- #1
nickadams
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Homework Statement
Let [itex]\vec{x}[/itex] and [itex]\vec{v}[/itex] be vectors in [itex]\mathbb{R}^3[/itex].
If A is a matrix such that A[itex]\vec{x}[/itex] gives the projection of [itex]\vec{x}[/itex] onto [itex]\vec{v}[/itex], 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 [itex]\vec{u}[/itex] such that A[itex]\vec{u}[/itex]=λ[itex]\vec{u}[/itex].
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[itex]\in{\mathbb{R}^3}[/itex] we can find a vector [itex]\vec{x}[/itex] such that A[itex]\vec{x}[/itex]=λ[itex]\vec{x}[/itex] because letting [itex]\vec{x}[/itex]=λ[itex]\vec{v}[/itex] let's us say Aλ[itex]\vec{v}[/itex]=λ[itex]\vec{v}[/itex] since A[itex]\vec{x}[/itex] just projects [itex]\vec{x}[/itex] onto [itex]\vec{v}[/itex] so projecting λ[itex]\vec{v}[/itex] onto [itex]\vec{v}[/itex] obviously just gives λ[itex]\vec{v}[/itex].
However, if for all [itex]k\in{\mathbb{R}}[/itex], [itex]\vec{u}\in{\mathbb{R}^3}\neq k\vec{v}[/itex] then [itex]\vec{u}[/itex] cannot possibly be an eigenvector. So this means the eigenspace is just all scalar multiples of [itex]\vec{v}[/itex].
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