To prove that the sum of the roots of the characteristic polynomial of a square matrix is equal to the trace of the matrix

Question

How do we prove that the sum of the roots of the characteristic polynomial of a square matrix is equal to the trace of the matrix ? I want a proof which does not use much computation or determinants ; please help , Thanks in Advance .

Answer 1

It is direct if you use Schur's decomposition

That is, for any square matrix $A$ with complex entries, there exist unitary matrix $U$ such that $U^{-1}A\ U$ is an upper triangular matrix with eigenvalues in the diagonal. Therefore the sum of eigenvalues = trace of $U^{-1}A\ U$ = trace of $A$

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If you don't know Schur's decomposition yet, try to prove it on your own! It's a lot of fun. Try using induction, similarly with the proof that every Hermitian matrix is diagonalizable.

Answer 2

Step 1. If $A$ and $B$ are two $n\times n$ matrices then ${\rm Tr}(AB)= {\rm Tr}(BA)$.

Step 2. If $A$ is an $n\times n$ complex matrix, then there is a triangular matrix $T$, and an invertible matrix $P$ such that $A=P^{-1}TP$, hence $${\rm Tr}(A)={\rm Tr}(P^{-1}TP)={\rm Tr}(P P^{-1}T )={\rm Tr}(T)$$ But the diagonal elements of $T$ are the eigenvalues of $A$.

Remark. This works on any field if the characteristic polynomial can be split into the product of first degree polynomials. And the result is wrong otherwise.

Answer 3

Sketch of an elementary answer.

Let $p$ be the characteristic polynomial of $A=(a_{ij})_{i,j=1}^n$. Then $$ p(\lambda)=\left|\begin{matrix} a_{11}-\lambda & a_{12} &\cdots & a_{1n} \\ a_{21} & a_{22}-\lambda &\cdots & a_{2n} \\ \vdots & & & \vdots \\ a_{n1} & a_{n2} &\cdots & a_{nn}-\lambda \end{matrix}\right|=(-1)^n\lambda^n+b_{n-1}\lambda^{n-1}+\cdots+b_0 $$ The sum of the roots of $p$ is $(-1)^{n-1}b_{n-1}$. Note that $$ p^{(n-1)}(\lambda)=(-1)^nn!\lambda+(n-1)!b_{n-1}. $$ Now try to find the $(n-1)-$derivative of the determinant $$\left|\begin{matrix} a_{11}-\lambda & a_{12} &\cdots & a_{1n} \\ a_{21} & a_{22}-\lambda &\cdots & a_{2n} \\ \vdots & & & \vdots \\ a_{n1} & a_{n2} &\cdots & a_{nn}-\lambda \end{matrix}\right|$$ After some tedious calculations we get the the answer.

Note that the first derivative of the determinant is $$ \left|\begin{matrix} -1 & 0 &\cdots & 0 \\ a_{21} & a_{22}-\lambda &\cdots & a_{2n} \\ \vdots & & & \vdots \\ a_{n1} & a_{n2} &\cdots & a_{nn}-\lambda \end{matrix}\right| +\cdots+ \left|\begin{matrix} a_{11}-\lambda & \cdots & a_{1n-1} & a_{1n} \\ \vdots & & & \vdots \\ a_{n-1,1} & \cdots & a_{n-1,2}-\lambda & a_{n-1,n} \\ 0 & \cdots &0 & -1 \end{matrix}\right| $$