nickadams
- 182
- 0
Homework Statement
Let \vec{x} and \vec{v} be vectors in \mathbb{R}^3.
If A is a matrix such that A\vec{x} gives the projection of \vec{x} onto \vec{v}, then what are the eigenvalues of A and what are their algebraic multiplicities?
Homework Equations
Eigenvalue: A real number λ is an eigenvalue of A if there exists a nonzero vector \vec{u} such that A\vec{u}=λ\vec{u}.
Algebraic Multiplicity: The algebraic multiplicity of an eigenvalue λ of A is determined by the largest integer k>0 such that (x-λ)k divides the characteristic polynomial of A, pA(x).
The Attempt at a Solution
I feel like for any λ\in{\mathbb{R}^3} we can find a vector \vec{x} such that A\vec{x}=λ\vec{x} because letting \vec{x}=λ\vec{v} let's us say Aλ\vec{v}=λ\vec{v} since A\vec{x} just projects \vec{x} onto \vec{v} so projecting λ\vec{v} onto \vec{v} obviously just gives λ\vec{v}.
However, if for all k\in{\mathbb{R}}, \vec{u}\in{\mathbb{R}^3}\neq k\vec{v} then \vec{u} cannot possibly be an eigenvector. So this means the eigenspace is just all scalar multiples of \vec{v}.
Another thing I think may be significant is that each eigenvalue has only one eigenvector corresponding to it; does this mean the algebraic multiplicity of every eigenvalue is 1? I can't think of a way to find out if (x-λ)k divides the characteristic polynomial of A... is there another way to find out the algebraic multiplicity of an eigenvalue?
Thanks