## Linear Algebra and Its Applications, Exercise 3.1.7

Exercise 3.1.7. For the matrix $A = \begin{bmatrix} 1&2&1 \\ 2&4&3 \\ 3&6&4 \end{bmatrix}$

find vectors $x$ and $y$ such that $x$ is orthogonal to the row space of $A$ and $y$ is orthogonal to the column space of $A$>

Answer: The nullspace of $A$ is orthogonal to the row space of $A$. We can therefore find a suitable vector $x$ by solving the system $Ax = 0$.

To solve the system we perform Gaussian elimination. We start by subtracting 2 times row 1 from row 2: $\begin{bmatrix} 1&2&1 \\ 2&4&3 \\ 3&6&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&1 \\ 0&0&1 \\ 3&6&4 \end{bmatrix}$

and then subtract 3 times row 1 from row 3: $\begin{bmatrix} 1&2&1 \\ 0&0&1 \\ 3&6&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&1 \\ 0&0&1 \\ 0&0&1 \end{bmatrix}$

Finally we subtract row 2 from row 3: $\begin{bmatrix} 1&2&1 \\ 0&0&1 \\ 0&0&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&1 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}$

The echelon matrix has 2 pivots in columns 1 and 3, so $x_1$ and $x_3$ are basic variables and $x_2$ is a free variable.

Setting $x_2 = 1$ from row 2 we have $x_3 = 0$ and from row 1 we have $x_1 + 2x_2 + x_3 = x_1 + 2 + 0 = 0$ or $x_1 = -2$. So the vector $x = (-2, 1, 0)$ is a solution to the system, a basis for the nullspace of $A$, and a vector orthogonal to the row space of $A$.

The left nullspace of $A$ is orthogonal to the column space of $A$. We can therefore find a suitable vector $y$ by solving the system $A^Ty = 0$.

To solve the system we perform Gaussian elimination. We start by subtracting 2 times row 1 from row 2: $\begin{bmatrix} 1&2&3 \\ 2&4&6 \\ 1&3&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&3 \\ 0&0&0 \\ 1&3&4 \end{bmatrix}$

and then subtract 1 times row 1 from row 3: $\begin{bmatrix} 1&2&3 \\ 0&0&0 \\ 1&3&4 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&3 \\ 0&0&0 \\ 0&1&1 \end{bmatrix}$

Finally we exchange rows 2 and row 3: $\begin{bmatrix} 1&2&3 \\ 0&0&0 \\ 0&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&2&3 \\ 0&1&1 \\ 0&0&0 \end{bmatrix}$

The resulting echelon matrix has 2 pivots in columns 1 and 2, so $y_1$ and $y_2$ are basic variables and $y_3$ is a free variable.

Setting $y_3 = 1$ from row 2 we have $y_2 + y_3 = y_2 + 1 = 0$ or $y_2 = -1$. From row 1 we have $y_1 + 2y_2 + 3y_3 = y_1 - 2 + 3 = 0$ or $y_1 = -1$. So the vector $y = (-1, -1, 1)$ is a solution to the system, a basis for the left nullspace of $A$, and a vector orthogonal to the column space of $A$.

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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