Page 207 - Linear Algebra Done Right

Block Upper-Triangular Matrices
the matrix of T has no eigenvalues (by 9.1). Thus regardless of whether
T has eigenvalues, we have the desired conclusion when dim V = 2.
Suppose now that dim V> 2 and the desired result holds for all real 197
vector spaces with smaller dimension. If T has an eigenvalue, let U be a
one-dimensional subspace of V that is invariant under T; otherwise let
U be a two-dimensional subspace of V that is invariant under T (5.24
guarantees that we can choose U in this fashion). Choose any basis
of U and let A 1 denote the matrix of T| U with respect to this basis. If
A 1 is a 2-by-2 matrix, then T has no eigenvalues (otherwise we would
have chosen U to be one-dimensional) and thus T| U has no eigenvalues.
Hence if A 1 is a 2-by-2 matrix, then A 1 has no eigenvalues (see 9.1).
Let W be any subspace of V such that

V = U ⊕ W;

2.13 guarantees that such a W exists. Because W has dimension less
than the dimension of V, we would like to apply our induction hypoth-
esis to T| W . However, W might not be invariant under T, meaning that
T| W might not be an operator on W. We will compose with the pro-
jection P W,U to get an operator on W. Specifically, define S ∈L(W) Recall that if
by v = w + u, where
Sw = P W,U (Tw) w ∈ W and u ∈ U,
then P W,U v = w.
for w ∈ W. Note that
Tw = P U,W (Tw) + P W,U (Tw)
9.6 = P U,W (Tw) + Sw

for every w ∈ W.
By our induction hypothesis, there is a basis of W with respect to
which S has a block upper-triangular matrix of the form
 
A 2 ∗
 . 
 . .  ,
 
0 A m
where each A j is a 1-by-1 matrix or a 2-by-2 matrix with no eigenvalues.
Adjoin this basis of W to the basis of U chosen above, getting a basis
of V. A minute’s thought should convince you (use 9.6) that the matrix
of T with respect to this basis is a block upper-triangular matrix of the
form 9.5, completing the proof.