## Linear Algebra and Its Applications, Exercise 3.1.12

Exercise 3.1.12. For the matrix

$A = \begin{bmatrix} 1&0&2 \\ 1&1&4 \end{bmatrix}$

find a basis for the nullspace and show that it is orthogonal to the row space. Take the vector $x = (3, 3, 3)$ and express it as the sum of a nullspace component $x_n$ and a row space component $x_r$.

Answer: We use Gaussian elimination to solve the system $Ax = 0$. We start by subtracting 1 times row 1 from row 2:

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

The resulting echelon matrix has two pivots, in columns 1 and 2, so $x_1$ and $x_2$ are basic variables and $x_3$ is a free variable. Setting $x_3 = 1$ from the second row we have $x_2 + 2x_3 = x_2 + 2 = 0$ or $x_2 = -2$. From the first row we then have $x_1 + 2x_3 = x_1 + 2 = 0$ or $x_1 = -2$. The vector $(-2, -2, 1)$ is therefore a solution to $Ax = 0$ and a basis for the nullspace.

Since $(-2, -2, 1)$ is in the nullspace so is its negative, $(2, 2, -1)$, which is an alternative basis for the nullspace. Note that the inner product of $u = (2, 2, -1)$ with $v = (1, 0, 2)$ is

$u^Tv = 2 \cdot 1 + 2 \cdot 0 + (-1) \cdot 2 = 2 - 2 = 0$

and that the inner product of $u = (2, 2, -1)$ with $w = (1, 1, 4)$ is

$u^Tw = 2 \cdot 1 + 2 \cdot 1 + (-1) \cdot 4 = 2 + 2 - 4 = 0$

The basis of the nullspace $u = (2, 2, -1)$ is thus orthogonal to each of the basis vectors for the row space, $v = (1, 0, 2)$ and $w = (1, 1, 4)$. The row space as a whole consists of all linear combinations $c_1v+c_2w$ of those basis vectors, where $c_1$ and $c_2$ are scalars.

We then have

$u^T(c_1v+c_2w) = c_1u^Tv + c_2u^Tw = c_1 \cdot 0 + c_2 \cdot 0 = 0$

so that the basis $u = (2, 2, -1)$ of the nullspace is orthogonal to all vectors in the row space.

Note that we have

$x = (3, 3, 3) = (2, 2, -1) + (1, 1, 4) = u + w$

Since the vector $u (2, 2, -1)$ is in the nullspace and the vector $w = (1, 1, 4)$ is in the row space (being the second row of $A$) we have $x = x_n + x_r$ where $x_n = u = (2, 2, -1)$ and $x_r = w = (1, 1, 4)$.

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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