Do you think that's what they were getting at in the Wikipedia article?
If we suppose the existence of complex numbers, or allow them at any rate, is it safe to say that a square matrix of size n will always have n eigenvalues (counting multiplicities)?
Reading more from Wikipedia:
To me, it would seem that there must be n roots (counting multiplicities) for the characteristic polynomial for every square matrix of size n. In other words, every square matrix of size n must have n eigenvalues (counting multiplicities, i.e., eigenvalues are...
That's true, but a 2D rotation matrix still has eigenvalues, they just aren't real eigenvalues. But the eigenvalues still exist.
Moreover, the 2D rotation matrix isn't symmetric/Hermitian. It's usually of the form:
T = \left(\begin{array}{cc} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi...
I'm reading from Wikipedia:
I thought linear operators always had eigenvalues, since you could always form a characteristic equation for the corresponding matrix and solve it?
Is that not the case? Are there linear operators that don't have eigenvalues?
Making a change of basis in the matrix representation of a linear operator will not change the eigenvalues of that linear operator, but could making such a change of basis affect the geometric multiplicities of those eigenvalues?
I'm thinking that the answer is "no", it cannot..
Since if...
Is an ideal always a linear space?
I'm reading a proof, where the author is essentially saying: (1) since x is in the ideal I, and (2) since y is in the ideal I; then clearly x-y is in the ideal I.
In other words, if we have two elements belonging to the same ideal, is their linear...
It's true that in general, the rotation matrix does not have real eigenvalues.. however, for the general 2-dimensional rotation matrix:
\left[\begin{array}{cc} \cos \phi & -\sin \phi \\ \sin \phi & \cos \phi\end{array}\right]
it will, in general have the two (complex) eigenvalues:
\lambda_1 =...
hkBattousai, I think HallsOfIvy is correct..
In the example he gave, the matrix has only one distinct eigenvalue (which is 1, w/ algebraic multiplicity of 2), and there is only one eigenvector corresponding to this eigenvalue (so the geometric multiplicity of the eigenvalue is 1).
I...
Is there such a thing as a square matrix with no eigenvectors?
I'm thinking not ... since even if you have:
\left[\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right]
you could just as well say that the eigenvalue(s) are 0 (w/ algebraic multiplicity 2) and the eigenvectors are:
u_1 =...
Suppose we have a linear transformation/matrix A, which has multiple left inverses B1, B2, etc., such that, e,g,:
B_1 \cdot A = I
Can we conclude from this (i.e., from the fact that A has multiple left inverses) that A has no right inverse?
If so, why is this?
I just want to test/verify my knowledge of change of basis in a linear operator.. (it's not a homework question).
Suppose I have linear operator mapping R^2 into R^2, and expressed in the canonical basis (1,0), (0,1). Suppose (for the sake of discussion) that the linear operator is given by...
Yes, thanks..
So algebraic multiplicity is the number of times the eigenvalue appears as a root in the characteristic equation.
Geometric multiplicity is the dimension of the subspace formed by the eigenvectors of that particular eigenvalue..
And the algebraic multiplicity is always...
Suppose I have a linear operator of dimension n, and suppose that this operator has a non-trivial null space. That is:
A \cdot x = 0
Suppose the dimension of the null space is k < n, that is, I can find 0 < k linearly independent vectors, each of which yields the 0 vector when the linear...