Page 218 - Advanced Linear Algebra

202 Advanced Linear Algebra

< = of is <-invariant if it is -invariant for every  < . It is often of interest
to know whether a family < of linear operators on = has a common
eigenvector, that is, a single vector #= that is an eigenvector for every
<  (the corresponding eigenvalues may be different for each operator,
however).

21. A pair of linear operators Á B²= ³ is simultaneously diagonalizable if
there is an ordered basis for for which = ´ µ 8 and ´ µ 8 are both diagonal,
8
that is, is an ordered basis of eigenvectors for both and . Prove that
8

two diagonalizable operators and are simultaneously diagonalizable if

and only if they commute, that is, ~ . Hint : If ~ , then the
eigenspaces of are invariant under .

22. Let Á B²= ³ . Prove that if and commute, then every eigenspace of

is -invariant. Thus, if is a commuting family, then every eigenspace
<
of any member of is -invariant.
<
<
<
23. Let be a family of operators in ²= ³ with the property that each operator
B
in < has a full set of eigenvalues in the base field - , that is, the
-
characteristic polynomial splits over . Prove that if < is a commuting
<
family, then has a common eigenvector #= .
24. What do the real matrices

(~ > ? ) ~ > and ?
c c
have to do with the issue of common eigenvectors?
Geršgorin Disks
It is generally impossible to determine precisely the eigenvalues of a given
² ³
complex operator or matrix ( Cd , for if ‚ , then the characteristic

equation has degree and cannot in general be solved. As a result, the

approximation of eigenvalues is big business. Here we consider one aspect of
this approximation problem, which also has some interesting theoretical
consequences.

² ³
Let (  Cd and suppose that (# ~ # where # ~ ² ÁÃÁ ³ ! . Comparing

th rows gives

( ~

~
which can also be written in the form

² c ( ³ ~ (

~
£

If has the property that (( ‚ ( ( for all , we have