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How to prove that $\operatorname{Tr}(A)=a+b$ and $\det(A)=ab$ if characteristic polynomial of $A$ is degreed 2 and with $a$ and $b$ as eigenvalue?

If $a$ and $b$ are distinct eigenvalues, then $A$ is diagonalizable, so that $\operatorname{Tr}(A)=a+b$ and $\det(A)=ab$ since there's a similar diagonal matrix. However, how are we making sure that there are no zeros in $A$ as we don't know the size of $A$?

Michael Hardy
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    http://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues –  Dec 21 '15 at 02:30

1 Answers1

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Write down the characteristic polynomial:

$$p(\lambda)=|A-\lambda I| = \lambda^2 - (a_{11}+a_{22}) \lambda + (a_{11} a_{22} - a_{12} a_{21} ) =\lambda^2 - tr(A) \lambda +|A|$$

On the other side, because the eigenvalues are the roots of $p(\lambda)$, (and because we already know the highest degree coefficient is one) we have:

$$p(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)=\lambda^2 -(\lambda_1+\lambda_2)\lambda +\lambda_1 \lambda_2 $$

Comparing both expressions, we get the desired result.

The beauty of this is that is generalizes to any size.

leonbloy
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