Computation of Eigenvalues
For a square matrix A of order n, the number
is an eigenvalue if and only if there exists a non-zero vector C such that
Using the matrix multiplication properties, we obtain
This is a linear system for which the matrix coefficient is
We also know that this system has one solution if and only if the matrix coefficient is invertible, i.e.
Since the zero-vector is a solution and C is not the zero vector, then we must have
Example. Consider the matrix
which is equivalent to the quadratic equation
Solving this equation leads to
In other words, the matrix A has only two eigenvalues.
In general, for a square matrix A of order n, the equation
will give the eigenvalues of A. This equation is called the characteristic equation or characteristic polynomial of A. It is a polynomial function in
of degree n. So we know that this equation will not have more than n roots or solutions. So a square matrix A of order n will not have more than n eigenvalues.
Example. Consider the diagonal matrix
Its characteristic polynomial is
So the eigenvalues of D are a, b, c, and d, i.e. the entries on the diagonal.
This result is valid for any diagonal matrix of any size. So depending on the values you have on the diagonal, you may have one eigenvalue, two eigenvalues, or more. Anything is possible.
Remark. It is quite amazing to see that any square matrix A has the same eigenvalues as its transpose AT because
For any square matrix of order 2, A, where
the characteristic polynomial is given by the equation
The number (a+d) is called the trace of A (denoted tr(A)), and clearly the number (ad-bc) is the determinant of A. So the characteristic polynomial of A can be rewritten as
Let us evaluate the matrix
B = A2 - tr(A) A + det(A) I2.
We leave the details to the reader to check that
In other word, we have
This equation is known as the Cayley-Hamilton theorem. It is true for any square matrix A of any order, i.e.
is the characteristic polynomial of A.
We have some properties of the eigenvalues of a matrix.
Theorem. Let A be a square matrix of order n. If
is an eigenvalue of A, then:
is an eigenvalue of Am, for
- If A is invertible, then
is an eigenvalue of A-1.
- A is not invertible if and only if
is an eigenvalue of A.
is any number, then
is an eigenvalue of
- If A and B are similar, then they have the same characteristic polynomial (which implies they also have the same eigenvalues).
The next natural question to answer deals with the eigenvectors. In the next page, we will discuss the problem of finding eigenvectors..
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