How can I find the dimension of the eigenspace? The matrix A=[9−117] has one eigenvalue of multiplicity 2. Find this eigenvalue and the dimension of the eigenspace. So we found the eigenvalue by doing A−λI to get: λ=8 But how exactly do we find the dimension of the eigenspace?

Question

How can I find the dimension of the eigenspace?  The matrix A=[9−117] has one eigenvalue of multiplicity 2. Find this eigenvalue and the dimension of the eigenspace.  So we found the eigenvalue by doing A−λI to get:  λ=8  But how exactly do we find the dimension of the eigenspace?

sonSnubsreose6v · Accepted Answer

The dimension of the eigenspace is given by the dimension of the nullspace of A−8I=(1−11−1) , which one can row reduce to (1−100), so the dimension is 1.  Note that the number of pivots in this matrix counts the rank of A−8I. Thinking of A−8I as a linear operator from R2→R2, the dimension of the nullspace of A is given by dim(R2) −rank(A)=2−1=1 by the so-called rank-nullity theorem. Of course, one can be more explicit: it is straightforward to see that the nullspace of A−8I is spanned by the vector (1,1), and hence has dimension 1.

Wendy Boykin · Accepted Answer

By definition, an eigenvector v with eigenvalue λ satisfies Av=λv, so we have Av−λv=Av−λIv=0, where I is the identity matrix. Thus, (A−λI)v=0, and v is in the nullspace of A−λI.   Since the eigenvalue in your example is λ=8, to find the eigenspace related to this eigenvalue we need to find the nullspace of A−8I, which is the matrix   [1−11−1]  We can row-reduce it to obtain  [1−100]  This corresponds to the equation   x - y = 0.  so x=y for every eigenvector associated to the eigenvalue λ=8. Therefore, if (x,y) is an eigenvector, then (x,y)=(x,x)=x(1,1), meaning that the eigenspace is W=[(1,1)], and its dimension is 1.

How can I find the dimension of the eigenspace? The matrix \[A= \b

Answered question

Answer & Explanation

New Questions in Linear algebra