$\beginbmatrix1&0&2\\0&1&-1\\0&0&0\\0&0&0\\0&0&0\endbmatrix$, however I"m not sure what to do next to give the "basis" of the column an are of A



The columns equivalent to the pivots of your initial matrix will be a basis for the tower space.

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The column room is the span of the column vectors of $A$, The original matrix. As got it--thanks commented, heat reduction does not keep the tower space.So the column an are is:

$\sirhenryjones-museums.orgrmspan\bigg(\beginbmatrix 3 \\ 3 \\ -2 \\ -2 \\ 4 \endbmatrix, \beginbmatrix 3 \\ 5 \\ 4 \\ -4 \\ 9 \endbmatrix, \beginbmatrix 3 \\ 1 \\ -8 \\ 0 \\ -1 \endbmatrix\bigg)$.

To find an yes, really basis because that the shaft space, we need to mitigate this list to a linearly elevation list, if that is not already.

In fact, girlfriend can present that these three vectors space not linearly independent. Particularly, the 3rd can be written as a linear combination of the an initial two.

Since the an initial two space linearly elevation (which you have to verify), we deserve to write $\bigg(\beginbmatrix 3 \\ 3 \\ -2 \\ -2 \\ 4 \endbmatrix, \beginbmatrix 3 \\ 5 \\ 4 \\ -4 \\ 9 \endbmatrix\bigg)$ together a basis for the column space.

edited Jul 12 "16 in ~ 1:53
answered Jul 11 "16 in ~ 19:03

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Fact. allow $A$ be a matrix. The nonzero rows the $\Declaresirhenryjones-museums.orgOperatorrrefrref\rref(A^\top)$ type a communication of the column room of $A$.

In our situation we have$$A=\left<\beginarrayrrr3 & 3 & 3 \\3 & 5 & 1 \\-2 & 4 & -8 \\-2 & -4 & 0 \\4 & 9 & -1\endarray\right>$$Row-reducing $A^\top$ gives$$\rref(A^\top)=\left<\beginarrayrrrrr1 & 0 & -\frac113 & \frac13 & -\frac76 \\0 & 1 & 3 & -1 & \frac52 \\0 & 0 & 0 & 0 & 0\endarray\right>$$The fact above implies that\beginalign*\langle1, 0, -11/3, 1/3, -7/6 \rangle&&\langle0,1, 3, -1, 5/2 \rangle\endalign*forms a basis of the column room of $A$.

answered Jul 12 "16 at 1:23

Brian FitzpatrickBrian Fitzpatrick
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$\begingroup$ due to the fact that the an initial two rows space independent, then the very first two columns room independent. Could we simply write the communication in terms of the live independence rows? choose in the accepted answer? $\endgroup$
Jun 13 "19 in ~ 11:56
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Using heat operations preservation the heat space, yet destroys the shaft space. Instead, what you want to perform is to usage column work to placed the procession in column diminished echelon form. The resulting procession will have actually the same shaft space, and also the nonzero columns will certainly be a basis.

It"s possibly too strong to say heat operations destroy the column an are — rather they have the right to be assumed of together performing a change of coordinates. The is why the technique of the other answers work: in the brand-new coordinates, it"s basic to find a maximal linearly independent collection of columns — i.e. A basis. And also such a thing is a communication no issue what coordinate depiction you use, so the equivalent columns native the original matrix kind a communication in the initial coordinates.

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reply Jul 11 "16 at 19:42
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