## 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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