Linear algebra Diagonalization

Question

Suppose a 3x3 real matrix A has only two (real) distinct eigenvalues. Suppose that trace(A) = 1 and det(A) = -45 . Find the eigenvalues of A with their algebraic multiplicities.

smaller eigenvalue = ?

has mulitiplicity = ?

larger eigenvalue = ?

has mulitiplicity = ?

Answer 1

We'll denote the trace of A as tr(A) and the determinant of A as det(A).

For this question it is helpful to know the following properties of a matrix's trace, determinant, and characteristic polynomial:

1. tr(A)=λ₁+λ₂+λ₃ where λ₁, λ₂, and λ₃ are the eigenvalues of A.
2. If A has eigenvalues λ₁, λ₂, and λ₃ then A² has eigenvalues λ₁², λ₂², and λ₃².
3. The eigenvalues λ₁, λ₂, and λ₃ are roots of the characteristic polynomial

det(A-λI) = -λ³+tr(A)λ²-(1/2)[(tr(A))²-tr(A²)]λ+det(A).

(If you want to know where these came from, feel free to ask and I'll be happy to answer those separately.)

Solution:

We are given that A has real eigenvalues λ₁ and λ₂=λ₃. So we have tr(A)=λ₁+λ₂+λ₃ = λ₁+2λ₂ = 1. The eigenvalues of A are roots of the characteristic polynomial

-λ³+tr(A)λ²-(1/2)[(tr(A))²-tr(A²)]λ+det(A)

= -λ³+λ²-(1/2)[1-tr(A²)]λ-45

= -λ³+λ²-(1/2)[1-(λ₁²+λ₂²+λ₃²)]λ-45 (because tr(A²)=λ₁²+λ₂²+λ₃²)

= -λ³+λ²-(1/2)[1-(λ₁²+2λ₂²)]λ-45 (because λ₂=λ₃)

= -λ³+λ²-(1/2)[1-((1-2λ²)²+2λ₂²)]λ-45 (because λ₁=1-2λ²)

=...simplify inside the square brackets...

= -λ³+λ²-(1/2)(4λ₂-6λ₂²)λ-45

= -λ³+λ²+(3λ₂²-2λ₂)λ-45

Now use the fact that λ₂ is an eigenvalue, hence a root of this polynomial. Therefore

-λ₂³+λ₂²+(3λ₂²-2λ₂)λ₂-45=0. Simplifying gives 2λ₂³-λ₂²-45=0. By the rational roots theorem, the possible rational roots of this polynomial are positive and negative factors of 45. This list of possibilities is 1, -1, 3, -3, 5, -5, 9, -9, 45, -45. By experimentation, we find that λ₂=3 is a solution.

This allows us to factor 2λ₂³-λ₂²-45 = (λ₂-3)(2λ₂²+5λ₂+15). Using the quadratic formula, we see that the quadratic has no real roots, so λ₂=3 is the only real root.

Therefore we have λ₂= λ₃=3 and we use tr(A)=λ₁+λ₂+λ₃=1 to find that λ₁=-5.

Our final answer is that A has eigenvalue -5 with algebraic multiplicity 1 and eigenvalue 3 with algebraic multiplicity 2.