Page 98 - Linear Algebra Done Right

Upper-Triangular Matrices
Extend (u 1 ,...,u m ) to a basis (u 1 ,...,u m ,v 1 ,...,v n ) of V. For
each k, we have
Tv k = (T − λI)v k + λv k . 85
The definition of U shows that (T − λI)v k ∈ U = span(u 1 ,...,u m ).
Thus the equation above shows that
5.15 Tv k ∈ span(u 1 ,...,u m ,v 1 ,...,v k ).

From 5.14 and 5.15, we conclude (using 5.12) that T has an upper-
triangular matrix with respect to the basis (u 1 ,...,u m ,v 1 ,...,v n ).

How does one determine from looking at the matrix of an operator
whether the operator is invertible? If we are fortunate enough to have
a basis with respect to which the matrix of the operator is upper tri-
angular, then this problem becomes easy, as the following proposition
shows.

5.16 Proposition: Suppose T ∈L(V) has an upper-triangular matrix
with respect to some basis of V. Then T is invertible if and only if all
the entries on the diagonal of that upper-triangular matrix are nonzero.

Proof: Suppose (v 1 ,...,v n ) is a basis of V with respect to which
T has an upper-triangular matrix

 
λ 1 ∗
 
 λ 2 
 
5.17 M T, (v 1 ,...,v n ) =  .  .
 . . 
 
0 λ n
We need to prove that T is not invertible if and only if one of the λ k ’s
equals 0.
First we will prove that if one of the λ k ’s equals 0, then T is not
invertible. If λ 1 = 0, then Tv 1 = 0 (from 5.17) and hence T is not
invertible, as desired. So suppose that 1 <k ≤ n and λ k = 0. Then,
as can be seen from 5.17, T maps each of the vectors v 1 ,...,v k−1 into
span(v 1 ,...,v k−1 ). Because λ k = 0, the matrix representation 5.17 also
implies that Tv k ∈ span(v 1 ,...,v k−1 ). Thus we can define a linear map

S : span(v 1 ,...,v k ) → span(v 1 ,...,v k−1 )