Proof that the trace of a matrix is the sum of its eigenvalues

I have looked extensively for a proof on the internet but all of them were too obscure. I would appreciate if someone could lay out a simple proof for this important result. Thank you.

Answer

These answers require way too much machinery. By definition, the characteristic polynomial of an $n\times n$ matrix $A$ is given by
$$p(t) = \det(A-tI) = (-1)^n \big(t^n – (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$
On the other hand, $p(t) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$, where the $\lambda_j$ are the eigenvalues of $A$. So, comparing coefficients, we have $\text{tr}A = \lambda_1 + \dots + \lambda_n$.

Attribution
Source : Link , Question Author : JohnK , Answer Author : Ted Shifrin

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