Eigenvalues question

**worc3247** · February 9th 2011, 08:51 AM

Let $A \in M_{n \times n}(\mathbb{R})$ . Show that if A has n eigenvalues $\lambda_1,\lambda_2,...,\lambda_n$ in $\mathbb{R}$ then $\lambda_1+\lambda_2+...+\lambda_n = traceA$ and $\lambda_1 \lambda_2 ... \lambda_n = detA$ .

Not really sure where i'm going with this question. I am thinking of using the fact that for the polynomial equation obtained from $det(\lambda I_n - A)=0$ has a coefficient of trace(A) for the $\lambda^{n-1}$ term, but not really sure how this is useful.

**Ackbeet** · February 9th 2011, 08:54 AM

The trace and determinant are unchanged under an orthogonal transformation. If you have, as I suspect, n distinct eigenvalues, then you're guaranteed to be able to orthogonally diagonalize A. If you can do that, and you know that the trace and determinant are unchanged by such a transformation, then you're done.

**FernandoRevilla** · February 9th 2011, 09:18 AM

An alternative:

As $A\in \mathbb{R}^{n\times n}$ has $n$ real eigenvalues (including multiplicities), then $A$ is similar to its canonical form of Jordan $J$ . As similar matrices have the same trace and the same determinant, then:

$\textrm{tr}A=\textrm{tr}J=\lambda_1+\ldots+\lambda _n$

$\det A=\det J=\lambda_1\cdot \ldots \cdot \lambda_n$

Fernando Revilla

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