only invertible if a homomorphism maps it to an invertible element of the field).

So from an algebraic point of view, can anything interesting be said about the determinant?

Some fairly specific technical questions:

  1. How does it fit into the big picture of all homomorphisms $M_n(\Bbb K)\to\Bbb K^\star$? Is it the only one? How are the others related to it? The invertibility condition is one of the most useful things about the determinant, but any homomorphism would give the same condition, so why use the determinant?
  2. Are there any ring homomorphisms $M_n(\Bbb K)\to\Bbb K^\star$? How does the determinant relate to them?

Some vaguer, softer questions:

  1. Are there any intuitive proofs for the formulae for calculating determinants based on the fact that it's a homomorphism?
  2. This one's pretty out there. The characteristic polynomial is a polynomial with coefficients in a field which is the homomorphic image of a polynomial with coefficients in the ring of matrices. Is there any way of explaining the various relationships between a matrix and its characteristic polynomial based on the structure-preserving properties of the determinant?
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    $\begingroup$ math.stackexchange.com/a/21617/1650 $\endgroup$ Commented Nov 2, 2013 at 23:00
  • $\begingroup$ @MartinBrandenburg Good god that's abstract. A lot (all) of that answer is beyond me, but thanks for the link, I'll try and read it some time. $\endgroup$
    – Jack M
    Commented Nov 2, 2013 at 23:03
  • $\begingroup$ @JackM What's interesting is that the determinant's existence is predicted by algebraic geometry. Namely, "elimination of quantifiers", which in algebraic geometry language says that the structure map $\mathbb{P}^n_A\to \text{Spec }A$ is closed (where $A=k[a,b,c,d,e,f,g,h,i]$ for example), tells us that there should be some polynomial $D(a,b,c,d,e,f,g,h,i)$ such that the system $\{ax+by+cz,dx+ey+fz,gx+hy+iz\}$ has a non-trivial solution if and only if $D(a,b,c,d,e,f,g,h,i)= 0$. Of course, this polynomial is just $\det\begin{pmatrix}a & b & c\\ d & e & f\\ g & h & i\end{pmatrix}$. $\endgroup$ Commented Nov 2, 2013 at 23:17

2 Answers 2

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Let's consider $M_n(\mathbb K)$ and $\mathbb K$ as monoids under multiplication. Here are fairly complete answers to your technical questions:

  1. Every monoid homomorphism $f$ from $M_n(\mathbb{K})$ to $\mathbb{K}$ is of the form $f=\phi \circ \text{det}$ for some monoid endomorphism $\phi:\mathbb{K}\to\mathbb{K}$. In other words, the determinant has the universal property that every monoid homomorphism from $M_n(\mathbb{K})$ to $\mathbb{K}$ factors through it. To make things more concrete, take, for example, $\mathbb{K}=\mathbb{R}$: if we restrict to continuous homomorphisms, the result implies that the only such homomorphisms are $f=(\text{sgn}(\text{det} A))^{\varepsilon}\cdot|\text{det} A|^r$ for $\varepsilon\in\{0,1\}$ and $r\in\mathbb{R}$.

Here is how we can prove this: Let $f : M_n(\mathbb{K}) \to \mathbb{K}$ be a monoid homomorphism. First, we claim that either $f(A)=1$ for all $A$, or $f(A)=0$ for all non-invertible $A$: We have $f(0)=f(00)=f(0)^2$, so either $f(0)=0$ or $f(0)=1$. If $f(0)=1$, then for all $A\in M_n(\mathbb{K})$, $f(A)=f(A)f(0)=f(A0)=f(0)=1$. So assume $f(0)=0$. Let $J$ be a nilpotent matrix of rank $n-1$, e.g., take $J$ to be a single Jordan block with zeros on the diagonal. Then $J^n=0$, so $f(J)^n=f(J^n)=f(0)=0$, hence $f(J)=0$. Now, any matrix $A$ of rank $n-1$ may be written $A=SJT$ for suitable matrices $S$ and $T$, and it follows that $f(A)=f(S)f(J)f(T)=0$. Finally, since any non-invertible matrix $A$ may be written as a product of several rank $n-1$ matrices, we obtain $f(A)=0$ for all non-invertible $A$, which proves the claim.

