4.

The kernel of A = [1 2 3] is the set of solutions of
x+2y+3z=0. Note that A is already in rref. The variables
y, z are free and the general solution is

-2y-3z

{ y : y, z in R}

z

which you can write as

-2 -3

y 1 + z 0

0 1

The column vectors

-2

1

0

and

-3

0

1

span the kernel, clearly.

16.

With A =

1 2 3

1 2 3

1 2 3

the image is spanned by

1

1

1

-- or any of the other columns. Note that the second colum is twice the first and the third column is three times the first, so they are not needed.

44.

a. Yes, A and B = rref(A) have the same kernel. The kernel of
A is the set of solutions of AX = 0, and the kernel of B is the
set of solutions of BX = 0. One can get from A to B by
elementary row operations, and so one can get from the equations
AX = 0 to BX = 0 by elementary row operations. We have known since
the first week that elementary row operations do not affect the set
of solutions of a system of equations.

b. No, in general the images of A and rref(A) are different. For example, if we take A =

0

1

then rref(A)
1

0

has a different image (column space). The first matrix has column space Span{e_2} and the second has column space Span{e_1}. These are two different lines.

Section 3.2.

6.

a. The intersection of two subspaces V, W of R^n IS always a subspace. Note that since 0 is in both V, W it is in their intersection. Second, note that if z, z' are two vectors that are in the intersection then their sum is in V (because V is a subspace and so closed under addition) and their sum is in W, similarly. Thus, their sum is in the intersection, and so the intersection is closed under addition. Finally, if z is in both V, W and c is a scalar then cz is in V because V is a subspace and it is in W similarly. Thus, cz is in the interection. The intersection is therefore a subsspace of R^n.

b. In general, the union of two subspaces of R^n is not a subspace. For example, the union of the span of e_1 and the span of e_2 in R^2 consists of all vectors that are on one coordinate axis or the other, and does not contain e_1 + e_2, which is not on either axis. Since the union is not closed under vector addition, it is not a subspace. (More generally, the union of two subspaces is not a subspace unless one is contained in the other. One can check that if v is in V and not in W and w is in W and not in V, then v + w is not in either V or W, i.e., it is not in the union.)

16.

Make the three given vectors into the columns of a matrix A:

1 4 7

2 5 8

3 6 9

To check for independence, one needs to check whether there is a nonzero relation on the vectors, i.e., where AX = 0 has any nonzero solutions, where X =

x

y

z

The most straightforward method is to find the rref:
1 4 7

0 -3 -6

0 -6 -12

1 4 7

0 1 2

0 -6 -12

1 0 -1

0 1 2

0 0 0

All that is really needed here is to say that since the rank is 2, there are nonzero solutions, and so the vectors are linearly DEPENDENT. However, here are some more detailed comments.

Looking at the rref one sees that the general solution of the equations is given by x = z, y = -2z, i.e. the solution set is

z

{-2z : z in R}

z

which is spanned by

1

-2

1

Thus the vectors are linearly dependent, and not linearly independent. (In general, the columns of a square n by n matrix are independent if and only if the rank is n, which is equivalent to saying that the matrix is invertible or that rref is the n by n identity matrix.)

28.

We want to find a basis for the column space of the matrix A =

1 1 1

1 2 5

1 3 7

We proceed by finding the rref as follows:

1 1 1

0 1 4

0 2 6

1 0 -3

0 1 4

0 0 -2

(one can now see that the rank is 3 and quit, but here's the rest anyway):

1 0 -3

0 1 4

0 0 1

1 0 0

0 1 0

0 0 1

Since the matrix is invertible, the kernel is 0, which means that the columns are independent. Therefore, they give a basis for the image, which must be all of R^3. (The answer to a question like this is not unique. For example, in this problem any three independent vectors in R^3 will be a basis for the image. Thus, one could also use the standard basis e_1, e_2, e_3.)