Give a basis for $(Col(A))\bot$ where $A$ is defined as follows:
$A =\begin{pmatrix}-1&3&1\\ 2&2&1\\0&8&3\end{pmatrix} -> \begin{pmatrix}1&0&1/8\\ 0&1&3/8\\0&0&0\end{pmatrix}$
My plan:
- put matrix in a row reduced echelon form and create a basis for the nullspace:
$basis = \begin{pmatrix}-1\\-3\\8 \end{pmatrix}$
- Then I wanted to find a vector that is orthogonal and has no 0 in it. I was thinking of $basis\bot = \begin{pmatrix}2\\2\\1 \end{pmatrix}$
this is apparently not a good way.
can someone help me and explain to me what steps I have to make. i choose for this one because $basis \cdot basis\bot = 0$
the given answer for this question is:$\begin{pmatrix}2\\1\\-1 \end{pmatrix}$
answer from the professor ( this question is an exam question):
2 Answers
$\begingroup$Since $\operatorname{Col}(A)$ cannot be $0$-dimensional ($A\neq0$) and it cannot be $1$-dimensional (that would happen only if the columns were all a multiple of the same vector), $\dim\operatorname{Col}(A)=2$ or $\dim\operatorname{Col}(A)=3$. But $\det A=0$ and therefore we cannot have $\dim\operatorname{Col}(A)=3$. So, $\dim\operatorname{Col}(A)=2$.
We can try to write the third column as a linear combination of the other two:$$\left\{\begin{array}{l}-a+3b=1\\2a+2b=1\\8b=3.\end{array}\right.$$And this works: you can take $a=\frac18$ and $b=\frac38$. So, $\operatorname{Col}(A)=\operatorname{span}\left\{(-1,2,0)^T,(3,2,8)^T\right\}$, and therefore$$\operatorname{Col}(A)^\perp=\operatorname{span}\left\{(-1,2,0)^T\times(3,2,8)^T\right\}=\operatorname{span}\left\{(16,8,-8)^T\right\}.$$
$\endgroup$ 7 $\begingroup$Your original idea doesn’t quite work because the null space of a matrix doesn’t in general have any particular relation to its column space. That’s more obvious when the matrix isn’t square, say $n\times m$ with $n\ne m$: the null space is a subset of an $m$-dimensional space, but the column space is a subset of an $n$-dimensional space.
Recall that the null space of a matrix is the orthogonal complement of its row space. Thus, what you really did was to find an element of $A$’s row space. What you need to do instead, then, is to find a basis for the null space of $A^T$.
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