[amp_mcq option1=”1, 1, -1″ option2=”1, 0.5 + j0.866, 0.5 – j0.866″ option3=”1, -0.5 + j0.866, -0.5 – j0.866″ option4=”0, 1, -1″ correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{A. }1, 1, -1}$.<\/p>\n
Let $v$ be an eigenvector of $P$ corresponding to the eigenvalue $\\lambda$. Then, $P v = \\lambda v$. Multiplying both sides by $P$, we get $P^2 v = P \\lambda v = \\lambda^2 v$. Continuing in this way, we get $P^3 v = \\lambda^3 v$. But since $P^3 = P$, we have $\\lambda^3 v = P v = \\lambda v$. Therefore, $\\lambda^3 = \\lambda$.<\/p>\n
Since $P$ is a 3 by 3 matrix, its eigenvalues are roots of the characteristic polynomial $p(x) = |xI – P|$. The characteristic polynomial of $P$ is $p(x) = x^3 – x$. The roots of $p(x)$ are $x = 1, 1, -1$. Therefore, the eigenvalues of $P$ are $1, 1, -1$.<\/p>\n
Option B is incorrect because it has three distinct eigenvalues. Option C is incorrect because it has two complex eigenvalues. Option D is incorrect because it has one zero eigenvalue and two distinct eigenvalues.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”1, 1, -1″ option2=”1, 0.5 + j0.866, 0.5 – j0.866″ option3=”1, -0.5 + j0.866, -0.5 – j0.866″ option4=”0, 1, -1″ correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[489],"tags":[],"class_list":["post-19987","post","type-post","status-publish","format-standard","hentry","category-linear-algebra","no-featured-image-padding"],"yoast_head":"\n