characteristic values and vectors (of a matrix)
Let us set for multiplicity of (). We will now prove the following theorem.
If , , and is a scalar polynomial, then , .
Let be an arbitrary scalar polynomial. We want to find the characteristic values of . For this purpose we split into linear factors
On substitution , we have
because on substitution in (1). Next we commute the binomial by introducing into the product signs and also we note that , so that
and we may use (2) for to obtain
Finally we substitute the polynomial by , where is an arbitrary parameter, getting for (4)
This proves the theorem. ∎
As an important particular case we have: , (), .
Connection between the characteristic polynomial and the adjugate matrix of .
As it is well known, the adjugate matrix of a matrix there corresponds to the algebraic complement or cofactor matrix of the transpose of . From this definition we have
Let us suppose is given by
which by comparing it with (6) we conclude that
where ( in (8))
which can also be obtained from the recurrence equation
What is more,
(13) as well as (14) follow inmediately from (6) if we equate the coefficients of equal powers of on both sides. Also, if we substitute from (12), into (14), we get (Cayley-Hamilton), an implicit consequence of generalized Bézout theorem. On the other hand, by setting in (7) we obtain , whenever be non- singular. From this and from (14) follow that
Let now be a characteristic value of , then and (6) becomes
Let us assume that and denote by an arbitrary non-zero column of this matrix. From (16) we have . That is,
Therefore every non-zero column of determines a characteristic vector corresponding to the characteristic value . Moreover, if to the characteristic value there correspond linearly independent characteristic vectors, will be the rank of and so the rank of does not exceed . In particular, if only one characteristic vector there corresponds to , then in the elements of any two columns will be proportional (In such a case , hence the rank of will be ).
In conclusion: If the coefficients of the characteristic polynomial are known, then the adjugate matrix may be found by (10). In addition, if the given matrix is non-singular, then the inverse matrix can be found from (15). Also if is a characteristic value of , the non-zero columns of are characteristc vectors of A for .
Example. We find out the characteristic values and vectors from the matrix
Comparing with (7), we have
Next we use (8),
so that from (11)
We will now evaluate and by using (12) and (13), respectively.
Also and is obtained from (15), i.e.
We notice the eigenvalue possesses multiplicity and also that all the entries of the adjugate are divisible by the binomial (, i.e. annihilates it), therefore it can be reduced which makes instructive this problem. Thus,
which for it becomes
From this we get the charactreristic vectors by multiplying the first colum by , and also , both correponding to . Third column is a linear combination of the first two (subtract it). Likewise we find for the another characteristic value
whence we get the eigenvector , being the remaining two columns clearly proportional to the first one.
|Title||characteristic values and vectors (of a matrix)|
|Date of creation||2013-03-22 17:43:58|
|Last modified on||2013-03-22 17:43:58|
|Last modified by||perucho (2192)|