Eigenvalues and Eigenvectors

The transpose of a matrix is formed by interchanging the rows and columns.

If a matrix is equal to its own transpose, it is said to be symmetric.

Eigenvalues and corresponding eigenvectors can be found for all symmetric matrices.The eigenvalues for the matrix M above are 1.928, 5.624, 11.099, and -10.652. Thus,there are 4 eigenvalue-eigenvector equations which may be written:

These eigenvectors can be combined into one matrix which has the same dimension asthe original symmetric matrix.

This matrix whose columns are the eigenvectors of the original symmetric matric has aninteresting property. The inverse of this matrix is identical to its transpose.

When the inverse of a matrix is equal to the transpose of that matrix, the matrix is said tobe orthogonal. In addition, if one multiples the original symmetric matrix on the right bythis orthogonal matrix and on the left by its inverse, a new diagonal matrix is generatedwhich has the eigenvalues of the original symmetric matrix on the diagonal.

This diagonalization of M to yield the eigenvalues of M is called a similaritytransformation.

If a matrix is equal to the transpose of its complex conjugate then it is said to beHermitian. In addition to real symmetric matrices, all Hermitian matrices can also bediagonalized to find their eigenvalues.

The eigenvalues of C are -3.374 and 10.374. Note that the eigenvalues are real eventhough C is complex. In addition, each eigenvalue has a corresponding eigenvector:

As before, the eigenvectorscan be combined into amatrix with the samedimension as the originalmatrix.

In this case, the matrix formed is not orthogonal. However, the inverse of the matrixformed from the eigenvectors is equal to the transpose of its complex conjugate.

When the inverse of a matrix is equal to the transpose of its complex conjugate, thenthat matrix is said to be unitary, and the similarity transformation becomes a unitarytransformation.

How does one find the eigenvalues of a symmetric or Hermitian matrix without a

computer? Consider a single eigenvalue-eigenvector equation: in which A isan n by n symmetric or Hermitian matrix, 8 is the eigen value, and v is the n by 1eigenvector or column matrix. If one wishes to subtract the right side of this equationfrom the left, the eigenvalue must be multiplied by an n by n identity matrix because

matrices must be the same dimension to add or subtract: where is

the n by n identity matrix and is the n by 1 zero matrix.

Let and

Then

or

Now, if one considered the different elements of the eigenvector as variables, a trivialsolution would be true if v were also an n by 1 zero matrix. However, the trivial solutionis uninteresting. How can one find the nontrivial solutions? There will be nontrivialsolutions only if the coefficients form a linearly dependent set. This will be true only ifthe determinant of the coefficient matrix is equal to zero. Thus the determinant of

must be zero.

This new equation: is called the secular equation. Fron the secular

equation, one obtains an nth-order polynomial equation in 8, which can be solved to yieldthe n values of 8.

This determinant yields the following cubic equation:

which can be factored into

The solutions to this cubic equation are

which are the eigenvalues of the originalsymmetric matrix A.

To find the eigenvectors corresponding to these eigenvalues, the eigenvalues must beinserted back into the original matrix equation one at a time. Let us begin with the firsteigenvalue, 8 = 1

The matrix multiplication reveals that c1 + c3 = 0 (so c1 = -c3) and that c2 = 0.

Thus, one form of the eigenvector corresponding to the eigenvalue of 1 would be , but this is not normalized. The magnitude of this vector is

Thus, dividing by yields the normalized eigenvector.

Now repeat the process for the other eigenvalues. For we get

The matrix multiplication reveals that

Thus, one form of the eigenvector corresponding to the eigenvalue of would be

However, this vector is not normalized. The magnitude of this vector is

Therefore, dividing by 2 yields the normalized eigenvector:

Finally, for we get

The matrix multiplication reveals that

Thus, one form of the eigenvector corresponding to the eigenvalue of would be

However, this vector is not normalized. The magnitude of this vector is

Therefore, dividing by 2 yields the normalized eigenvector: