If a matrix $A$ is Diagonalizable, then $\exists$ a Non singular matrix $P$ such that

$$D=P^{-1}AP$$ Now taking Trace on both sides


Now since $D$ is Diagonal matrix with diagonal elements as eigen values we have

$Tr(A)$ as Sum of eigen values of $A$.

But how to prove this if $A$ is not diagonalizable?

Umesh shankar
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  • By extending the underlying field if necessary, we may replace the diagonal matrix $D$ with a matrix in normal Jordan form. Such a matrix is upper triangular, so again its diagonal entries are its eigenvalues. – Travis Willse Jul 01 '17 at 17:04

1 Answers1


What you can do is realize that the sum of eigen values is -(the sum of coefficients of $x^{n-1}$) in the characterestic polynomial. Now just see what the coeeficient of $x^{n-1}$ is while expanding the determinant of $xI-A$.( I assume A is a $n\times n$ matrix)

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