## Transcript

Once one has defined a linear mapping L from Cn to Cm, and found its m-by-n matrix bracket-[L] = A, we can find the following subspaces of Cn and Cm.

The nullspace of an m-by-n matrix A is Null(A), and is equal to the set of all z in Cn such that Az equals the 0 vector.

If we let A be an m-by-n matrix, and z1, z2, through zn be the columns of A, then the columnspace of A, written Col(A), is the span of the vectors {z1 through zn}. Recall that the columnspace of A is the same as the range of the linear mapping Az, or the corresponding linear mapping L(z).

Given an m-by-n matrix A, the rowspace of A is the subspace spanned by the rows of A (regarded as vectors) and is denoted Row(A).

We can find a basis for each of these subspaces the same as we did in Rn.

So let’s look at an example. To find a basis for the rowspace, columnspace, and nullspace of the matrix A seen here, we first need to find the reduced row echelon form of A. So first, I’ll replace row 2 with (row 2 + (2i)(row 1)), and replace row 3 with (row 3 + (1 – i)(row 1)) to get this matrix. Next, I can multiply row 2 by (-1 – i)/2 to get this matrix. Then I can replace row 3 with (row 3 + (-3 – i)(row 2)) to get this matrix. Then I can multiply row 3 by 1/3 to get this matrix. Now I can replace row 1 with (row 1 + (-1 – 2i)(row 3)), and replace row 2 with (row 2 – 2(row 3)) to get this matrix. And last, I’ll replace row 1 with (row 1 – i(row 2)) to get this matrix. Now we have a matrix in reduced row echelon form.

To find a basis for the rowspace of A, we first need to recall that, for any matrix B that is row equivalent to A, the rowspace of B is the same as the rowspace of A. So in particular, the rowspace of the reduced row echelon form of A is the same as the rowspace of A. And the structure of any reduced row echelon form matrix ensures that the non-zero rows are linearly independent. So just look at the entries that correspond to the leading 1—only one row has a non-zero entry. Since our particular reduced row echelon form does not have any rows of all zeroes, we find that the rows of the reduced row echelon form of A form a basis for the rowspace of A. So we’ll see it here.

To find a basis for the columnspace of A, we need to determine which columns of A are linearly independent, and which are dependent members. Now if we think of A as the coefficient matrix for the equation seen here, then the RREF of A tells us that alpha1 + 2(alpha3) will equal 0, alpha2 + 3(alpha3) – i(alpha4) would equal 0, and alpha5 would equal 0. If we replace the variable alpha3 with the parameter beta, and the variable alpha4 with the parameter gamma, then we see that the general solution to this equation is that our alpha vector will equal (beta)[-2; -3; 1; 0; 0] + (gamma)[0; i; 0; 1; 0].

If we plug in beta = 1 and gamma = 0, we see that the vector [-2; -3; 1; 0; 0] is a solution, which means that (-2 times our first column) – (3 times our second column) + (our third column) + (0 times our fourth column) + (0 times our fifth column) equals the 0 vector. And this means that our third column equals (2 times our first column) + (3 times our second column). As such, our third column is a dependent member. By plugging in beta = 0 and gamma = 1, we would see that [0; i; 0; 1; 0] is a solution to our equation, and going through a similar process, it means that our fourth column equals (-i) times our second column, and so our fourth column is also a dependent member.

So thus, we know that the span of our five columns is simply the span of the first, second, and fifth columns. To see that this latter set is independent, we would row reduce the matrix with just those three columns, but we would use the exact same steps as before, and we would end up with the identity matrix. Since the rank of this matrix is the same as the number of columns, our set is linearly independent. As such, we see that the columns of A that correspond to the columns of the reduced row echelon form that have a leading 1 will form a basis for the columnspace of A. Again, this is the exact same technique we used when looking for the columnspace in the real numbers.

Now, to find a basis for the nullspace, we need to find the general solution to the equation Az = 0. Now luckily, we’ve already done this. In our efforts to find a basis for the columnspace, we found that the general solution to Az = 0 is that z will equal (beta)[-2; -3; 1; 0; 0] + (gamma)[0; i; 0; 1; 0], and this means that the set {[-2; -3; 1; 0; 0], [0; i; 0; 1; 0]} is a spanning set for the nullspace. And as these vectors are clearly not multiples of each other, even in the complex numbers, this set is linearly independent, and so this set is a basis for the nullspace.

In this example, I have reviewed the theory behind finding these bases as well as the technique. You will not need to restate this entire process on the assignment and on the exam. Simply finding the reduced row echelon form and obtaining the correct basis from the RREF will suffice. On the other hand, if you need more of a refresher on these techniques, refer back to Section 3.4 in the text.

In general, much of what we learned about real vector spaces will translate directly into complex vector spaces, since much of it depends on the fundamental properties of vector spaces, and not the real or the complex numbers. Instead of continuing to focus on all the ways that complex vector spaces are the same as real vector spaces, we instead will turn our attention to the ways in which they are different, perhaps taking for granted a few similarities not yet stated along the way.