Question #290596

. 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.


1
Expert's answer
2022-01-26T16:19:32-0500

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 [132]\begin{bmatrix} 1 & 3 & -2 \end{bmatrix}[x1x2x3]\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} = [0].

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

Thus, vector x = [2s3tts]\begin{bmatrix} 2s - 3t \\ t \\ s \end{bmatrix} = [310]\begin{bmatrix} -3 \\ 1 \\ 0 \end{bmatrix}t + [201]\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}s.

This is the null space.

The basis for the null space is {[310],[201]}\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 {[310],[201]}\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 {[310],[201]}\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 [310201]\begin{bmatrix} -3 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix}[x1x2x3]\begin{bmatrix} x1 \\ x2 \\ x3 \end{bmatrix} = [0].

Write it in matrix form [31002010]\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.

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