PREREQUISITE:Each eigenvector of T having multiplicity

(if there are any) spans an eigenspace of dimension d = r.

ASSERTION:Then it is possible to find a basis

of eigenvectors.

PROOF:Because of the prerequisite, it is possible to find a

set of n eigenvectors. What remains to be shown is that these n eigenvectors constitute a

basis. According to Exercise [13.28], this is tantamount to the linear independence of the

.

To show this, a particular numbering of the eigenvectors is first introduced, in which any of the (

) eigenvalues

is associated with the eigenvectors

. The indices

shall cover the whole index range from 1 to n.

The linear independence of the eigenvectors is now proven indirectly by assuming the opposite:

ASSUMPTION: There are coefficients

which do not all vanish such that:

............................................(1)

The last line uses the definition

. Note that the

are eigenvectors with

.

Applying T to eq.(1) yields:

............(2)

It can be assumed without loss of generality that all eigenvalues besides possibly

are nonzero (renumber the eigenvalues if another one should be zero). Eq.(2) can then be resolved for

:

......................(3)

With eq.(3), it is possible to eliminate vector

from eq.(1), yielding:

..........(1')

The last line uses the definition

. Note that all

are nonzero (because all eigenvalues are different from each other) and that the

are eigenvectors with

.

The above reasoning starting at eq.(1) can now be repeated with eq.(1') to eliminate vector

, and so on, yielding the following sequence of equations:

..........(1'')

...

....................(1''')

..............................(1'''')

As all coefficients

are nonzero, it follows successively

that

(from eq.(1'''')),

then that

(the above and eq.(1''')),

etc. until

.

For short, all

.

For all eigenvalues of multiplicity 1 we have

(cf. definition of

with

). The above finding

hence implies

for these eigenvalues.

Nonzero coefficients

can hence only appear in

s belonging to eigenvalues of multiplicity

. However, if

would be such a vector, this and the above finding that

are in contradiction to the prerequisite that all corresponding eigenvectors

are linearly independent.

Thus we have the desired contradiction which proves that the ASSUMPTION was wrong.