Exercise [13.31]

jbeckmann · 18 May 2010, 15:51

PREREQUISITE:
Each eigenvector of T having multiplicity $r \geq 2$ (if there are any) spans an eigenspace of dimension d = r.

ASSERTION:
Then it is possible to find a basis $\{e_1, e_2, ... e_n\}$ of eigenvectors.

PROOF:

Because of the prerequisite, it is possible to find a set $\{e_1, ... e_n\}$ 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 e_j

.

To show this, a particular numbering of the eigenvectors is first introduced, in which any of the ( $m \leq n$ ) eigenvalues $\lambda_k$ is associated with the eigenvectors $e_{r_k}, ... e_{R_k}$ . The indices 1=r_1, ... R_1, r_2, ... R_2, ... r_k, ... R_k, ... r_m, ... R_m=n

1=r_1, ... R_1, r_2, ... R_2, ... r_k, ... R_k, ... r_m, ... R_m=n

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 $\alpha_j$ which do not all vanish such that:

$0 = \alpha_1 e_1 + ... \alpha_n e_n$
$0 = (\alpha_{r_1} e_{r_1} + ... \alpha_{R_1} e_{R_1}) + ... (\alpha_{r_m} e_{r_m} + ... \alpha_{R_m} e_{R_m})$
0 = E_1 + ... E_m

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

The last line uses the definition $E_j := (\alpha_{r_j} e_{r_j} + ... \alpha_{R_j} e_{R_j})$ . Note that the E_j

are eigenvectors with $T(E_j) = \lambda_j E_j$ .

Applying T to eq.(1) yields:

$0 = T \left( E_1 + ... E_m \right) = \lambda_1 E_1 + ... \lambda_m E_m$ ............(2)

It can be assumed without loss of generality that all eigenvalues besides possibly $\lambda_1$ are nonzero (renumber the eigenvalues if another one should be zero). Eq.(2) can then be resolved for E_m

:

$E_m = -\frac{\lambda_1}{\lambda_m}E_1 - ... \frac{\lambda_{m-1}}{\lambda_m}E_{m-1}$ ......................(3)

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

from eq.(1), yielding:

$0 = (1-\frac{\lambda_1}{\lambda_m})E_1 - ... (1-\frac{\lambda_{m-1}}{\lambda_m})E_{m-1}$

$0 = a_1 E_1 + a_2 E_2 + ... a_{m-2}E_{m-2} + a_{m-1}E_{m-1}$ ..........(1')

The last line uses the definition $a_j := (1-\frac{\lambda_j}{\lambda_m})$ . Note that all a_j

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

are eigenvectors with $T(a_j E_j) = \lambda_j \cdot a_j E_j$ .

The above reasoning starting at eq.(1) can now be repeated with eq.(1') to eliminate vector $a_{m-1}E_{m-1}$ , and so on, yielding the following sequence of equations:

$0 = b_1 E_1 + b_2 E_2 + ... b_{m-2}E_{m-2}$ ..........(1'')
...
0 = g_1 E_1 + g_2 E_2

....................(1''')
0 = h_1 E_1

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

As all coefficients b_j, ... g_j, h_1

are nonzero, it follows successively
that E_1 = 0

(from eq.(1'''')),
then that E_2 = 0

(the above and eq.(1''')),
etc. until E_m = 0

.
For short, all E_j = 0

.

For all eigenvalues of multiplicity 1 we have $E_k = \alpha_{r_k} e_{r_k}$ (cf. definition of E_k

with

). The above finding E_k = 0

hence implies $\alpha_k = 0$ for these eigenvalues.

Nonzero coefficients $\alpha_k$ can hence only appear in E_k

s belonging to eigenvalues of multiplicity $\geq 2$ . However, if $E_k = (\alpha_{r_k} e_{r_k} + ... \alpha_{R_k} e_{R_k})$ would be such a vector, this and the above finding that E_k = 0

are in contradiction to the prerequisite that all corresponding eigenvectors $e_{r_k}, ... e_{R_k}$ are linearly independent.

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