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The question;

$U = \{x |Ax = 0\}$ If $ A = \begin{bmatrix}1 & 2 & 1 & 0 & -2\\ 2 & 1 & 2 & 1 & 2\\1 & 1 & 0 & -1 & -2\\ 0 & 0 & 2 & 0 & 4\end{bmatrix}$

Find a basis for $U$.

To make it linearly independent, I reduce the rows of $A$;

$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 & x_2 & x_3 & x_4\end{bmatrix} = 0$

So I figure I just have to list the rows out as vectors;

Basis vectors = $\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix} , \begin{pmatrix}0\\1\\0\\0\\-2\end{pmatrix} , \begin{pmatrix}0\\0\\1\\0\\2\end{pmatrix} , \begin{pmatrix}0\\0\\0\\1\\0\end{pmatrix}$

(can't figure out how to surround it all with curly brackets in MathJax).

Am I doing this properly? It feels too simple...

2 Answers 2

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Observe that row operations doesn't change the solution for the system of linear equations. From the last column of the reduced row echelon form (RREF) of $A$, $$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{bmatrix} = 0,\tag{*}\label{*}$$ we see that if $x = (x_1,\dots,x_5)^T \in U$, $Ax = 0$, so $\sum\limits_{j=1}^5 a_i x_i = 0$, where $a_i$ denotes the $i$-th column of $A$. From \eqref{*}, we observe that $(x_1,x_2,x_3,x_4,x_5) = (0,-2,2,0,-1)$ is a possible solution. By the rank-nullity theorem, since $\mathop{\mathrm{rank}}(A) = 4$, so $\dim U = \dim\mathsf{N}(A) = 5 - 4 = 1$. Hence, we conclude that $x = (0,-2,2,0,-1)^T$ is a basis for $U$.

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    Ah, thank you very much. I see exactly where I went wrong - I completely forgot about the equation part :). As a general question, if the dimension of the null space was i.e 2, then would that mean I have to search for a second vector (and the basis would then be a set containing both vectors)? – 2017-02-04
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    @KookieMonster Oops, typo: missed "dim" before "N(A)". Yes, but I use this theorem for explaining only. In actual calculations (say, in GNU Octave), one reads from the columns *with non-elementary column vectors* of `rref(A)` to deduce the linear combination of columns (i.e. basis for $U$). – 2017-02-04
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$U$ is the set of vectors $\vec x$, whose image under $f_A$ is $\vec 0$.

Since your matrix $A$ is rank 4 and $f_A:K^5 \rightarrow K^4$, it must have a $1$ dimensional Kernel.

So your answer should be subspace of $K^5$ spanned by 1 vector. What vector could that be from your reduced matrix?