# Homework Help: Basis for the orthogonal complement.

1. Nov 12, 2011

### -Dragoon-

1. The problem statement, all variables and given/known data
Let W be the plane $3x + 2y - z = 0$ in R3. Find a basis for $W^{\perp}$

2. Relevant equations
N/A

3. 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.

2. Nov 12, 2011

### ehild

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

ehild

3. Nov 12, 2011

### -Dragoon-

Its dimension is 3...

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

4. Nov 12, 2011

### ehild

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

ehild

5. Nov 12, 2011

### I like Serena

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?

6. Nov 12, 2011

### HallsofIvy

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.

Last edited by a moderator: Nov 12, 2011
7. Nov 12, 2011

### -Dragoon-

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

8. Nov 12, 2011

### I like Serena

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!

9. Nov 12, 2011

### -Dragoon-

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?

10. Nov 12, 2011

### -Dragoon-

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

11. Nov 12, 2011

### HallsofIvy

Since this is in R3, use the cross product.

12. Nov 12, 2011

### I like Serena

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.