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 Exercise [13.31] 
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Joined: 22 Apr 2010, 15:52
Posts: 43
Location: Olpe, Germany
Post Exercise [13.31]
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 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 r_k=R_k). The above finding E_k = 0 hence implies \alpha_k = 0 for these eigenvalues.

Nonzero coefficients \alpha_k can hence only appear in E_ks 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.


18 May 2010, 15:51

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [13.31]
Here's an alternative proof:

Construct any set of n eigenvectors of T \{e_i\} such that one eigenvector is chosen for each of the n eigenvalues (counting multiple eigenvalues more than once), ensuring that the subset of eigenvectors chosen for each multiple eigenvalue is linearly independent. This is always possible given the assumption about T stated in the text.

Now suppose \alpha_1 e_1 + ... + \alpha_n e_n = 0 (1)

If there are any \alpha which are non-zero in this equation, we can re-write it including only the non-zero terms (and re-numbering), to get:

\alpha_1 e_1 + ... + \alpha_m e_m = 0 where no \alpha is zero, and where 1 \leq m \leq n

Then pre-multiplying both sides by T n times (for any n \in \{0,1,2,...\} yeilds:

\alpha_1 {\lambda_1}^n e_1 + ... + \alpha_m {\lambda_m}^n e_m = 0

Without loss of generality, assume the numbering of the \alpha's and e's is chosen so that |\lambda_1| \geq |\lambda_i|, for all i>1. Then dividing by {\lambda_1}^n yeilds:

\alpha_1 e_1 + \alpha_2 (\frac{\lambda_2}{\lambda_1})^n e_2 + ... + \alpha_m (\frac{\lambda_m}{\lambda_1})^n e_m = 0

Now as this is true for all n in {0,1,2,...}, it is true in the limit as n approaches infinity.
(If there's a problem with the validity of this step, please let me know!)

But in this limit, all terms j for which |\lambda_j| < |\lambda_1| vanish, leaving only the terms k for which \lambda_k = \pm \lambda_1 (k>1).

There are then three cases to consider.

Case 1: There are no such \lambda_k's. Then we are left with \alpha_1 e_1 = 0. This is a contradiction because eigenvectors cannot be zero.

Case 2: There are only \lambda_k's equal to \lambda_1. Then the (\frac{\lambda_k}{\lambda_1})^n are all equal to 1, and the resulting equation is:

\alpha_1 e_1 + \alpha_k e_k + \alpha_l e_l + ... = 0

where all the e_i's are eigenvectors of the (multiple) eigenvalue \lambda_1. This is a contradiction because we assumed we picked our eigenvectors so that the subset of eigenvectors chosen for any multiple eigenvalue would be linearly independent.

Case 3: There is at least one term r for which \lambda_r = -\lambda_1. For these terms, (\frac{\lambda_r}{\lambda_1})^n is 1 when n is even and -1 when n is odd. Considering n \rightarrow \infty for only odd n, and also for only even n, we have:

\alpha_1 e_1 + (\mbox{other} \lambda_1 \mbox{terms}) + \pm(\alpha_r e_r + \alpha_s e_s + ...) = 0

But clearly this equation can only hold true when the \pm(...) term equates to zero, and so after crossing it out, this case reduces to either case 1 or case 2; or else we again have a contradiction.

Thus in each case there is a contradiction.

Therefore, there can be no \alpha's which are non-zero in equation (1).
Therefore, our chosen \{e_i\} are linearly independent, and are thus a basis for the n-dimensional vector space V on which T operates.

QED


31 Jul 2010, 17:54
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