Review exercise 2.26. State whether the following statements are true or false:

a) For every subspace of there exists a matrix for which the nullspace of is .

b) For any matrix , if both and its transpose have the same nullspace then is a square matrix.

c) The transformation from to that transforms into (for some scalars and ) is a linear transformation.

Answer: a) For the statement to be true, for any subspace of we must be able to find a matrix such that . (In other words, for any vector in we have and for any such that the vector is an element of .)

What would such a matrix look like? First, since is a subspace of any vector in has four entries: . If we have then the matrix must have four columns; otherwise would not be able to multiply from the left side.

Second, note that the nullspace of and the column space of are related: If the column space has dimension then the nullspace of has dimension where is the number of columns of . Since has four columns (from the previous paragraph) the dimension of is .

Put another way, if is the dimension of and is the nullspace of then we must have or . So if the dimension of is then we have , if the dimension of is then we have , and so on.

Finally, if the matrix has rank then has both linearly independent columns and linearly independent rows. As noted above, the number of columns of (linearly independent or otherwise) is fixed by the requirement that the nullspace of should be a subspace of . So must always have four columns, and must have at least rows.

We now have five possible cases to consider, and can approach them as follows:

- has dimension 0 or 4. In both these cases there is only one possible subspace and we can easily find a matrix meeting the specified criteria.
- has dimension 1, 2, or 3. In each of these cases there are many possible subspaces (an infinite number, in fact). For a given we do the following:
- Start with a basis for . (Any basis will do.)
- Consider the effect on the basis vectors if they were to be in the nullspace of some matrix meeting the criteria above.
- Take the corresponding system of linear equations, re-express it as a system involving the entries of as unknowns and the entries of the basis vectors as coefficients, and show that we can solve the system to find the unknowns.
- Show that all other vectors in are also in the nullspace of .
- Show that any vector in the nullspace of must also be in .

We now proceed to the individual cases:

has dimension . We then have . (If has dimension 4 then its basis has four linearly independent vectors. If then there must be some vector in but not in , and that vector must be linearly independent of the vectors in the basis of . But it is impossible to have five linearly independent vectors in a 4-dimensional vector space, so we conclude that .)

We then must have for any vector in . This is true when is equal to the zero matrix. As noted above must have exactly four columns and at least rows. So one possible value for is

(The matrix could have additional rows as well, as long as they are all zeros.)

We have thus found a matrix such that any vector in is also in . Going the other way, any vector in must have four entries (in order for to be able to multiply it) so that any such vector is also in .

So if is a 4-dimensional subspace (namely ) then a matrix exists such that is the nullspace of .

has dimension . The only 0-dimensional subspace of is that consisting only of the zero vector . (If contained only a single nonzero vector then it would not be closed under multiplication, since multiplying that vector times the scalar 0 would produce a vector not in . If were not closed under multiplication then it would not be a subspace.)

In this case the matrix would have to have rank . If then all four columns of would have to be linearly independent and would have to have at least four linearly independent rows. Suppose we choose the four elementary vectors through as the columns, so that

If we then have

so that the only solution is . We have thus again found a matrix for which .

(Note that any other matrix of rank would have worked as well: The product is a linear combination of the columns of , with the coefficients being through . If the columns of are linearly independent then that linear combination can be zero only if all the coefficients through are zero.)

Having disposed of the easy cases, we now proceed to the harder ones.

has dimension . In this case we are looking for a matrix with rank such that . The matrix thus must have only one linearly independent column and (more important for our purposes) only one linearly independent row. We need only find a matrix that is 1 by 4. (If desired we can construct suitable matrices that are 2 by 4, 3 by 4, etc., by adding additional rows that are multiples of the first row.)

Since the dimension of is 3, any three linearly independent vectors in form a basis for ; we pick an arbitrary set of such vectors , , and . For to be equal to we must have , , and . We are looking for a matrix that is 1 by 4, so these equations correspond to the following:

These can in turn be rewritten as the following system of equations:

This is a system of three equations in four unknowns, equivalent to the matrix equation where

Since the vectors , , and are linearly independent (because they form a basis for the 3-dimensional subspace ) and the vectors in question form the rows of the matrix , the rank of is . We also have , the number of rows of . Since this is true, per 20Q on page 96 there exists at least one solution to the above system . But is simply the first and only row in the matrix we were looking for, so this in turn means that we have found a matrix for which , , and are in the nullspace of .

If is a vector in then can be expressed as a linear combination of the basis vectors , , and for some set of coefficients , , and . We then have

So any vector in is also an element in the nullspace of .

Suppose that is a vector in the nullspace of and is not in . Since is not in it cannot be expressed as a linear combination solely of the basis vectors , , and ; rather we must have

where is some vector that is linearly independent of , , and .

If is in the nullspace of then we have so

If then either or . If then we have

so that is actually an element of , contrary to our supposition. If then is an element of the nullspace of . But has dimension 3 and already contains the three linearly independent vectors , , and . The fourth vector cannot be both an element of and also linearly independent of , , and .

Our assumption that is in the nullspace of but is not in has thus led to a contradiction. We conclude that any element of is also in . We previously showed that any element of is also in , so we conclude that .

For any 3-dimensional subspace of we can therefore find a matrix such that is the nullspace of .

has dimension . In this case we are looking for a matrix with rank such that . The matrix thus must have only two linearly independent columns and only two linearly independent rows. We thus look for a matrix that is 2 by 4.

Since the dimension of is 2, any two linearly independent vectors in form a basis for ; we pick an arbitrary set of such vectors and . For to be equal to we must have and . We are looking for a matrix that is 2 by 4, so these equations correspond to the following:

These can in turn be rewritten as the following system of four equations in eight unknowns:

or where

Since the four rows are linearly independent (this follows from the linear independence of and ) we have so that the system is guaranteed to have a solution . The entries in are just the entries of so we have found a matrix for which the basis vectors and are members of the nullspace.

Since the basis vectors of are in all other elements of are in also. By the same argument as in the 3-dimensional case, any vector in must be in also; otherwise a contradiction occurs. Thus we conclude that .

For any 2-dimensional subspace of we can therefore find a matrix such that is the nullspace of .

has dimension . In this case we are looking for a matrix with rank such that . The matrix thus must have only three linearly independent columns and only three linearly independent rows. We thus look for a matrix that is 3 by 4.

Since the dimension of is 1, any nonzero vector in forms a basis for . For to be equal to we must have . We are looking for a matrix that is 3 by 4, so this equation corresponds to the following:

This can be rewritten as a system of three equations with twelve unknowns (), a system which is guaranteed to have at least one solution due to the linear independence of the three rows of . The solution then forms the entries of , so that we have . Since is a basis for any other vector in is also in the nullspace of .

By the same argument as in the 3-dimensional case, any vector in must be in also; otherwise a contradiction occurs. Thus we conclude that .

For any 1-dimensional subspace of we can therefore find a matrix such that is the nullspace of .

Any subspace of must have dimension from 0 through 4. We have thus shown that for any subspace of we can find a matrix such that . The statement is true.

b) Suppose that for some by matrix both and its transpose have the same nullspace.

The rank of is also the rank of . The rank of is then , and the rank of is . Since we then have so that .

The number of rows of is the same as the number of columns of so that (and thus ) is a square matrix. The statement is true.

c) If represents the transformation in question, with , we have

These two quantities are not the same unless , so for the transformation is not linear. The statement is false.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.