Page 99 - Linear Algebra Done Right

Chapter 5. Eigenvalues and Eigenvectors
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by Sv = Tv for v ∈ span(v 1 ,...,v k ). In other words, S is just T
restricted to span(v 1 ,...,v k ).
Note that span(v 1 ,...,v k ) has dimension k and span(v 1 ,...,v k−1 )
has dimension k − 1 (because (v 1 ,...,v n ) is linearly independent). Be-
cause span(v 1 ,...,v k ) has a larger dimension than span(v 1 ,...,v k−1 ),
no linear map from span(v 1 ,...,v k ) to span(v 1 ,...,v k−1 ) is injective
(see 3.5). Thus there exists a nonzero vector v ∈ span(v 1 ,...,v k ) such
that Sv = 0. Hence Tv = 0, and thus T is not invertible, as desired.
To prove the other direction, now suppose that T is not invertible.
Thus T is not injective (see 3.21), and hence there exists a nonzero
vector v ∈ V such that Tv = 0. Because (v 1 ,...,v n ) is a basis of V,we
can write
v = a 1 v 1 +· · ·+ a k v k ,
where a 1 ,...,a k ∈ F and a k = 0 (represent v as a linear combination
of (v 1 ,...,v n ) and then choose k to be the largest index with a nonzero
coefficient). Thus

0 = Tv
0 = T(a 1 v 1 + ··· + a k v k )
= (a 1 Tv 1 + ··· + a k−1 Tv k−1 ) + a k Tv k .

The last term in parentheses is in span(v 1 ,...,v k−1 ) (because of the
upper-triangular form of 5.17). Thus the last equation shows that
a k Tv k ∈ span(v 1 ,...,v k−1 ). Multiplying by 1/a k , which is allowed
because a k = 0, we conclude that Tv k ∈ span(v 1 ,...,v k−1 ). Thus
when Tv k is written as a linear combination of the basis (v 1 ,...,v n ),
the coefficient of v k will be 0. In other words, λ k in 5.17 must be 0,
completing the proof.

Powerful numeric Unfortunately no method exists for exactly computing the eigenval-
techniques exist for ues of a typical operator from its matrix (with respect to an arbitrary
finding good basis). However, if we are fortunate enough to find a basis with re-
approximations to the spect to which the matrix of the operator is upper triangular, then the
eigenvalues of an problem of computing the eigenvalues becomes trivial, as the following
operator from its proposition shows.
matrix.
5.18 Proposition: Suppose T ∈L(V) has an upper-triangular matrix
with respect to some basis of V. Then the eigenvalues of T consist
precisely of the entries on the diagonal of that upper-triangular matrix.