Thus, if we ignore the trivial case $f=1$ and restrict attention to the case where $f(A)=0$ for all non-invertible $A$, then $f$ is completely determined by its restriction to $GL_n(\mathbb{K})$, which is a group homorphism into the multiplicative group $K^*$, whose kernel is a normal subgroup $N \unlhd GL_n(\mathbb{K})$. If we exclude the exceptional situation where $n=2$ and $\mathbb{K}=\mathbb{F}_2$ or $\mathbb{F}_3$, then either $N$ is a subgroup of scalar matrices or $N$ contains $SL_n(\mathbb{K})$. In the former case, $GL_n(\mathbb{K})/N$ has the group $PGL_n(\mathbb{K})$ as a homomorphic image, which is non-abelian (for $n\geq 2$), in contradiction to the fact that the First Isomorphism Theorem applied to $f$ provides an embedding of $GL_n(\mathbb{K})/N$ into the abelian group $\mathbb{K}^*$. Therefore, $N$ must contain $SL_n(\mathbb{K})$, which is the kernel of the determinant homomorphism $\text{det} : GL_n(\mathbb{K}) \to \mathbb{K}^*$. It follows that $f$ factors through the determinant homomorphism, i.e., for $A\in GL_n(\mathbb{K})$ we have $f(A) = \phi(\text{det} A)$ for some group endomorphism $\phi$ of $\mathbb{K}^*$. If we extend $\phi$ to a monoid endomorphism on $\mathbb{K}$ by setting $\phi(0)=0$, then the equation $f(A)=\phi(\text{det} A)$ also holds for non-invertible matrices $A$ since in this case both $f(A)=0$ and $\phi(\text{det}(A))=\phi(0)=0$. Thus $f = \phi \circ \text{det}$.

In the exceptional situation that $n=2$ and $\mathbb K=\mathbb F_2$ or $\mathbb F_3$, it is straightforward to check that although there are some additional normal subgroups of $GL_2(\mathbb K)$, none of them give rise to any new group homomorphisms from $GL_2(\mathbb K)$ into $\mathbb K^*$.

  1. There are no ring homomorphisms $M_n(\mathbb K) \to \mathbb K$ except for $n=1$, in which case the ring homomorphisms are just the field automorphisms of $\mathbb K$.

This is just a consequence of the fact that $M_n(\mathbb K)$ is a simple ring.

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  • $\begingroup$ Nice answer! Very minor correction at the end: the homomorphisms are the field endomorphisms of $K$, not necessarily invertible. $\endgroup$ Commented Jun 26 at 18:40
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Q1:

The invertibility condition is one of the most useful things about the determinant, but any homomorphism would give the same condition, so why use the determinant?

There is a lot more to say about this! Brent's nice answer already explains that the determinant has a certain universal property, but the property Brent describes is also satisfied by $\varphi \circ \det$ where $\varphi : K \to K$ is any automorphism. For example over $\mathbb{C}$ it is also satisfied by $\overline{\det}$. So it still does not pick out the determinant itself uniquely.

Here are some further observations about the determinant which do lead to conditions that pick it out uniquely.

  1. The determinant is a polynomial in the entries of a matrix with integer coefficients, and this polynomial is independent of the field $K$.
  2. #1 implies that the determinant not only makes sense over every field $K$, but even over every commutative ring $R$, because a polynomial with integer coefficients can be evaluated over every commutative ring. Furthermore, it has the same property as the usual determinant over a field does, although it has to be stated a bit differently: a matrix $X \in M_n(R)$ is invertible iff $\det X \in R$ is invertible. For example if $R = \mathbb{Z}$ the condition is that a matrix in $M_n(\mathbb{Z})$ is invertible (in the sense that its inverse is another matrix in $M_n(\mathbb{Z})$) iff its determinant is $\pm 1$.
  3. #1 also implies that the determinants over different rings are related by the following condition, which loosely speaking says that they are all "compatible": if $f : R \to S$ is a ring homomorphism, then applying it entrywise to a matrix $X \in M_n(R)$ produces a matrix $f(X) \in M_n(S)$, and the determinants of these matrices are related by $f(\det X) = \det(f(X))$. For example, if $X$ is a matrix with real coefficients, and you interpret it as a matrix with complex coefficients (formally, applying the inclusion $f : \mathbb{R} \to \mathbb{C}$), then the determinant doesn't change.

Maybe surprisingly, condition 1 turns out to be equivalent to the combination of condition 2 and condition 3! That is, requiring that the determinant be defined in a compatible way over every commutative ring $R$ is equivalent to requiring that it be defined by a single polynomial with integer coefficients. This is a consequence of a fundamental result in category theory called the Yoneda lemma.

Philosophically, you can say that what this means is that it's unnecessary to think of the determinant as this infinite collection of functions $\det_K : M_n(K) \to K$ indexed by all fields $K$ (or all commutative rings $R$). In fact the determinant is a single object $\det \in \mathbb{Z}[x_{ij}]$, namely this single polynomial (given by the Leibniz formula) which computes it over all commutative rings.

Concretely, this condition suggests that we try to single out the determinant among all the polynomial homomorphisms which give compatible families of maps $M_n(R) \to R$. This is a much more restrictive condition than just requiring that it be multiplicative and rules out the weird examples like $\overline{\det}$ and $|\det|^r$. Now we have the following uniqueness result.

