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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