# Finding an orthogonal basis

I am currently working on a problem that involves finding an orthogonal basis. It says this:

Consider the matrix A =

|1  1|
|1 -1|
|1 -3|

So I started out doing this:

B =

|1 |       |1|   | 2|
|-1| - -3/3|1| = | 0|
|-3|       |1|   |-2|

I am stuck here. I know the answer is:

|1/sqrt(3) 2/sqrt(8)|
|1/sqrt(3) 0        |
|1/sqrt(3) 2/sqrt(8)|

How would I get from where I am to the final answer? Do I do another step to the gram schmidt?

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Is the question asking for orthogonal or orthonormal? – Alec Dec 11 '13 at 5:59
The question states: Find an orthogonal basis of the column space of A. – user104536 Dec 11 '13 at 6:06

I think you have the right answer already. If you are looking for an orthogonal basis, you have correctly applied Gram-Schmidt to get the set $$\left\{\left(\begin{matrix}1\\1\\1\end{matrix}\right),\left(\begin{matrix}2\\0\\-2\end{matrix}\right)\right\}$$ Now, comparing this to the answer you say is supposed to be correct: their first vector is the same as yours, except normalized (hence my question about orthonormal). Their second vector I disagree with outright. In fact, it is NOT in the range of the transformation of the matrix. We can check this fact by attempting to solve the system: $$\begin{matrix}x+y=\frac{2}{\sqrt{8}}\\x-y=0\\x-3y=\frac{2}{\sqrt{8}}\end{matrix}$$ From the first two equations we get that $x=y=\frac{1}{\sqrt{8}}$, but this contradicts the third equation. So the second vector is not even in the range.