For question 15 of section 6.1, page 295 of the text, it asks for two other lambdas (eigenvalues) for the permutation matrix P. All P's have an eigenvalue of 1, because they do not change the vector x = (1,1,1...). However, my answer does not correspond to the answer in the back of the book: See work below.

Question

For question 15 of section 6.1, page 295 of the text, it asks for two other lambdas (eigenvalues) for the permutation matrix P.  All P's have an eigenvalue of 1, because they do not change the vector x = (1,1,1...). However, my answer does not correspond to the answer in the back of the book: See work below.

datanewb · Answer

attachment

datanewb · Answer

For the matrix P = $$\left[\begin{matrix}0 & 1&0 \ 0 & 0&1\1&0&0\end{matrix}\right]$$
I get $$\lambda = 1$$
as the only root of $$1-\lambda^3 = 0$$

datanewb · Answer

The answer in the book is $$\lambda = 1/ 2\left( -1 \pm i \sqrt{3}\right)$$ in addition to the $$\lambda = 1$$ value that is true for all permutation matrices.

anonymous · Answer

You are totally correct with the equation $$1-\lambda ^{3}=0$$
So, the eigenvalues are the cube-roots of unity. You can find a good deal of explanation in any book on calculus I suppose. In short, the obvious root (which is 1) is factored out to express the equation as $$(\lambda - 1)(\lambda^2+\lambda+1)=0$$
And, the answer you see in the book just follows. Hope this helps

datanewb · Answer

Thank you so much.  I was able to read about the roots of unity and understand now.  I'm currently self-studying calculus I.