## 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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