## Linear Algebra and Its Applications, Exercise 2.4.4

Exercise 2.4.4. For the matrix

$A = \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}$

describe each of its four associated subspaces.

Answer: We first consider the column space $\mathcal R(A)$. The matrix $A$ has two pivots (in the second and third columns) and therefore rank $r = 2$; this is the dimension of $\mathcal R(A)$. The pivot columns

$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$

serve as a basis for $\mathcal R(A)$. The column space is the x-y plane in $\mathbf R^3$.

The row space $\mathcal{R}(A^T)$ also has dimension $r = 2$. The two nonzero rows of $A$, the vectors

$\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \qquad \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

form a basis for $\mathcal{R}(A)$. The row space is the y-z plane in $\mathbf R^3$.

We now turn to the nullspace $\mathcal N(A)$ consisting of the solutions to $Ax = 0$. Since the pivots of $A$ are in the second and third columns we have $x_2$ and $x_3$ as basic variables and $x_1$ as the free variable. For $Ax = 0$ we must have

$\begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = 0$

From the second row of the above system we have $x_3 = 0$ and from the first row we have $x_2 = 0$. Setting the free variable $x_1$ to 1 gives us

$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$

as a solution to $Ax = 0$ and a basis for the null space of $A$. The null space is the x-axis in $\mathbf R^3$.

Finally we turn to the left nullspace $\mathcal N(A^T)$ consisting of all solutions to $A^Ty = 0$ or $y^TA = 0$. In general we can find the left nullspace by looking at the operations on the rows of $A$ needed to produce zero rows in the echelon matrix produced by Gaussian elimination.

In this case $A$ is already in echelon form, with the third row already zero. In terms of the elimination process this corresponds to taking the third row of $A$ as is, with no contributions from the first and second rows; the coefficients for this operation are thus 0 (for the first row), 0 (for the second row), and 1 (for the third). The vector

$\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

is therefore a basis for the left nullspace $\mathcal N(A^T)$ (which has dimension 1). We can test this by multiplying $A$ on the left by the transpose of this vector:

$\begin{bmatrix} 0&0&1 \end{bmatrix} \begin{bmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 0&0&0 \end{bmatrix}$

We can also compute the left nullspace $\mathcal N(A^T)$ by solving the system $A^Ty$ or

$\begin{bmatrix} 0&0&0 \\ 1&0&0 \\ 0&1&0 \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0&0 \\ 0&0 \end{bmatrix}$

Since the pivots of $A^T$ are in the first and second columns we have $y_1$ and $y_2$ as basic variables and $y_3$ as the free variable. From the second row of the above system we have $y_1 = 0$ and from the third row we have $y_2 = 0$. Setting the free variable $y_3 = 1$ then gives us the vector

$\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$

as a basis for the left nullspace of $A$. The left nullspace is the z-axis in $\mathbf R^3$.

Note that the column space $\mathcal R(A)$ (the x-y plane) is perpendicular to the left nullspace $\mathcal N(A^T)$ (the z-axis), while the row space $\mathcal R(A^T)$ (the y-z plane) is perpendicular to the nullspace $\mathcal N(A)$ (the x-axis). This is a foreshadowing of the discussion in section 3.1 on orthogonal subspaces.

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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