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Let $A = \begin{bmatrix}3&3&3\\3&5&1\\-2&4&-8\\-2&-4&0\\4&9&-1\end{bmatrix}$

Give a basis for the column space of A

So what I've done so far is put it in RREF (which was a task itself) and got

$\begin{bmatrix}1&0&2\\0&1&-1\\0&0&0\\0&0&0\\0&0&0\end{bmatrix}$, but I'm not sure what to do next to give the "basis" of the column space of A

Community
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Shua
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4 Answers4

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The columns corresponding to the pivots of your original matrix will be a basis for the column space.

got it--thanks
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    Got it, thanks. – Shua Jul 11 '16 at 19:00
  • Can you explain why? Because row operation changes the column space! – Hosein Rahnama Jul 11 '16 at 19:04
  • @Shua: I just can get that why the columns corresponding to the pivots of your original matrix are for sure linearly independent!? – Hosein Rahnama Jul 11 '16 at 19:08
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    @H.R. Elementary row operations don't switch any columns. That implies that if we delete the last column of $A$ we get the RREF which is the first two columns of $RREF(A)$. Which has rank $2$. Thus the first two columns of $A$ are linearly independent. Since $RREF(A)$ also has rank $2$ (and thus $A$ does), those two columns must form a basis for the column space. – got it--thanks Jul 11 '16 at 19:09
  • @gotit--thanks: That was convincing. :) I think you should add this to your answer. – Hosein Rahnama Jul 11 '16 at 19:12
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The column space is the span of the column vectors of $A$, The original matrix. As got it--thanks commented, row reduction does not preserve the column space. So the column space is:

$\mathrm{span}\bigg(\begin{bmatrix} 3 \\ 3 \\ -2 \\ -2 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ 5 \\ 4 \\ -4 \\ 9 \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ -8 \\ 0 \\ -1 \end{bmatrix}\bigg)$.

To find an actual basis for the column space, we need to reduce this list to a linearly independent list, if it is not already.

In fact, you can show that these three vectors are not linearly independent. Particularly, the third can be written as a linear combination of the first two.

Since the first two are linearly independent (which you should verify), we can write $\bigg(\begin{bmatrix} 3 \\ 3 \\ -2 \\ -2 \\ 4 \end{bmatrix}, \begin{bmatrix} 3 \\ 5 \\ 4 \\ -4 \\ 9 \end{bmatrix}\bigg)$ as a basis for the column space.

Christian
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  • Will the column space always be some set of the component vectors? – Aaron Franke Apr 04 '19 at 08:17
  • @AaronFranke Yes, if you have a set of vectors that span a vector space, there exists a subset of that set of vectors that is a basis (still spanning, but also linearly independent). – markasoftware Feb 11 '20 at 01:34
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Fact. Let $A$ be a matrix. The nonzero rows of $\DeclareMathOperator{rref}{rref}\rref(A^\top)$ form a basis of the column space of $A$.

In our case we have $$ A= \left[\begin{array}{rrr} 3 & 3 & 3 \\ 3 & 5 & 1 \\ -2 & 4 & -8 \\ -2 & -4 & 0 \\ 4 & 9 & -1 \end{array}\right] $$ Row-reducing $A^\top$ gives $$ \rref(A^\top)= \left[\begin{array}{rrrrr} 1 & 0 & -\frac{11}{3} & \frac{1}{3} & -\frac{7}{6} \\ 0 & 1 & 3 & -1 & \frac{5}{2} \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] $$ The fact above implies that \begin{align*} \langle1, 0, -{11}/{3}, {1}/{3}, -{7}/{6} \rangle && \langle0,1, 3, -1, 5/2 \rangle \end{align*} forms a basis of the column space of $A$.

Brian Fitzpatrick
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  • since the first two rows are independent, then the first two columns are independent. Could we just write the basis in terms of the independent rows? like in the accepted answer? –  Jun 13 '19 at 11:56
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    Yes, but this technique works for any matrix, while the one used in the accepted answer would stutter badly with more than three columns. Already in the accepted answer, the fact that the third column is a linear combination of the first two, though not hard, is not obvious either; it follows from the row-reduction performed by the OP. – Martin Argerami Feb 17 '21 at 20:05
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Using row operations preserves the row space, but destroys the column space. Instead, what you want to do is to use column operations to put the matrix in column reduced echelon form. The resulting matrix will have the same column space, and the nonzero columns will be a basis.

It's maybe too strong to say row operations destroy the column space — instead they can be thought of as performing a change of coordinates. That is why the method of the other answers work: in the new coordinates, it's easy to find a maximal linearly independent set of columns — i.e. a basis. And such a thing is a basis no matter what coordinate representation you use, so the corresponding columns from the original matrix form a basis in the original coordinates.