Theorem: Every polynomial with integer coefficients defining a compatible family of multiplicative maps $M_n(R) \to R$ for all commutative rings $R$ must be a non-negative integer power $\det^n, n \ge 0$ of the determinant. So the monoid of all such homomorphisms is isomorphic to $\mathbb{N}$, and $\det$ is the unique generator of this monoid.

Also it suffices to replace "all commutative rings" with "all fields" above. For a proof see here. Note that this theorem depends on standard facts about the determinant and does not give an independent construction of it. However I think it does clarify a sense in which it is "inevitable" that we study the determinant.


Q4:

This one's pretty out there. The characteristic polynomial is a polynomial with coefficients in a field which is the homomorphic image of a polynomial with coefficients in the ring of matrices. Is there any way of explaining the various relationships between a matrix and its characteristic polynomial based on the structure-preserving properties of the determinant?

What we've learned from the above discussion is that it's very significant that the determinant is a polynomial with integer coefficients in the entries of a matrix, because this is equivalent to saying that it can be defined in a compatible way for matrices over all commutative rings $R$. Since the characteristic polynomial is defined as a certain determinant, it also has this property: that is, all of its coefficients are also polynomials with integer coefficients in the entries of a matrix, and they can also be defined in a compatible way over all $R$!

This means the coefficients of the characteristic polynomial are some special polynomials in the entries of a matrix. What is it that makes them special exactly? Well, the determinant is conjugation invariant in the sense that for every commutative ring $R$, every $X \in M_n(R)$, and every invertible $P \in GL_n(R)$, it satisfies $\det(PXP^{-1}) = \det(X)$. This follows from multiplicativity (although it is weaker) and says that the determinant is coordinate invariant; that is, although we've been talking about determinants of matrices here, the determinant actually does not depend on a choice of basis and makes sense for abstract linear maps. The coefficients of the characteristic polynomial, such as the trace $\text{tr}(X)$, inherit this feature from the determinant: they are also conjugation invariant. So, this means they are all capturing some feature of a matrix which is invariant under change of coordinates (and we know what that feature is, namely it is a symmetric polynomial in the eigenvalues).

Theorem: Every polynomial $f \in \mathbb{Z}[x_{ij}]$ with integer coefficients in the entries of a matrix which is conjugation invariant in the above sense is a polynomial in the coefficients of the characteristic polynomial.

So the characteristic polynomial captures, loosely speaking, "all polynomial conjugation invariant information" about a matrix (take care to note that the two uses of the word "polynomial" here are different!). Since the data of the characteristic polynomial is the same as the data of its eigenvalues, more or less, this is saying the eigenvalues are the only "polynomial conjugation invariant information" about a matrix. So polynomials cannot detect the refined Jordan block information of the full Jordan normal form.

Sketch. The same Zariski density argument as in the proof of the previous result applies here and shows that $f$ must be a polynomial in the eigenvalues. Since the eigenvalues can be freely permuted, $f$ must be a symmetric polynomial in the eigenvalues. Now the result follows from the fundamental theorem of symmetric polynomials. $\Box$

The same Zariski density argument can also be used to prove Cayley-Hamilton, although arguably it is more complicated than necessary. In any case, in all of these arguments it is crucial that the determinant and all the other coefficients of the characteristic polynomial are themselves polynomials.

Edit: Here is a different fact about the characteristic polynomial which uses the fact that the determinant is multiplicative, together with a cute trick involving the fact that its coefficients are polynomials again. Conjugation invariance implies that if $X, Y$ are two square matrices such that either $X$ or $Y$ is invertible then $XY, YX$ are conjugate so they have the same characteristic polynomial. However it is in fact true that they have the same characteristic polynomial in general, with no invertibility hypotheses (even though they can fail to be conjugate in this case). This follows from a Zariski density argument as above but there's another way to do it:

Let $X = [x_{ij}], Y = [y_{ij}]$ be two matrices whose entries live in a polynomial algebra $\mathbb{Z}[x_{ij}, y_{ij}]$; this is the "universal pair of matrices." Then

$$\begin{align*} \det Y \det (t - XY) &= \det(tY - YXY) \\ &= \det(t - YX) \det Y. \end{align*}$$

Now the fun observation is that, because the polynomial algebra is an integral domain, we can cancel $\det Y$ from both sides of this equation, and we conclude that $\det (t - XY) = \det(t - YX)$. This means that the two matrices having the same characteristic polynomial is a polynomial identity and so holds everywhere, for any two matrices over any commutative ring, despite the fact that we cannot directly apply this proof to a specific pair of matrices if $Y$ does not specialize to an invertible matrix. This is actually equivalent to the Zariski density argument but phrased in a way that doesn't require that concept.

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