Question

Linear Algebra I:
Let A be a 2x2 matrix and suppose that A has eigenvalues -1 and 3. Let I denote the 2x2 identity matrix. Prove that A² - 4A + 3 = 0

Answer 1

My attempt:
I tried working backwards to the eigenvalues (A-3)(A-1) = 0 but doesn't this imply eigen values are 3, 1 and not 3,-1? Any help would be appreciated

Answer 2

(A-3)(A-1) = 0 isn't a valid matrix equation as for instance A - 3 doesn't make sense, a matrix minus a number. But even if you write
(A-3I)(A-I)
you're still not going to get exactly what you want.

Yet you are right that there's a problem with this question. The sign of one of the eigenvalues is wrong or a sign in the quadratic is wrong.

Answer 3

Okay so lets we assumed the eigenvalues were 1,3, how would be go about solving the problem then?

Answer 4

A is similar to the diagonal matrix of eigenvalues. Suppose D =
3 0
0 1

Then there exists an invertible matrix P such that
\[ A = P^{-1}DP \]

Now given that, you can show A satisfies the quadratic.

Answer 5

Ohhh so because of the eigenvalues we know the diagonal matrix D is. So if we assume A is dionilizable and there exist a invertible matrix P, then (P^-1 D P)² - 4(P^-1 D P) + 3I = 0, if we show this we are done, right?

Answer 6

We don't have to assume A is diagonalizable. Given two distinct eigenvalues it's a theorem that A is diagonalizable. Note that
\[ A^n = P^{-1}D^nP \]
for all \( n \geq 1 \) and therefore given any polynomial p(A),
\[ p(A) = P^{-1}p(D)P \]

Answer 7

Write, i agree with that

Answer 8

so if you show D satisfies your quadratic, it follows A satisfies it.

Answer 9

ohhh I get it so we just gotta show D² - 4D + 3I = 0 is satisfied, then that proves A must be satisfied as well? right?

Answer 10

Yes. Just write it out carefully.

Answer 11

Got it, thank you very much :)