Linear Algebra and Its Applications, Review Exercise 2.8

Review exercise 2.8. Do the following:

a) Find a matrix whose nullspace contains the vector x = (1, 1, 2).

b) Find a matrix whose left nullspace contains the vector y = (1, 5).

c) Find a matrix with column space spanned by (1, 1, 2) and row space spanned by (1, 5).

d) Given an arbitrary set of three vectors in \mathbb{R}^6 and a second arbitrary set of three vectors in \mathbb{R}^5 determine whether a 6 by 5 matrix exists for which the first three vectors span the column space and the second three vectors span the row space.

Answer: a) One way to find a matrix satisfying this criterion is simply to try different combinations of the entries of (1, 1, 2). For example, if we add the first two entries of (1, 1, 2) and subtract the third entry we get zero. Similarly, if we multiply the second element of (1, 1, 2) by 2 and subtract the third entry we also get zero. So if we have the matrix

A = \begin{bmatrix} 1&1&-1 \\ 0&2&-1 \end{bmatrix}

then the vector (1, 1, 2) is in the nullspace of A:

\begin{bmatrix} 1&1&-1 \\ 0&2&-1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

b) We are looking for a matrix B for which (1, 5) is in the left nullspace of B. This means that (1, 5) would be in the nullspace of B^T. An easy way to find a matrix B is thus to find a matrix (call it B^T) for which (1, 5) is in the nullspace and then transpose that matrix to obtain B.

Note that if we multiply the first entry in (1, 5) by 5 and then subtract the second entry we get 0. We also get zero if we multiply the first entry by 10 and subtract 2 times the second entry. So if we have

B^T = \begin{bmatrix} 5&-1 \\ 10&-2 \end{bmatrix}

then the vector (1, 5) is in the nullspace of B^T and thus is in the left nullspace of

B = \begin{bmatrix} 5&10 \\ -1&-2 \end{bmatrix}

c) The easiest way to construct the desired matrix is to make the first row of the matrix (1, 5) and the first column of the matrix (1, 1, 2); since the first entry of each vector is 1 this works out:

C = \begin{bmatrix} 1&5 \\ 1&? \\ 2&? \end{bmatrix}

We then fill out the other entries of C so that the second and third rows are multiples of the first row, and the second column is a multiple of the first column:

C = \begin{bmatrix} 1&5 \\ 1&5 \\ 2&10 \end{bmatrix}

This means that the column space and row space of C depend solely on the first column and first row respectively.

d) There are at least two possible ways to approach this problem: To find a way to construct a 6 by 5 matrix that has the desired property, no matter the choice of vectors, or to find a counter-example that shows that for at least some sets of vectors we cannot construct such a matrix. We’ll take the latter approach.

Note that since the sets of vectors are arbitrary we can choose vectors that are linearly independent or linearly dependent. For example, suppose from \mathbb{R}^6 we choose three linearly independent vectors to span the column space of the matrix. Also suppose that from \mathbb{R}^5 we choose three vectors that are linearly dependent (for example, with the second two vectors multiples of the first vector) and wish to have these vectors span the row space of the matrix.

But this is impossible: If the column space of the matrix is spanned by the first set of three linearly independent vectors then the dimension of the column space will be 3. (Recall that a set of linearly independent vectors that spans a space forms a basis for the space, with the dimension of the space equal to the number of basis vectors.) On the other hand, if the row space of the matrix is spanned by the second set of three linearly dependent vectors then the dimension of the row space will be less than 3. (For example, if the second and third vectors in the set are multiples of the first one then there is only one linearly independent vector in the set. If the set spans the row space then that first vector forms a basis for the space, and the dimension of the space will be 1.)

Since the dimension of the row space of a matrix must equal the dimension of the column space, we have a contradiction.

Here is a concrete example of such a 6 by 5 matrix, with a set of three linearly independent vectors chosen to be columns 1, 2, and 3, and a second set of three linearly dependent vectors chosen to be rows 4, 5, and 6:

\begin{bmatrix} 1&0&0&0&0 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 1&1&1&0&0 \\ 2&2&2&0&0 \\ 3&3&3&0&0 \end{bmatrix}

Note that the three linearly independent columns do indeed span the column space, since the last two columns are zero and thus are linearly dependent on the first three. However the last three rows do not span the row space, since the first three rows cannot be expressed as linear combinations of the last three rows. (Instead rows 4 through 6 can be expressed as linear combinations of rows 1 through 3; rows 1 through 3 are linearly independent and form a basis for the row space. The dimension of the row space is thus 3, the same as the dimension of the column space.)

We conclude that we cannot construct a 6 by 5 matrix whose column space is spanned by an arbitrary set of three vectors from \mathbb{R}^6 and whose row space is spanned by an arbitrary set of three vectors from \mathbb{R}^5.

UPDATE: Rewrote the answer to (d) to make it more clear.

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.

 Buy me a snack to sponsor more posts like this!

This entry was posted in linear algebra and tagged , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s