What happens if the eigenvalues are complex?
What happens if the eigenvalues are complex?
If the n × n matrix A has real entries, its complex eigenvalues will always occur in complex conjugate pairs. This is very easy to see; recall that if an eigenvalue is complex, its eigenvectors will in general be vectors with complex entries (that is, vectors in Cn, not Rn).
Can complex eigenvalues have real eigenvectors?
If α is a complex number, then clearly you have a complex eigenvector. But if A is a real, symmetric matrix ( A=At), then its eigenvalues are real and you can always pick the corresponding eigenvectors with real entries. Indeed, if v=a+bi is an eigenvector with eigenvalue λ, then Av=λv and v≠0.
Can eigen value be a complex number?
If c is any complex number, then cx is a complex eigenvector corresponding to the eigenvalue λ. Moreover, since the eigenvalues of A are the roots of the characteristic polynomial of A, the complex eigenvalues come in conjugate pairs and λ is an eigenvalue.
Can a matrix have real and complex eigenvalues?
In general, a real matrix can have a complex number eigenvalue. In fact, the part (b) gives an example of such a matrix.
Are complex eigenvalues Diagonalizable?
In general, if a matrix has complex eigenvalues, it is not diagonalizable.
Can a matrix have both real and imaginary eigenvalues?
If each entry of an n×n matrix A is a real number, then the eigenvalues of A are all real numbers. False. In general, a real matrix can have a complex number eigenvalue.
Is zero a real eigenvalue?
Eigenvalues may be equal to zero. We do not consider the zero vector to be an eigenvector: since A 0 = 0 = λ 0 for every scalar λ , the associated eigenvalue would be undefined.
Do all matrices have complex eigenvalues?
If the scalar field is the field of complex numbers, then the answer is YES, every square matrix has an eigenvalue. This stems from the fact that the field of complex numbers is algebraically closed.
How are complex eigenvalues related to differential equations?
With complex eigenvalues we are going to have the same problem that we had back when we were looking at second order differential equations. We want our solutions to only have real numbers in them, however since our solutions to systems are of the form,
How to solve ODE system with complex eigenvalues?
The expression (2) was not written down for you to memorize, learn, or even use; the point was just for you to get some practice in seeing how a calculation like the one in our example looks when written out in general. To actually solve ODE systems having complex eigenvalues, imitate the procedure in the following example. 2. Worked Example 1 2
Which is the formula for the eigenvalue λ?
λ= a + bi, λ¯ = a − bi . Let’s start with the eigenvalue a + bi. According to the solution method described in the note Eigenvectors and Eigenvalues, (from earlier in this ses sion) the next step would be to find the corresponding eigenvector v, by solving the equations (a −λ)a 1 + ba 2 = 0 ca 1 + (d −λ)a 2 = 0 for its components a 1 and a
Why is the eigenvector vcorresponding to λ complex?
Since λis complex, the a iwill also be com plex, and therefore the eigenvector vcorresponding to λwill have complex components. Putting together the eigenvalue and eigenvector gives us for mally the complex solution x= e(a+bi)tv. (1) Naturally, we want real solutions to the system, since it was real to start with.