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Let A be the matrix: $$\begin{pmatrix} 1&2&3&2&1&0\\2&4&5&3&3&1\\1&2&2&1&2&1 \end{pmatrix}$$.

Show that {$\bigl( \begin{smallmatrix} 1 \\ 4\\3\end{smallmatrix} \bigr)$, $\bigl( \begin{smallmatrix} 3\\4\\1 \end{smallmatrix} \bigr)$} is a basis for the column space of A. Find a "nice basis for the column space of A.

So far, I have row reduced A to $$\begin{pmatrix} 1&2&0&-1&4&3\\0&0&1&1&-1&-1\\0&0&0&0&0&0 \end{pmatrix}$$ where the pivots occur in column 1 and column 3, so {(1,2,1),(3,5,2)} should be a "nice" column space? I do not see where {$\bigl( \begin{smallmatrix} 1 \\ 4\\3\end{smallmatrix} \bigr)$, $\bigl( \begin{smallmatrix} 3\\4\\1 \end{smallmatrix} \bigr)$} come from though.

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  • $\begingroup$ IMO, you usually get a “nicer” basis by column-reducing the matrix instead. In this case, doing that produces $(1,0,-1)^T$ and $(0,1,1)^T$. $\endgroup$
    – amd
    Mar 13, 2017 at 17:48
  • $\begingroup$ You’ve found a basis for the column space, but you haven’t shown that the given vectors are a basis. One way to do the latter is to show that all of the columns of $A$ are linear combinations of those two vectors. Any ideas on how you might do that? $\endgroup$
    – amd
    Mar 13, 2017 at 17:50
  • $\begingroup$ @amd I am not familiar with "column-reducing" the matrix, could you explain that a little further? Thanks for the other help though, I was able to show the vectors are a basis. $\endgroup$
    – AmaC
    Mar 13, 2017 at 20:20
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    $\begingroup$ It’s just like row reduction, except that you operate on the columns of the matrix instead. If that makes your head hurt, think of it as row-reducing the transpose, and then transposing the result. When you’re done, the non-zero columns of the reduced matrix give you a basis for the column space. $\endgroup$
    – amd
    Mar 13, 2017 at 23:03
  • $\begingroup$ That makes sense, thank you much! $\endgroup$
    – AmaC
    Mar 13, 2017 at 23:37

3 Answers 3

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You can just show, that rank of matrix $A$ is equal to $2$ (you actually did it by reducing $A$). So if rank is $2$ the dimension of a column space is also two. It means that there are two elements in the basis of $A$ column space. And you only need to notice that vectors $(1, 4, 3)^T$ and $(3, 4, 3)^T$ are linearly independent, so they form a basis of a column space.

Note that you can easily get this two vectors via simple matrix transformations (working with columns). For example $(3, 4, 3)^T = (3, 5, 2)^T - (0, 1, 1)^T$ and $(1, 4, 3)^T = (1, 2, 1)^T + 2(0, 1, 1)^T$ and so on you can show that after all this matrix transformations, you will reduce it to $$ \begin{pmatrix} 0 & 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 4 & 4 \\ 0 & 0 & 0 & 0 & 3 & 1 \end{pmatrix} $$

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Building on the insights of @puhsu, you know you have 2 vectors that span the image, columns 1 and 3. Can you construct the target vectors in this basis? $$ \left[ \begin{array}{r} 1 \\ 4 \\3 \end{array} \right] = \alpha \left[ \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right] + \beta \left[ \begin{array}{r} 3 \\ 5 \\ 2 \end{array} \right], \quad \alpha = 7, \ \beta = -2. $$ $$ \left[ \begin{array}{r} 3 \\ 4 \\ 1 \end{array} \right] = \alpha \left[ \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right] + \beta \left[ \begin{array}{r} 3 \\ 5 \\ 2 \end{array} \right], \quad \alpha = -3, \ \beta = 2. $$

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By finding the rref of $A$ you’ve determined that the column space is two-dimensional and the the first and third columns of $A$ for a basis for this space. The two given vectors, $(1,4,3)^T$ and $(3,4,1)^T$ are obviously linearly independent, so all that remains is to show that they also span the column space. You might be able to spot that the given vectors are linear combinations of some of the columns of $A$, which shows that they’re elements of the column space. Taken together with the other facts you have, that’s enough to show that they’re a basis. There’s a more systematic way to go about this, though.

In his answer, dantopa suggests trying to express these two vectors in the basis that you’ve found. If so, then they span the same space and you’re done. I suggest turning that around and instead seeing if you can express the basis that you’ve found in terms of the two given vectors. That is, solving the system $$\begin{align}a\pmatrix{1\\4\\3}+b\pmatrix{3\\4\\1}&=\pmatrix{1\\2\\1} \\ c\pmatrix{1\\4\\3}+d\pmatrix{3\\4\\1}&=\pmatrix{3\\5\\2}\end{align}$$ which can be written as the single equation $$\pmatrix{1&3\\4&4\\3&1}\pmatrix{a&c\\b&d}=\pmatrix{1&3\\2&5\\1&2}.$$ This can be solved by forming an augmented matrix and row-reducing, just as you’re used to doing. If the system has a solution, you’ll end up with pivots in the first two columns and the required coefficients in the other two.

Observe, though that you could have done this with the matrix $A$ in the first place and skipped the intermediate step of finding another basis for its column space. That is, prepend the two vectors to $A$ and row-reduce as usual. If those two vectors form a basis for the column space, then you’ll have pivots in the first two columns and nowhere else. For the problem at hand, you’d create the augmented matrix $$\left(\begin{array}{cc|cc}1&3 & 1&2&3&2&1&0 \\ 4&4 & 2&4&5&3&3&1 \\ 3&1 & 1&2&2&1&2&1 \end{array}\right)$$ and row-reduce it as usual.

As for finding a “nice” basis for the column space, that really depends on what “nice” means. That said, I find that the basis that you get by taking the columns that correspond to pivot columns in the rref doesn’t usually produce a “useful” basis. You can, however, column-reduce the matrix instead, which is more or less what puhsu does in his answer. Column-reduction is like row-reduction, except that you operate on the columns of the matrix instead of its rows. If you like, you can think of it as row-reducing the transpose and then transposing the result. The non-zero columns of the matrix produced by this process are a basis for the column space. You can see why this works if you remember that the non-zero rows of the rref of a matrix form a basis for its row space, and that the row space of a matrix is equal to the column space of its transpose.

For the matrix in this problem, we end up with $$\left(\begin{array}{rc}1&0 & 0&\cdots&0 \\ 0&1 & 0&\cdots&0 \\ -1&1 & 0&\cdots&0 \end{array}\right)$$ so a “nice” basis for the column space might be $(1,0,-1)^T$ and $(0,1,1)^T$. In general, the vectors for a basis computed this way will be sparse, i.e., they will have $r-1$ zeros as components, where $r=\operatorname{rank}A$, and another of the components of each vector will be $1$.

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