Image
Kernal
Linear Independence
Basis
Dimension
Rank-Nullity Theorem
Coordinates How to Examples Exercise
How to:
In the kernal section, we learned that for each column without a leading one, we would have another vector spanning our kernal.
In the image section, we learned that all the vectors of a matrix span it, but only the vectors with leading one's are linearly independent and form a basis for the image. Also, the number of leading one's is the rank of the matrix.
So, if we have an MxN matrix, A, all its columns are going to either have leading one's or not (there isn't another choice!), and are going to contribute to the kernal or to the image. So, we have the following:
dim(kernal(A)) + dim(image(A)) = N
But, the dim(image(A)) is always equal to the rank. Also, we are going to refer to the dim(kernal(A)) as the nullity(A). So, now we have our theorem:
rank(A) + nullity(A) = N
Examples:
Let's work again with R = [3 -1 4; 4 -2 6; 5 -3 7]. We've already found out that in RREF it is: rref(R) = [1 -1 0; 0 0 1; 0 0 0] and that the basis of the image are the vectors [3; 4; 5] and [4; 6; 7]. Let's now find the kernal for R. We know:
x1 - x2 = 0
x1 = x2
x3 = 0
So, our kernal is span([1; 1; 0]). Obviously, this has a dimension of 1.
Let's check that the theorem holds here:
nullity(A) + rank(A) = N
dim(kernal(A)) + dim(image(A)) = 3
1 + 2 = 3
3 = 3
This matrix is not a special case, this theorem will hold with any matrix imaginable.
Exercise: