Linear Algebra and Its Applications, Exercise 2.4.19

Exercise 2.4.19. For the following matrix

\begin{bmatrix} 0&1&2&3&4 \\ 0&1&2&4&6 \\ 0&0&0&1&2 \end{bmatrix} = \begin{bmatrix} 1&0&0 \\ 1&1&0 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 0&1&2&3&4 \\ 0&0&0&1&2 \\ 0&0&0&0&0 \end{bmatrix}

find a basis for each of the four associated subspaces.

Answer: Per the above equation the matrix A on the left side can be factored into a lower triangular matrix L with unit diagonal and an upper triangular matrix U.

The matrix U is what is obtained from A through the process of Gaussian elimination (with the matrix L containing the multipliers). The row space of A is therefore the same as the row space of U. The two nonzero rows of U

\begin{bmatrix} 0 \\ 1 \\ 2 \\3 \\ 4 \end{bmatrix} \quad \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 2 \end{bmatrix}

are a basis for the row space \mathcal R(A^T) = \mathcal R(U^T).

The matrix U has pivots in the second and fourth columns; those columns are linearly independent and form a basis for the column space \mathcal R(U). The corresponding second and fourth columns of A

\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} \quad \begin{bmatrix} 3 \\ 4 \\ 1 \end{bmatrix}

are also linearly independent and form a basis for the column space \mathcal R(A).

The nullspace of A is the same as the nullspace of U, which contains all solutions to

Ux = \begin{bmatrix} 0&1&2&3&4 \\ 0&0&0&1&2 \\ 0&0&0&0&0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = 0

Since U has pivots in the second and fourth columns the variables x_2 and x_4 are basic variables with the others being free variables. From the second row above we have x_4 + 2x_5 = 0 or x_4 = -2x_5. Substituting into the first row we have

x_2 + 2x_3 + 3x_4 +4x_5

= x_2 + 2x_3 - 6x_5 +4x_5

= x_2 + 2x_3 - 2x_5 = 0

or x_2 = -2x_3 + 2x_5.

Setting each of the free variables to 1 in turn and the others to zero, a solution for Ux = 0 and thus Ax = 0 is

x = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 2 \\ 0 \\ -2 \\ 1 \end{bmatrix}

The vectors

\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 2 \\ 0 \\ -2 \\ 1 \end{bmatrix}

are thus a basis for the nullspace \mathcal N(A) = \mathcal N(U).

Finally we consider the left nullspace of A. Since U has two pivots the rank of U and therefore A is r = 2. The dimension of \mathcal N(A^T) is therefore m - r = 3 - 2 = 1.

Since A = LU we have L^{-1}A = U. The last row of L^{-1} is a basis for the (1-dimensional) left nullspace \mathcal N(A^T) consisting of those y such that A^Ty = 0 or y^TA = 0. However we do not need to compute L^{-1}.

Instead to find a suitable y we can find the coefficients that make the rows of A combine to form the zero row of U. Gaussian elimination on A proceeds by subtracting the first row of A from the second row:

\begin{bmatrix} 0&1&2&3&4 \\ 0&1&2&4&6 \\ 0&0&0&1&2 \end{bmatrix} \Rightarrow \begin{bmatrix} 0&1&2&3&4 \\ 0&0&0&1&2 \\ 0&0&0&1&2 \end{bmatrix}

and then subtracting the second row of the resulting matrix from the third row:

\begin{bmatrix} 0&1&2&3&4 \\ 0&0&0&1&2 \\ 0&0&0&1&2 \end{bmatrix} \Rightarrow \begin{bmatrix} 0&1&2&3&4 \\ 0&0&0&1&2 \\ 0&0&0&0&0 \end{bmatrix} = U

The zero row in U therefore is composed of 1 times the first row of A plus -1 times the second row of A plus 1 times the third row of A. The vector

\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}

is therefore a basis for the left nullspace \mathcal N(A^T).

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. 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