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My problem says: "In the matrix $A \in M_{3\times3}$ the columns $c_1, c_2$ are linearly independent and $c_3=c_1+c_2$. Determine a basis in the $\operatorname{Null}(A)$."

Does this have an unique approach? I would solve this by choosing two independent vectors in $R^3$ and then computing the $c_3$. Is it the right approach?

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What is $Null(A)$? –  Rasmus Jan 18 '12 at 16:53
@Rasmus Null space of A –  Andrew Jan 18 '12 at 16:53
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up vote 1 down vote accepted

Well, how about computing the nullspace of $A$? An element of the nullspace is a vector $x = (x_1, x_2, x_3)$ such that $Ax = 0$. What is $Ax$? It's $x_1 c_1 + x_2 c_2 + x_3 c_3 = (x_1 + x_3) c_1 + (x_2 + x_3) c_2$. This means we have an element of the nullspace if and only if $x_1 + x_3 = x_2 + x_3 = 0$, which means $x_3$ is arbitrary and $x_1 = x_2 = -x_3$. There you go! A basis of the nullspace would be the vector $(1,1,-1)$.

Hope that helps!

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A basis for $\mathrm{Null}(A)$ is just $(1, 1, -1) \in \mathbb{R}^3$. Because:

$$ A \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = c_1 + c_2 - c_3 = 0 \ . $$

Hence, $(1, 1, -1) \in \mathrm{Null}(A)$.

On the other hand,

$$ \mathrm{dim}\ \mathrm{Null}(A) = 3 - \mathrm{rank}\ A = 3 -2 = 1 \ , $$

because $A$ has just two columns which are linearly independent. So

$$ \mathrm{span}\ \left\{ (1,1,-1) \right\} = \mathrm{Null}(A) \ . $$

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