# The Unapologetic Mathematician

## Extracting the Determinant from the Characteristic Polynomial

This one’s a pretty easy entry. If we know the characteristic polynomial of an endomorphism $T$ on a vector space $V$ of finite dimension $d$, then we can get its determinant from the constant term.

First let’s look at the formula for the characteristic polynomial in terms of the matrix entries of $T$: $\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^d(\lambda\delta_k^{\pi(k)}-t_k^{\pi(k)})$

Now we’re interested in $\det(T)$, which is exactly what we calculate to determine if the kernel of $T$ is nontrivial. But the kernel of $T$ is the eigenspace corresponding to eigenvalue ${0}$, so this should have something to do with the characteristic polynomial at $\lambda=0$. So let’s see what happens. $\displaystyle\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^d(-t_k^{\pi(k)})=\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)(-1)^d\prod\limits_{k=1}^dt_k^{\pi(k)}=(-1)^d\sum\limits_{\pi\in S_d}\mathrm{sgn}(\pi)\prod\limits_{k=1}^dt_k^{\pi(k)}$

This is just $(-1)^d$ times our formula for the determinant of $T$. But of course we know the dimension ahead of time, so we know whether to flip the sign or not. So just take the characteristic polynomial, evaluate it at zero, and flip the sign if necessary to get the determinant.

There’s one thing to note here, even though it doesn’t really tell us anything new. We’ve said that $T$ is noninvertible if and only if its determinant is zero. Now we know that this will happen if and only if the constant term of the characteristic polynomial is zero. In this case, the polynomial must have a root at $\lambda=0$, which means that the ${0}$-eigenspace of $T$ is nontrivial. But this is just the kernel of $T$ is nontrivial. Thus (as we already know) a linear transformation is noninvertible if and only if its kernel is nontrivial.

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January 29, 2009 Posted by | Algebra, Linear Algebra | 1 Comment