Linear Algebra

6.2 Some Applications Of Eigenvalues And Eigenvectors

Recall that n×n matrices can be considered as linear transformations. If F is a 3 × 3 real matrix having positive determinant, it can be shown that F = RU where R is a rotation matrix and U is a symmetric real matrix having positive eigenvalues. An application of this wonderful result, known to mathematicians as the right polar decomposition, is to continuum mechanics where a chunk of material is identified with a set of points in three dimensional space.

The linear transformation, F in this context is called the deformation gradient and it describes the local deformation of the material. Thus it is possible to consider this deformation in terms of two processes, one which distorts the material and the other which just rotates it. It is the matrix U which is responsible for stretching and compressing. This is why in continuum mechanics, the stress is often taken to depend on U which is known in this context as the right Cauchy Green strain tensor. This process of writing a matrix as a product of two such  matrices, one of which preserves distance and the other which distorts is also important in applications to geometric measure theory an interesting field of study in mathematics and to the study of quadratic forms which occur in many applications such as statistics. Here I am emphasizing the application to mechanics in which the eigenvectors of U determine the principle directions, those directions in which the material is stretched or compressed to the maximum extent.

It is nice to be given the eigenvalues. The largest eigenvalue is 3 which means that in the direction determined by the eigenvector associated with 3 the stretch is three times as large. The smallest eigenvalue is 1∕2 and so in the direction determined by the eigenvector for 1∕2 the material is compressed, becoming locally half as long. It remains to find these directions. First consider the eigenvector for 3. It is necessary to solve

( ( ) ( ) ) ( ) ( ) 1 0 0 2191 611 611- x 0 |(3 |( 0 1 0 |) − |( 6- 41 19-|) |) |( y |) = |( 0 |) 161 4149 4441- 0 0 1 11 44 44 z 0

Thus the augmented matrix for this system of equations is

( ) 411 − 611 −161 0 |( − 6- 91 − 19 0 |) − 116- −4419 4914 0 11 44 44

The row reduced echelon form is

( ) | 1 0 − 3 0 | ( 0 1 − 1 0 ) 0 0 0 0

and so the principle direction for the eigenvalue 3 in which the material is stretched to the maximum extent is

( ) 3 |( 1 |) . 1

A direction vector in this direction is

( √ -- ) 3∕ 11 |( 1∕√11- |) . 1∕√11-

You should show that the direction in which the material is compressed the most is in the direction

( ) | 0√- | ( − 1∕√ 2 ) 1∕ 2

Note this is meaningful information which you would have a hard time finding without the theory of eigenvectors and eigenvalues.

Another application is to the problem of finding solutions to systems of differential equations. It turns out that vibrating systems involving masses and springs can be studied in the form

′′ x = Ax (6.8)

(6.8)

where A is a real symmetric n×n matrix which has nonpositive eigenvalues. This is analogous to the case of the scalar equation for undamped oscillation, x^′′ + ω²x = 0. The main difference is that here the scalar ω² is replaced with the matrix −A. Consider the problem of finding solutions to 6.8. You look for a solution which is in the form

x(t) = veλt (6.9)

(6.9)

and substitute this into 6.8. Thus

′′ 2λt λt x = v λ e = e Av

and so

λ2v = Av.

Therefore, λ² needs to be an eigenvalue of A and v needs to be an eigenvector. Since A has nonpositive eigenvalues, λ² = −a² and so λ = ±ia where −a² is an eigenvalue of A. Corresponding to this you obtain solutions of the form

x (t) = v cos(at),vsin(at).

Note these solutions oscillate because of the cos

and sin in the solutions. Here is an example.

According to the above, you can find solutions by looking for the eigenvectors. Consider the eigenvectors for −3. The augmented matrix for finding the eigenvectors is

( 4 1 1 ) | − 3 3 3 0 | ( 13 − 56 − 56 0 ) 13 − 56 − 56 0

and its row echelon form is

( ) 1 0 0 0 |( 0 1 1 0 |) . 0 0 0 0

Therefore, the eigenvectors are of the form

( ) | 0 | v = z( − 1 ) . 1

It follows

( ) ( ) 0 (√ - ) 0 (√ -) |( − 1|) cos 3t ,|( − 1 |) sin 3t 1 1

are both solutions to the system of differential equations. You can find other oscillatory solutions in the same way by considering the other eigenvalues. You might try checking these answers to verify they work.

This is just a special case of a procedure used in differential equations to obtain closed form solutions to systems of differential equations using linear algebra. The overall philosophy is to take one of the easiest problems in analysis and change it into the eigenvalue problem which is the most difficult problem in algebra. However, when it works, it gives precise solutions in terms of known functions.