complicated matrix question.?

Question

Let A be an n x n matrix with A^2-4A+5I=0. how do I show that n must be even??

thanks for the help!

Answer 1

Suppose that A²-4A+5I=0. Since all the matrices in this polynomial commute with each other, we may factor it to produce (A-2I)²+I=0, or (A-2I)²=-I. Now let λ be any eigenvalue of A-2I and let v be a nonzero eigenvector associated with that eigenvalue. Then right-multiplying both sides by v, we obtain (A-2I)²v=λ²v=-v. Thus, λ²=-1, meaning λ=±i. But, since this is true for an arbitrary eigenvalue of A-2I, it must be true for all of them, meaning that all the roots of (A-2I)'s characteristic polynomial are imaginary. However, every polynomial of odd degree has at least one real root, so (A-2I)'s characteristic polynomial is of even degree, so A-2I is an n×n matrix with n even, which means that A must also be an n×n matrix with n even. Q.E.D.

Answer 2

Any matrix satisfies its characteristic polynomial. in case you be conscious that the eigenvalues of A are 3, 2, and 5, you will see that that the characteristic polynomial is (x-5)(x-2)(x-3) = (x^2 - 7x + 10)(x-3) = x^3 -10x^2 + 31 x - 30 = f(x). for this reason f(A) = 0.

Answer 3

wow... it is even more complicated than linear algebra.. i've never see this type of question.