Download Problem. Let ψ(x) = x 3 +1 be the minimal polynomial of matrix A

Problem. Let ψ(x) = x3 + 1 be the minimal polynomial of matrix A ∈ R4×4 . Find the minimal
polynomial, trace, and determinant of (I − A−1 ).
Solution:
ψ(A) = 0 ⇒ A3 + I = 0
⇒ −A3 = (A)(−A2 ) = (−A2 )(A) = I
⇒ A−1 = −A2
A is a real matrix. Thus, the coefficients of its characteristic polynomial (∆(x)) are real. This
makes complex eigenvalues of A to appear in conjugate pairs.
deg(φ) = 3
ψ(x) = x3 + 1
= (x + 1)(x2 − x + 1)
deg(∆) = 4
∆(x) = ?
√
1+j 3
α=
2
= (x + 1)(x − α)(x − α)
Considering conjugacy, and the fact ψ(x)|∆(x), we conclude that
∆(x) = (x + 1)(x3 + 1) = (x + 1)2 (x2 − x + 1).
So, the Jordan canonical form of A is


−1 0 0 0
 0 −1 0 0 

J=
0
0 α 0
0
0 0 α
and
A = PJP−1,
for some nonsingular P.
I − A−1 = I + A2
= I + PJ2 P−1
= PP−1 + PJ2 P−1
= P I + J2 P−1


1 + (−1)2
0
0
0

0
1 + (−1)2
0
0 

⇒ I + J2 = 
2

0
0
1+α
0 
0
0
0
1 + α2
Note that x2 − x + 1 = 0 for x = α and x = −α. Therefore,


2 0 0 0
0 2 0 0 

I + J2 = 
0 0 α 0 
0 0 0 α
= Jordan form of I − A−1 .
We know that similarity transformation1 preserves eigenvalues. Therefore, it preserves trace
and determinant since
n
n
X
Y
trace(A) =
λi
and
det(A) =
λi .
i=1
1
i=1
7→ PP−1
1
Let B denote I − A−1 . Readily, we have the following:
ψB (x) = (x − 2)(x − α)(x − α)
= (x − 2)(x2 − x + 1) = x3 − 3x2 + 3x − 2
trace(B) = 2 + 2 + α + α
= 4 + 2ℜ{α} = 5
det(B) = 2 × 2 × α × α
= 4|α|2 = 4
2