Finding Bases for Image and Kernel

Some Preliminary Programs

As usual, we'll start by inputting the matrix into Mathematica.

To find bases for the image and kernel, we first row reduce the matrix.

The theorem we stated in class says that the columns of A which correspond to a pivot column in the reduced row echelon form of A form a basis for the image of A.  Since the first, third, sixth, seventh and eighth columns of A are the pivot columns, this means the vectors

form a basis for the image of A.

What about a basis for the kernel of A?  In class yesterday we found a spanning set for the kernel by looking at the rows of rref(A); check out yesterday's Mathematica example if you don't remember how we did this.  We'll do the same procedure today by looking at the columns of rref(A) which don't contain a pivot.  You an do it either way in practice; whichever way works best for you is the one you should use.

Notice that the second column of rref(A) isn't a pivot column.  We generate a vector v in the kernel of A as follows.  First, since the 2nd column of rref(A) isn't a pivot column, the second entry in our vector v will be 1.  Now we notice that the 1st coordinate of the second column of rref(A) is 2.  Since the first row corresponds to the first pivot, which is in the 1st column, we'll put a -2 in the first entry of our vector v.  There are no other non-zero entries in the second column of rref(A), so the other coordinates of v will be 0's.  We're left with

We move on to the next non-pivot column of rref(A), and we'll build a new vector w in the kernel of A.  Since the fourth column of rref(A) is non-pivot, w will have a 1 in the 4th coordinate.  The 2 in the second entry of the 4th column of rref(A) tells me I should put a -2 in the 3rd entry of w.  Why the third entry?  Because the second pivot is in the 3rd column of rref(A)!  The 1 in the 1st entry of the 4th column of rref(A) tells me to put a -1 in the 1st entry of w.  Why the first entry?  Because the first pivot is in the 1st column of A. These are the only nonzero coefficients in the 4th column of rref(A), and so the other entries of w are 0.

We move on to the next non-pivot column of rref(A) and build a vector u in the kernel of A.  Since the fifth column of rref(A) is the next non-pivot, u will have a 1 in the 5th coordinate.  The 1 in the second coordinate of the fifth column of rref(A) tells me I should put a -1 in the 3rd entry of u (again, the third entry because the second pivot is the 3rd column of A).  The 1 in the second coordinate of the fifth column of rref(A) tells me I should put a -1 in the 1st entry of u (since the first pivot is in the first column).

The next (and last) non-pivot column is the ninth, and it will produce a vector z in the kernel of A as follows.  Since the ninth column of rref(A) is the non-pivot we're considering, z will have a 1 in the 9th position.  Since there is a 3 in the 5th position of the ninth column of rref(A), we will put a -3 in the 8th entry of z (the 8th entry because the 5th pivot is in the 8th column of rref(A)).  The other entries of the ninth column of rref(A) are zero, and so all other entries of z are zero.

We know that all these vectors are in the kernel of A, but the content of the theorem we stated in class is that these vectors actually form a basis for the kernel of A.  In summary, we have shown that a basis for the kernel of A is the collection of vectors