## Linear Algebra and Its Applications, Review Exercise 2.6

Review exercise 2.6. Given the matrices

$A = \begin{bmatrix} 1&2 \\ 3&6 \end{bmatrix} \qquad B = \begin{bmatrix} 0&0 \\ 1&2 \end{bmatrix} \qquad C = \begin{bmatrix} 1&1&0&0 \\ 0&1&0&1 \end{bmatrix}$

find bases for each of their four fundamental subspaces.

Answer: The second column of $A$ is equal to twice the first column so the rank of $A$ (and the dimension of the column space of $A$) is 1. The first column $(1, 3)$ is a basis for the column space of $A$.

The dimension of the row space of $A$ is also 1 (the rank of $A$), and the first row $(1, 2)$ is a basis for the row space of $A$.

We can reduce $A$ to echelon form by subtracting 3 times the first row from the second row:

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

The resulting echelon matrix has one pivot, with $x_1$ a basic variable and $x_2$ a free variable. Setting $x_2 = 1$ we have from the first row of the echelon matrix $x_1 + 2x_2 = x_1 + 2 \cdot 1 = 0$ or $x_1 = -2$. The vector $(-2, 1)$ is thus a basis for the nullspace of $A$.

The left nullspace of $A$ is the nullspace of

$A^T = \begin{bmatrix} 1&3 \\ 2&6 \end{bmatrix}$

We can reduce $A^T$ to echelon form by subtracting 2 times the first row from the second row:

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

The resulting echelon matrix has one pivot, with $x_1$ a basic variable and $x_2$ a free variable. Setting $x_2 = 1$ we have from the first row of the echelon matrix $x_1 + 3x_2 = x_1 + 3 \cdot 1 = 0$ or $x_1 = -3$. The vector $(-3, 1)$ is thus a basis for the nullspace of $A^T$ and for the left nullspace of $A$.

As with $A$, the second column of $B$ is equal to twice the first column, so the rank of $B$ (and the dimension of the column space of $B$) is 1. The first column $(0, 1)$ is a basis for the column space of $B$.

The dimension of the row space of $B$ (the rank of $B$) is also 1, and the second row $(1, 2)$ is a basis for the row space of $B$. Note that this is the same as the basis for the row space of $A$ so that the row spaces of $A$ and $B$ are identical.

We can transform the matrix $B$ to echelon form simply by exchanging the first and second rows:

$\begin{bmatrix} 0&0 \\ 1&2 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2 \\ 0&0 \end{bmatrix}$

The resulting echelon matrix has one pivot, with $x_1$ a basic variable and $x_2$ a free variable. Setting $x_2 = 1$ we have from the first row of the echelon matrix $x_1 + 2x_2 = x_1 + 2 \cdot 1 = 0$ or $x_1 = -2$. The vector $(-2, 1)$ is thus a basis for the nullspace of $B$. This is the same as the basis for the nullspace of $A$ so that the nullspaces of $A$ and $B$ are identical, like the row spaces.

The left nullspace of $B$ is the nullspace of

$B^T = \begin{bmatrix} 0&1 \\ 0&2 \end{bmatrix}$

We can reduce $B^T$ to echelon form by subtracting 2 times the first row from the second row:

$\begin{bmatrix} 0&1 \\ 0&2 \end{bmatrix} \Rightarrow \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$

The resulting echelon matrix has one pivot, with $x_2$ a basic variable and $x_1$ a free variable. Setting $x_1 = 1$ we have from the first row of the echelon matrix $x_2 = 0$. The vector $(1, 0)$ is thus a basis for the nullspace of $B^T$ and for the left nullspace of $B$.

The matrix $C$ is already in echelon form, with pivots in the first and second columns. It has rank 2 and the first and second columns $(1, 0)$ and $(1, 1)$ respectively form a basis for the column space. Note that since the second column is equal to the first column plus the fourth column, an alternate basis for the column space consists of the first and fourth columns $(1, 0)$ and $(0, 1)$ respectively.

The dimension of the row space of $C$ (the rank of $C$) is also 2, and the two rows $(1, 1, 0, 0)$ and $(0, 1, 0, 1)$ of $C$ are a basis for the row space of $C$.

In the system $Cx = 0$ we have $x_1$ and $x_2$ as basic variables and $x_3$ and $x_4$ as free variables. If we set $x_3 = 1$ and $x_4 = 0$ then from the second row of $C$ we have $x_2 + x_4 = x_2 + 0 = 0$ or $x_2 = 0$. From the first row of $C$ we then have $x_1 + x_2 = x_1 + 0 = 0$ or $x_1 = 0$. Thus one solution to $Cx = 0$ is $(0, 0, 1, 0)$.

If we set $x_3 = 0$ and $x_4 = 1$ then from the second row of $C$ we have $x_2 + x_4 = x_2 + 1 = 0$ or $x_2 = -1$. From the first row of $C$ we then have $x_1 + x_2 = x_1 - 1 = 0$ or $x_1 = 1$. Thus a second solution to $Cx = 0$ is $(1, -1, 0, 1)$. The two solutions $(0, 0, 1, 0)$ and $(1, -1, 0, 1)$ are a basis for the nullspace of $C$.

Finally, since $C$ has rank $r = 2$ and has $m = 2$ rows, the dimension of the left nullspace is $m - r = 2 - 2 = 0$. This means that the only vector in the left nullspace is the zero vector $(0, 0, 0, 0)$.

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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