Trace is the derivative of determinant

Posted on June 5, 2020 by rohanjoshi

A question I always had when learning linear algebra is, “what does the trace of a matrix mean conceptually?” For example, the determinant of a matrix is, roughly speaking, the factor by which the matrix expands the volume. The conceptual meaning of trace is not as straightforward, but one way to think about it is

trace is the derivative of determinant at the identity.

Roughly you can think of this in the following way. If you start at the identity matrix and move a tiny step in the direction of $M$ , say $M\epsilon$ where $\epsilon$ is a tiny number, then the determinant changes approximately by $\text{tr}(M)$ times $\epsilon$ . In other words, $\det(1 + M\epsilon) \approx 1 + \text{tr}(M)\epsilon$ . Here $1$ stands for the identity matrix.

One can be very precise about what it means to take the “derivative” of the determinant, so let me do some setup. Let $K$ be either $\mathbb{R}$ or $\mathbb{C}$ (so we are working with real or complex Lie groups; but of course, everything makes sense for algebraic groups over arbitrary fields). Then there is a morphism of Lie groups called the determinant $\det: \text{GL}_n(K) \to K^\cross$ , given by sending a matrix to its determinant. Since we are restricting to invertible matrices, the determinants are nonzero. To check that this is really a morphism of Lie groups (i.e. both a smooth map and a homomorphism of groups), note that the determinant map is a polynomial map in the entries of the matrix (and therefore smooth) and is a group homomorphism by the property that $\det(AB)=\det(A)\det(B)$ .

Now, given any smooth map of manifolds $f$ which maps point $p \mapsto f(p)$ , there is an induced linear map on from the tangent space of $p$ to the tangent space of $f(p)$ called the derivative of $f$ at $p$ . In particular, if $f$ is a Lie group homomorphism, then it maps the identity point to the identity point, and the derivative at the identity is furthermore a homomorphism of Lie algebras. What this means is that, in addition to being a linear map, it preserves the bracket pairing.

In the case of $\text{GL}_n$ , the Lie algebra at the identity matrix is called $\mathfrak{gl}_n$ . We can think of it as consisting of all $n \cross n$ matrices, and the bracket operation is defined by $[A, B] = AB - BA$ . The Lie algebra of $K^\cross$ at $1$ consists of the elements of $K$ ; since $K^\cross$ is abelian, the bracket is trivial.

The main claim, which I will prove subsequently, is that this map $\mathfrak{gl}_n(K) \to K$ , the derivative of the determinant at the identity, is actually the trace. That is, it sends a matrix to its trace, the sum of the entries on the diagonal. Note that since it is a homomorphism of Lie algebras, it preserves the bracket, and we recover the familiar property of trace $\text{tr}(AB - BA) = 0$ , so $\text{tr}(AB)=\text{tr}(BA)$ .

We can find the derivative of a smooth map on $\text{GL}_n(K)$ directly, since it is an open subset of a vector space. Let $\phi$ be a matrix; then the derivative at the identity evaluated at $\phi$ is

$\lim_{t \to 0} \frac{\det(1+t\phi) - 1}{t}.$

$\det(1+t\phi)$ is a polynomial in $t$ , and the number we’re looking for is the coefficient of the $t$ term.

We have

$\det(1 + t\phi) (e_1 \wedge \cdots \wedge e_n) = (e_1+\phi(e_1)t)\wedge(e_2+\phi(e_2)t)\wedge \cdots \wedge (e_n+\phi(e_n)t).$

Just to get a concrete idea of what this expands to, let’s look when $n=2$ . Then

$(e_1+\phi(e_1)t)\wedge(e_2+\phi(e_2)t)=e_1\wedge e_2 + (\phi(e_1)\wedge e_2 + e_1 \wedge \phi(e_2))t + (\phi(e_1)\wedge \phi(e_2)) t^2.$

When $n=3$ ,

$(e_1+\phi(e_1)t)\wedge(e_2+\phi(e_2)t)\wedge(e_3+\phi(e_3)t)$

$=e_1\wedge e_2\wedge e_3$

$+ (\phi(e_1)\wedge e_2 \wedge e_3 + e_1 \wedge \phi(e_2) \wedge e_3 + e_1 \wedge e_2 \wedge \phi(e_3))t$

$+ (\phi(e_1)\wedge \phi(e_2) \wedge e_3 + \phi(e_1)\wedge e_2) \wedge \phi(e_3) + e_1\wedge \phi(e_2) \wedge \phi(e_3)) t^2$

$+ (\phi(e_1)\wedge \phi(e_2) \wedge \phi(e_3)) t^3.$

In particular, the coefficient of $t$ is $\text{tr}(\phi)$ . (In fact, see if you can convince yourself that the coefficient of $t^i$ is $\text{tr}(\wedge^i \phi)$ .)

See some discussion of the meaning of trace.

Acknowledgements: Thanks to Ben Wormleighton for originally telling me the slogan “trace is the derivative of determinant”, and for teaching me about Lie groups and Lie algebras.

