The Unapologetic Mathematician

The Trace of a Linear Transformation

Again, let’s take a linear endomorphism $T$ on a vector space $V$ of finite dimension $d$. We know that its characteristic polynomial can be defined without reference to a basis of $V$, and so each of the coefficients of $V$ is independent of any choice of basis. The leading coefficient is always ${1}$, so that’s not very interesting. The constant term is the determinant, which we’d known from other considerations before. There’s one more coefficient we’re interested in, partly for the interesting properties we’ll explore, and partly for its ease of computation. This is the coefficient of $\lambda^{d-1}$.

So, let’s go back to our formula for the characteristic polynomial:

$\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^d(\lambda\delta_k^{\pi(k)}-t_k^{\pi(k)})$

Which terms can involve $\lambda^{d-1}$. Well, we can get one factor of $\lambda$ every time we have $k=\pi(k)$, and so we need this to happen at least $d-1$ times to get $d-1$ factors of $\lambda$. But if the permutation $\pi$ sends all but one index back to itself, then the last index must also be fixed, since there’s nowhere else for it to go! So we only have to look at the term corresponding to the identity permutation. This will simplify our lives immensely.

Now we’re considering the product

$\displaystyle\prod\limits_{k=1}^d(\lambda-t_k^k)$

When we multiply this out, we make $d$ choices. At each step we can either take the $\lambda$, or we can take the $t_k^k$. We’re interested in the terms where we take the $\lambda$ $d-1$ times. There are $d$ ways of making this choice, corresponding to which one of the indices we don’t take the $\lambda$. Incidentally, we could also think of this in terms of combinations, as $d=\binom{d}{d-1}$.

Anyhow, for each choice of one index to use the matrix entry instead of the variable, we’ll have a term $-t_k^k\lambda^{d-1}$. We add all of these up, summing over $k$ — as our notation suggests we should! And now we have the second coefficient of the characteristic polynomial. We drop the negative sign and call this the “trace” of $T$:

$\displaystyle\mathrm{Tr}(T)=\sum\limits_{k=1}^dt_k^k=t_k^k$

where in the last formula we’re using the summation convention again. Incidentally, “trace” should be read as referring to a telltale sign that $T$ has left behind, like a hunted animal’s.. um.. “leavings”.

Anyhow, we can now write out a few of the terms in the characteristic polynomial:

$\det(\lambda1_V-T)=\lambda^d-\mathrm{Tr}(T)\lambda^{d-1}+...+(-1)^d\det(T)$

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Posted by John Armstrong | Algebra, Linear Algebra

5 Comments	»

1. [...] we can define to be the trace of this matrix, and to be its determinant. If , we’ve got an eigenpair. Possibly related [...]

Pingback by | March 13, 2009 | Reply

2. [...] is the trace of an endomorphism. Given a matrix, it’s the sum of the diagonal entries. Since the [...]

Pingback by | October 13, 2010 | Reply

3. [...] is, the character is “the trace of the representation”. But why this is interesting is almost completely opaque at this [...]

Pingback by | October 14, 2010 | Reply

4. [...] with dimension and start with . Inside this, we consider the subalgebra of endomorphisms whose trace is zero, which we write and call the “special linear Lie algebra”. This is a subspace, [...]

Pingback by | August 8, 2012 | Reply

5. [...] for all in the cases of and the special linear Lie algebra — the latter because the trace is invariant under a change of [...]

Pingback by | August 18, 2012 | Reply

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