Page 210 - Linear Algebra Done Right

Chapter 9. Operators on Real Vector Spaces
200
Suppose V is a real vector space with dimension 2 and T ∈L(V) has
no eigenvalues. The last proposition shows that there is precisely one
monic polynomial with degree 2 that when applied to T gives 0. Thus,
though T may have different matrices with respect to different bases,
each of these matrices must have the same characteristic polynomial.
2
For example, consider T ∈L(R ) defined by
9.8 T(x 1 ,x 2 ) = (3x 1 + 5x 2 , −2x 1 − x 2 ).
2
The matrix of T with respect to the standard basis of R is

3 5
−2 −1 .

The characteristic polynomial of this matrix is (x − 3)(x + 1) + 2 · 5,
2
which equals x − 2x + 7. As you should verify, the matrix of T with

respect to the basis (−2, 1), (1, 2) equals

1 −6
1 1 .

The characteristic polynomial of this matrix is (x − 1)(x − 1) + 1 · 6,
2
which equals x − 2x + 7, the same result we obtained by using the
standard basis.
When analyzing upper-triangular matrices of an operator T on a
complex vector space V, we found that subspaces of the form

null(T − λI) dim V

played a key role (see 8.10). Those spaces will also play a role in study-
ing operators on real vector spaces, but because we must now consider
block upper-triangular matrices with 2-by-2 blocks, subspaces of the
form
2
null(T + αT + βI) dim V
will also play a key role. To get started, let’s look at one- and two-
dimensional real vector spaces.
First suppose that V is a one-dimensional real vector space and that
T ∈L(V).If λ ∈ R, then null(T − λI) equals V if λ is an eigenvalue
2
of T and {0} otherwise. If α, β ∈ R with α < 4β, then