# Eigenvalues of $Q=I+2P$

by user481975   Last Updated May 16, 2019 06:20 AM

I have tried to do it evaluated option (a). I think it is correct.Can not get the other options.Please help me.

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Part c) Suppose $$v$$ is an eigenvector of $$P$$ with eigenvalue $$\lambda$$, i.e. $$Pv = \lambda v$$. Then $$Qv = (I+2P)v = v+2Pv = v+2\lambda v = (1+2\lambda)v$$, i.e. $$v$$ is an eigenvector of $$Q = I+2P$$ with eigenvalue $$1+2\lambda$$. So if $$-1$$ and $$+1$$ are the possible eigenvalues of $$P$$, what are the possible eigenvalues of $$Q$$?

Part b) A matrix is invertible if and only if all the eigenvalues are non-zero. Use your answer to part c to answer this.

Part d) The determinant of a matrix is the product of its eigenvalues (counting multiplicity). So $$\det P$$ is the product of a bunch of $$-1$$'s and $$+1$$'s. Hence, $$\det P > 0$$ if $$P$$ has an even number of eigenvalues that are $$-1$$ and $$\det P < 0$$ if $$P$$ has an odd number of eigenvalues that are $$-1$$. Can you make a similar statement about $$Q$$? How are the eigenvalues of $$P$$ and $$Q$$ related?

Part a) If $$Q$$ is invertible (see part b), then since $$\{a_1,\ldots,a_n\}$$ is a basis, so is $$\{Qa_1,\ldots,Qa_n\}$$. But $$Qa_i = (I+2P)a_i = \cdots$$.

JimmyK4542
May 16, 2019 06:17 AM

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