## 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 .

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