Answer to Question #290596 in Linear Algebra for abed

Question

. Find a basis for the orthogonal complement of the vector v(1, 3,−2) of the euclidean vector space (R 3 , ·). Argue whether this basis is orthonormal or not.

Answer 1

Since every vector in the orthogonal complement should be orthogonal to every vector in the given subspace, we need to find the null space of [1, 3, −2]

To find the null space, solve the matrix equation $\begin{bmatrix} 1 & 3 & -2 \end{bmatrix}$$\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix}$ = [0].

If we take x2 = t, x3 = s, then x1 = 2s − 3t.

Thus, vector x = $\begin{bmatrix} 2s - 3t \\ t \\ s \end{bmatrix}$ = $\begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}$t + $\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$s.

This is the null space.

The basis for the null space is $\begin{Bmatrix} \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \end{Bmatrix}$.

This is the basis for the orthogonal complement.

Answer: The basis for the orthogonal complement is $\begin{Bmatrix} \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \end{Bmatrix}$.

To ague whether this basis is orthonormal or not we could find the orthogonal complement of the basis $\begin{Bmatrix} \begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} \end{Bmatrix}$ and compare it to the first vector v (1, 3,−2).

Firstly, solve the matrix equation $\begin{bmatrix} -3 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix}$$\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix}$ = [0].

Write it in matrix form $\begin{bmatrix} -3 & 1 & 0 & 0 \\ 2 & 0 & 1 & 0 \end{bmatrix}$.

So we have x1 = x1, x2 = 3x1, x3 = -2x1.

So vector x = (1, 3,−2) and equals the vector v (1, 3,−2).

It means that the basis for the orthogonal complement of the vector v was found correctly and is orthonormal.