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Mathematics 18 Online
OpenStudy (anonymous):

Find the eigenspaces of the matrix and describe them geometrically [15 6 -4; -40 -16 11; 0 0 -1] where λ=-1 and 0

OpenStudy (anonymous):

I got the eigenvector for -1 with (-3/8 1 0)

OpenStudy (anonymous):

Since it is a one dimensional eigenspace (its only 1 eigenvector), the eigenspace is a line through the origin in the direction of that vector

OpenStudy (anonymous):

er, strike that part about "only one eigenvector", i meant to to say because the eigenspace is only spanned by one vector

OpenStudy (anonymous):

yeah λ=0 also has one it's (-2/5 1 0) I think but the answer for the eigenspace is t(2, -5,0) and t( 3, -8 0) I don't know how they got that?

OpenStudy (anonymous):

How does that relate to the eigenvector because it's like the numerator and denominator of the fractions

OpenStudy (anonymous):

i guess they dont like fractions in their eigenvectors, so they multiply by the denominators to make all the numbers integers. Any multiple of an eigenvector is also an eigenvector because if x is an eigenvector and c is a constant: \[A(cx) = cAx = c\lambda x = \lambda (cx)\]

OpenStudy (anonymous):

But whats the difference between eigenvector and eigenspace are they one of the same?

OpenStudy (anonymous):

Because the 1 is not included in the eigenspace

OpenStudy (anonymous):

OH I get what you mean

OpenStudy (anonymous):

it seems like both of your eigenspaces are one dimensional, so they are both lines through the origin going in the direction of the vectors that created them.

OpenStudy (anonymous):

Oh I see so if they weren't one dimensional the eigenvectors and eigenspaces won't necessarily be the same

OpenStudy (anonymous):

yeah, lets say you had gotten 2 free variables and had a two dimensional eigenspace, then the geometric interpretation of that would be the plane spanned by the two vectors.

OpenStudy (anonymous):

then, as long as a vector is in that plane (as long as the vector is a linear combination of the 2 vectors that make the plane), it is an eigenvector.

OpenStudy (anonymous):

Alright make sense thanks a lot for your help. You sure know your stuff about eigenvectors etc.

OpenStudy (anonymous):

No prob :) Linear Algebra is my favorite subject, i wish more people would ask questions about it lol

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