Page 209 - Linear Algebra Done Right

The Characteristic Polynomial
(x − a)(x − d), when applied to T, gives 0 even when b = 0. We have
(T − aI)(T − dI)v 1 = (T − dI)(T − aI)v 1 = (T − dI)(bv 2 ) = bcv 1 199
and
(T − aI)(T − dI)v 2 = (T − aI)(cv 1 ) = bcv 2 .
Thus (T − aI)(T − dI) is not equal to 0 unless bc = 0. However, the
equations above show that (T − aI)(T − dI) − bcI = 0 (because this
operator equals 0 on a basis, it must equal 0 on V). Thus if q(x) =
(x − a)(x − d) − bc, then q(T) = 0.
Motivated by the previous paragraph, we define the characteristic
ac
polynomial of a 2-by-2 matrix bd to be (x − a)(x − d) − bc. Here
we are concerned only with matrices with real entries. The next re-
sult shows that we have found the only reasonable definition for the
characteristic polynomial of a 2-by-2 matrix.

9.7 Proposition: Suppose V is a real vector space with dimension 2 Part (b) of this
and T ∈L(V) has no eigenvalues. Let p ∈P(R) be a monic polynomial proposition would be
with degree 2. Suppose A is the matrix of T with respect to some basis false without the
of V. hypothesis that T has
no eigenvalues. For
(a) If p equals the characteristic polynomial of A, then p(T) = 0. example, define
2
(b) If p does not equal the characteristic polynomial of A, then p(T) T ∈L(R ) by
is invertible. T(x 1 ,x 2 ) = (0,x 2 ).
Take p(x) = x(x − 2).
Proof: We already proved (a) in our discussion above. To prove (b), Then p is not the
let q denote the characteristic polynomial of A and suppose that p = q. characteristic
2
2
We can write p(x) = x + α 1 x + β 1 and q(x) = x + α 2 x + β 2 for some polynomial of the
α 1 ,β 1 ,α 2 ,β 2 ∈ R. Now matrix of T with
respect to the standard
p(T) = p(T) − q(T) = (α 1 − α 2 )T + (β 1 − β 2 )I. basis, but p(T) is not
invertible.
If α 1 = α 2 , then β 1 = β 2 (otherwise we would have p = q). Thus if
α 1 = α 2 , then p(T) is a nonzero multiple of the identity and hence is
invertible, as desired. If α 1 = α 2 , then

β 2 − β 1
p(T) = (α 1 − α 2 )(T − I),
α 1 − α 2
which is an invertible operator because T has no eigenvalues. Thus (b)
holds.