To add: discussion of Jacobi’s formula, exponential map

Determinant of transpose

Posted on November 8, 2019 by rohanjoshi

An important fact in linear algebra is that, given a matrix $A$ , $\det A = \det {}^tA$ , where ${}^tA$ is the transpose of $A$ . Here I will prove this statement via explciit computation, and I will try to do this as cleanly as possible. We may define the determinant of $A$ by

$\det A = \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} A_{\sigma(1), 1} \cdots A_{\sigma(n), n}.$

Here $S_n$ is the set of permutations of the set $\{ 1, \dots n \}$ , and $sgn(\sigma)$ is the sign of the permutation $\sigma$ . This formula is derived from the definition of the determinant via exterior algebra. One can check by hand that this gives the familiar expressions for the determinant when $n = 2, 3$ .

Now, since $({}^tA)_{i, j} = A_{j, i}$ , we have

$\det {}^tA = \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} A_{1, \sigma(1)} \cdots A_{n, \sigma(n)}$

$= \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)} A_{\sigma^{-1}(\sigma(1)), \sigma(1)} \cdots A_{\sigma^{-1}(\sigma(n)), \sigma(n)}.$

The crucial observation here is that we may rearrange the product inside the summation so that the second indices are increasing. Let $b = \sigma(a)$ . Then the product inside the summation is

$\prod_{1 \leq a \leq n} A_{\sigma^{-1}(\sigma(a)), \sigma(a)} = \prod_{1 \leq b \leq n} A_{\sigma^{-1}(b), b}$

Combining this with the fact that $sgn(\sigma) = sgn(\sigma^{-1})$ , our expression simplifies to

$\sum_{\sigma \in S_n} (-1)^{sgn(\sigma^{-1})} A_{\sigma^{-1}(1), 1} \cdots A_{\sigma^{-1}(n), n}.$

Noticing that the sum is the same sum if we replace all $\sigma^{-1}$ s with $\sigma$ s, we see that this equals $\det A$ . So $\det A = \det {}^tA$ . $\square$

I wonder if there is a more conceptual proof of this? (By “conceptual”, I mean a proof based on exterior algebra, bilinear pairings, etc…)

Two proofs complex matrices have eigenvalues

Posted on July 16, 2019 by rohanjoshi

Today I will briefly discuss two proofs that every matrix $T$ over the complex numbers (or more generally, over an algebraically closed field) has an eigenvalue. Notice that this is equivalent to finding a complex number $\lambda$ such that $T - \lambda I$ has nontrivial kernel. The first proof uses facts about “linear dependence” and the second uses determinants and the characteristic polynomial. The first proof is drawn from Axler’s textbook [1]; the second is the standard proof.

Proof by linear dependence

Let $p(x) = a_nx^n + \dots a_1x + a_0$ be a polynomial with complex coefficients. If $T$ is a linear map, $p(T) = a_nT^n + \dots a_n T + a_0I$ . We think of this as “ $p$ evaluated at $T$ ”.

Exercise: Show $(pq)(T) = p(T)q(T)$ .

Proof: Pick a random vector $v \in V$ . Consider the sequence of vectors $v, Tv, T^2v, \dots T^nv.$ This is a set of $n+1$ vectors, so they must be linearly dependent. Thus there exist constants $a_0, \dots a_n \in \C$ such that $a_nT^nv + a_{n-1}T^{n-1}v + \dots a_1Tv + a_0v$ $= (a_nT^n + a_{n-1}T^{n-1} + \dots a_1T + a_0I)v = 0$ .

Define $p(x) = a_nx^n + \dots a_1x + a_0$ . Then, we can factor $p(x) = a_n(x-\lambda_1)\dots (x-\lambda_n).$ By the Exercise, this implies $a_n(T - \lambda_1I)\dots (T-\lambda_nI)v = 0$ . So, at least one of the maps $T - \lambda_iI$ has a nontrivial kernel, so $T$ has an eigenvalue. $\square$

Proof by the characteristic polynomial

Proof: We want to show that there exists some $\lambda$ such that $T - \lambda I$ has nontrivial kernel: in other words, that $T - \lambda I$ is singular. A matrix is singular if and only if its determinant is nonzero. So, let $\chi_T(x) = \det(T - xI)$ ; this is a polynomial in $x$ , called the characteristic polynomial of $T$ . Now, every polynomial has a complex root, say $\lambda$ . This implies $T - \lambda I$ , so $T$ has an eigenvalue. $\square$

Thoughts

To me, it seems like the determinant based proof is more straightforward, although it requires more machinery. Also, the determinant based proof is “constructive”, in that we can actually find all the eigenvalues by factoring the characteristic polynomial. On subject of determinant-based vs determinant-free approaches to linear algebra, see Axler’s article “Down With Determinants!” [3].

There is a similar situation for the problem of showing that the sum (or product) of two algebraic numbers is algebraic. Here there is a non-constructive proof using “linear dependence” (which I attempted to describe in a previous post) and a constructive proof using the characteristic polynomial (which will hopefully be the subject of a future blog post). A further advantage of the determinant-based proof is that it can be used more generally to show that the sum and product of integral elements over a ring are integral. In this more general context, we no longer have linear dependence available.

References

Sheldon Axler, Linear algebra done right. Springer 2017
Evan Chen, An Infinitely Large Napkin, available online
Sheldon Axler. Down with Determinants! The American Mathematical Monthly, 102(2), 139, 1995. doi:10.2307/2975348, available online

Arithmetic variety

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