1. Oct 30, 2012

### charlies1902

I am a little confused about dimensions, if the nullspace of a matrix is spanned by the 0 vector, does that mean the dimension of the nullspace of this matrix is 0?

In the problems I attached, both A and B reduced to the identity matrix.
Note (2) is supposed to be dim(N(B)) and dim(col(B)) instead of dim(N(A)) and dim(col(A)).

Since they both reduce to I, that means a pivot is in every column, which means that the dimension of N(A) and N(B) is 0 right?

Consequently, this means dim(Col(A)) and dim(col(B)) is 3 right?

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2. Oct 30, 2012

### Staff: Mentor

Yes.
Right.
Yes.

Let's look at this situation from the perspective of transformations. Your matrices are 3 X 3, meaning that they represent transformations from R3 to R3 (or possibly a lower dimension subspace of R3). Since both matrices reduce to I3, both matrices map a nonzero vector in R3 to some other nonzero vector. The only vector mapped to <0, 0, 0> is the zero vector itself. Another way to say this is N(A) = N(B) = {0}.

If it had turned out that the nullspace for one of the matrices was one-dimensional, then some vectors in R3 are getting mapped to 0. For this matrix, the range would be a two-dimensional subspace of R3. In other words, a plane embedded in space. An example of this is T(<x, y, z>) = <x, y, 0>. This transformation projects a vector v down to the x-y plane.

Last edited: Oct 30, 2012