Question

Let A be a 4x4 matrix such that A² - 3A + 2I = 0. Which two real numbers could possibly be eigenvalues of A?

Answer 1

The first think i taugth of is factoring into (A-1)(A-2) = 0 and saying 1 and 2 are the possible eigenvalues, but is that really it?

Answer 2

Yes, but not quite for that reason, because A - 2 doesn't make sense: a matrix minus a scalar

Answer 3

it should be (A-2I) right?

Answer 4

However 1 and 2 are the correct answers.

Answer 5

(A-I)(A-2I) = 0

Answer 6

More or less. Why does that equation imply 1 and 2 are eigenvalues?

Answer 7

Thats the part i was wondering :/ If A has two eigenvalues that it must be diagonlizable, so the diagonal matrix D will have the eigen values of it diagonal, so if we had (D-I)(D-2I) then D will have 0 entry in both the monomials (This is just a guess)

Answer 8

An eigenvalue \( \lambda \) is a solution of the equation
\[ \det(A - \lambda I) = 0 \]

Answer 9

correct

Answer 10

Now if (A-I)(A-2I) = 0, then det[ (A-I)(A-2I) ] = 0.

But the determinant of a product of square matrices is the product of the determinants. Thus
det(A-I).det(A-2I) = 0
and it follows that one or both of these determinants is zero. And therefore ...

Answer 11

Perfecttt, thank you soo much, that makes sense :)