How do I find the dimension of an eigenspace (so that I can determine geometric

Question

How do I find the dimension of an eigenspace (so that I can determine geometric multiplicity)? From Khan Academy, I understand that the dimension of a subspace is the number of elements in a basis for that subspace, or the number of non-pivot columns in the reduced row echelon form of the matrix (the number of columns that do not contain a leading one): http://www.youtube.com/watch?v=abYAUqs_n6I But what if ALL the columns contain leading ones? Is the dimension zero? Is there no basis then? I'm confused. What if the RREF is: 1 0 0 0 1 0 0 0 1 what is the dimension of that RREF? how many vectors are in the basis? Those would all be linearly independent and they would span R^3 --- so they would form a basis? But in the video, he says that only the columns without leading ones form a basis.... I don't understand.

Explanation / Answer

The dimension of the eigenspace corresponding to eigenvalue t is the nullity (dimension of the null space) of the matrix (A - tI). From the rank-nullity theorem, for an nxn matrix, this is n minus the rank. In general the rank of a matrix is the number of linearly independent columns, and can be computed by row reducing the matrix if you can't eyeball the number of linearly independent columns. In this case, the matrix you have is A = A - 0I, and it obviously has a single linearly independent column, so the rank is 1, and the nullity is 3-1 = 2. Thus the dimension of the eigenspace is 2.

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