# Basis for the orthogonal complement.

## Homework Statement

Let W be the plane $3x + 2y - z = 0$ in R3. Find a basis for $W^{\perp}$

N/A

## The Attempt at a Solution

Firstly, I take some arbitrary vector $u = \begin{bmatrix}a\\b\\c\end{bmatrix}$
that is in $W^{\perp}$. Then I note that W can be rewritten in terms of the natural basis for R3:
$W = 3\begin{bmatrix}1\\0\\0\end{bmatrix} + 2 \begin{bmatrix}0\\1\\0\end{bmatrix} - 1 \begin{bmatrix}0\\0\\1\end{bmatrix}$

Then, by the inner product: $\langle u,w\rangle = 3a + 2b - c = 0$. Thus,
$a = -\frac{2}{3}b + \frac{1}{3}c$ and hence the vector u in $W^{\perp}$ has the form:
$u = \begin{bmatrix}-\frac{2}{3}b+\frac{1}{3}c\\b\\c\end{bmatrix} = b\begin{bmatrix}\frac{-2}{3}\\1\\0\end{bmatrix} + c u = \begin{bmatrix}\frac{1}{3}\\0\\1\end{bmatrix}$

Thus, the set $\begin{Bmatrix} \begin{bmatrix}-\frac{2}{3}\\1\\0\end{bmatrix} \begin{bmatrix}\frac{1}{3}\\0\\1\end{bmatrix} \end{Bmatrix}$ spans $W^{\perp}$.

Since both vectors are linearly independent, that is, one cannot be written as a linear combination of the other, I conclude that the set forms a basis for $W^{\perp}$. After checking the solutions manual for the answer, the book came up with a completely different solution. What did I do wrong? Thanks in advance.

ehild
Homework Helper
What is the dimension of $W^{\perp}$?

ehild

What is the dimension of $W^{\perp}$?

ehild

Its dimension is 3...

So, do I need to start all over or can the incorrect process to the solution be fixed half way?

ehild
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The whole vector field has three dimensions. What is the dimension of a plane? What is the dimension of a straight line?

ehild

I like Serena
Homework Helper
You found the vector (-2/3, 1, 0) to be in $W^{\perp}$.

Can you verify that it is NOT in W?
That is, does it satisfy the equation 3x + 2y - z = 0 which is the equation for W?

HallsofIvy
Homework Helper
A plane, through the origin, is a two dimensional subspace of $R^3$. Its orthogonal complement is a line through the origin and has dimension 1. You should have only one vector in the basis.

Your error is in thinking that a vector <a, b, c>, in the orthogonal complement must satisfy 3a+ 2y- c= 0. That is the equation of the plane- any vector in the plane must satisfy that equation. Instead, write the equation as z= 3x+ 2y. Any vector in the plane must be of the form
$$\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ 3x+ 2y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 3\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}$$
Any vector in the orthogonal complement must be perpendicular to both of those vectors.

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You found the vector (-2/3, 1, 0) to be in $W^{\perp}$.

Can you verify that it is NOT in W?
That is, does it satisfy the equation 3x + 2y - z = 0 which is the equation for W?

I never realized the vector I mistakenly concluded to be in $W^{\perp}$ is actually in W.

I like Serena
Homework Helper
I never realized the vector I mistakenly concluded to be in $W^{\perp}$ is actually in W.

Yes.
Effectively you have solved the equation of W to find a basis of 2 vectors that span W.

But for $W^{\perp}$ you need a vector that is perpendicular to those 2 vectors.

Btw, if you try and find it, you may find that you recognize it!

A plane, through the origin, is a two dimensional subspace of $R^3$. Its orthogonal complement is a line through the origin and has dimension 1. You should have only one vector in the basis.

Your error is in thinking that a vector <a, b, c>, in the orthogonal complement must satisfy 3a+ 2y- c= 0. That is the equation of the plane- any vector in the plane must satisfy that equation. Instead, write the equation as z= 3x+ 2y. Any vector in the plane must be of the form
$$\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x \\ y \\ 3x+ 2y\end{bmatrix}= x\begin{bmatrix}1 \\ 0 \\ 3\end{bmatrix}+ y\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix}$$
Any vector in the orthogonal complement must be perpendicular to both of those vectors.
Ah, is this is because of one of the theorems (is unnamed in my text, only called "theorem 5.12") that states the sum of the dimensions of the subspace and the subspace's orthogonal complement must not be greater than the dimension Rn? In this case, R3?

Yes.
Effectively you have solved the equation of W to find a basis of 2 vectors that span W.

But for $W^{\perp}$ you need a vector that is perpendicular to those 2 vectors.

Btw, if you try and find it, you may find that you recognize it!

I see. So, can that vector now be found by using the Gram-Schmidt process?

HallsofIvy
Homework Helper
Since this is in R3, use the cross product.

I like Serena
Homework Helper
I see. So, can that vector now be found by using the Gram-Schmidt process?

You can.

I recommend you try to find a vector that has an dot product of zero with the first vector.
Once you have that, use the remaining unknown component to make the dot product with the second vector also zero.

This path is easy, since both your vectors have a zero in it.