Compute Orthonormal Basis from Transformation

Question

Given:

$$ W = \{(x, y, z, w) \in R^4 \ |\ x \ + 2y + w = 0 \ and \ y - w = 0 \ \} $$

And the use of the standard dot product. How do I find the orthonormal basis of W and then extend that to a four dimensional Basis of R ?

Answer 1

Let the matrix $A$ be,

$$A = \begin{pmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 0 & -1\end{pmatrix}$$

Then $W$ comprises of the vectors $v = (x,y,z,w)^{T} \in \mathbb{R}^4$ such that,

$$Av = 0$$

In other words, $W$ is nothing but the null space of $A$, $N(A)$. The basis of $N(A)$ comprises of vectors,

$$(-3,1,0,1)^{T} \text{ and } (0,0,1,0)^{T}$$

which are independent and happen to be orthogonal. Divide the two vectors by their norms and you will get orthonormal basis of $W$.

Now, to extend them to basis of $R^4$, you must realize that the row space of $A$ (i.e. $C(A^T)$) is orthogonal complement of $N(A)$. So, you need to find the basis of $C(A^T)$ and combine the vectors in it with above vectors to get basis of $R^4$. It turns out that the two rows of $A$ form the basis of the $C(A^{T})$ because they are linearly independent and span the $C(A^T)$.

P.S. I am revising Linear Algebra so there might be a better solution.

Answer 2

The space you are interested in is the null space of $$ A=\left[\begin{array}{rrrr} 1 & 2 & 0 & 1 \\ 0 & 1 & 0 & -1 \end{array}\right] $$ The reduced row-echelon form of $A$ is $$ \DeclareMathOperator{rref}{rref}\rref(A)= \left[\begin{array}{rrrr} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & -1 \end{array}\right] $$ This tells us that $\vec x\in W$ if and only if $$ \vec x= \left[\begin{array}{r} x \\ y \\ z \\ w \end{array}\right]= \left[\begin{array}{r} -3\,w \\ w \\ z \\ w \end{array}\right]= z \left[\begin{array}{r} 0 \\ 0 \\ 1 \\ 0 \end{array}\right]+w\left[\begin{array}{r} -3 \\ 1 \\ 0 \\ 1 \end{array}\right] $$ This shows that $$ \left\{\left[\begin{array}{r} 0 \\ 0 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} -3 \\ 1 \\ 0 \\ 1 \end{array}\right]\right\} $$ is a basis of $W$. To extend this basis to a basis of $\Bbb R^4$, consider the matrix $$ M=\left[\begin{array}{rr|rrrr} 0 & -3 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \end{array}\right] $$ The reduced row-echelon form of $M$ is $$ \rref(M)=\left[\begin{array}{rr|rrrr} 1 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 & -1 \end{array}\right] $$ The pivots in $\rref(M)$ are in the first four columns. Thus the first four columns of $M$ form a basis of $\Bbb R^4$.

Do you see how to use this to complete your problem?

Answer 3

Well, I think you want an basis of $\mathbb{R}^{4}$ which contains an orthonormal basis of W.

You could use the following procedure:

• First, compute a basis for W. That is, find a minimum set of vectors that generate the space $W$. Since your space is given by linear equality constraints, it can be computed as suggested by Brian Fitzpatrick i.e. compute a basis for the nullspace of A, where A is defined as $$A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 0 & -1\end{bmatrix}$$ Lets assume that the result gives you that Null$(A) = span\{v_1,v_2\}$ with $v_1,v_2\in\mathbb{R}^4$. Then, a basis $B$ for $W$ is given by $B=\{v_1,v_2\}$.
• Notice that $B$ is not necessarily an orthogonal basis. But, its easy to find one by applying the Gram-schmidt ortogonalization process, and obtain new vectors $\{u_1,u_2\}$.
• The last part is the extension of the basis $B$ to basis of the whole space $\mathbb{R}^4$, say $B'$. An easy an straightforward choice is just in front of you! The subspace $W$ is given by linear equations in the following way: $$W =\left\{(x,y,z,w)\in\mathbb{R}^4\bigg| \begin{split}x + 2y +0z + w = 0\\ 0x +y +0z -w = 0\end{split}\right\}$$ If you look at it carefully, you might notice that the vectors $(1,2,0,1)$ and $(0,1,0,-1)$ constitute a basis of a particular interesting subspace: the orthogonal complement of $W$, denoted as $W^{\perp}$. A classic result applies for the final resolution: $$W\oplus W^{\perp} = \mathbb{R}^4$$ $W$ and $W^{\perp}$ are always in direct sum, and using classical dimension theorems the result follows. That implies that you can construct a basis for $\mathbb{R}^4$ by simply "sticking" the basis of $W$ and $W^{\perp}$. And that's easy as: $$B'=\left\{\underbrace{u_1,u_2}_{\text{basis for } W},\underbrace{(1,2,0,1),(0,1,0,-1)}_{\text{basis for }W^{\perp}}\right\}$$

I hope my answer helps! If you need any theoretical detail, just let me know!

Cheers!