# How to solve this particular linear system?

So I have this linear system:

\begin{align*} -u + v &= y_1\\ u + v &= y_2 \\ 2u + v &= y_3 \end{align*}

After doing gaussian elimination, I get:

\begin{align*} u = y_3 - y_2 \\ v = 2y_2 - y_3 \\ y_1 -3y_2 + 2y_3 = 0 \end{align*}

I'm asked to find for what values of $y_1, y_2, y_3$ this system has a solution. The answer to that would be:

$$\left[ \begin{array}{ccc} 1 & -3 & 2 \\ \end{array} \right] * \left[ \begin{array}{ccc} y_1 \\ y_2 \\ y_3 \end{array} \right] = 0$$

i.e. the null space. How do I find a basis for the null space though?

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The easiest solution by far is just to guess two linearly independent vectors that satisfy the condition, such as $(3,1,0)$ and $(2,0,-1)$.

If that is not systematic enough for you, take $(1,-3,2)$ and append a known basis for the entire space, such as $\mathbf e_1, \mathbf e_2, \mathbf e_3$. Then use the Gram-Schmidt process on the resulting list of 4 vectors. You'll need to throw out one when it turns out it is not linearly independent of the precedent ones. In the resulting basis for $\mathbb R^3$, remove the first one (which will be proportional to $(1,-3,2)$); the other two are a basis for the orthogonal complement.

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Thanks for the description! I didn't realize it could be done that way, but it appears guessing seems to be the simplest approach. – John Doe Dec 13 '11 at 22:42

Clearly two independent vectors from the null space are $a_1 = (3, 1, 0)^t$ and $a_2 = (2, 0, -1)^t$.

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And we know that you only need two because the dimension of the null space is 3-1 = 2, right? – John Doe Dec 13 '11 at 22:40

If you're in 3 dimensions, then $y_1-3y_2+2y_3=0$ is the cartesian equation of a plane. Choose values for any two of the three parameters and that will determine the third one.

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