Let A be the matrix: $$eginpmatrix 1&2&3&2&1&0\2&4&5&3&3&1\1&2&2&1&2&1 endpmatrix$$.

Sexactly how that $igl( eginsmallmatrix 1 \ 4\3endsmallmatrix igr)$, $igl( eginsmallmatrix 3\4\1 endsmallmatrix igr)$ is a basis for the column area of A. Find a "nice basis for the column area of A.

So far, I have actually row lessened A to $$eginpmatrix 1&2&0&-1&4&3\0&0&1&1&-1&-1\0&0&0&0&0&0 endpmatrix$$ where the pivots occur in column 1 and also column 3, so (1,2,1),(3,5,2) have to be a "nice" column space? I execute not view where $igl( eginsmallmatrix 1 \ 4\3endsmallmatrix igr)$, $igl( eginsmallmatrix 3\4\1 endsmallmatrix igr)$ come from though.




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asked Mar 13 "17 at 17:41
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AmaCAmaC
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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 measurement of a column space is also 2. It suggests that there are 2 aspects in the basis of $A$ column room. And you just need to alert that vectors $(1, 4, 3)^T$ and $(3, 4, 3)^T$ are livirtually independent, so they develop a basis of a column area.

Keep in mind that you deserve to conveniently obtain this two vectors via basic matrix revolutions (working with columns). For instance $(3, 4, 3)^T = (3, 5, 2)^T - (0, 1, 1)^T$ and $(1, 4, 3)^T = (1, 2, 1)^T + 2(0, 1, 1)^T$ and also so on you have the right to present that after all this matrix revolutions, you will certainly mitigate it to$$eginpmatrix0 & 0 & 0 & 0 & 1 & 3 \0 & 0 & 0 & 0 & 4 & 4 \0 & 0 & 0 & 0 & 3 & 1endpmatrix$$


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edited Mar 13 "17 at 19:25
answered Mar 13 "17 at 17:56
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puhsupuhsu
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Building on the insights of
puhsu, you know you have 2 vectors that expectancy the photo, columns 1 and also 3. Can you construct the target vectors in this basis?$$left< eginarrayr 1 \ 4 \3 endarray ight>=alphaleft< eginarrayr 1 \ 2 \ 1 endarray ight> +etaleft< eginarrayr 3 \ 5 \ 2 endarray ight>, quad alpha = 7, eta = -2.$$$$left< eginarrayr 3 \ 4 \ 1 endarray ight>=alphaleft< eginarrayr 1 \ 2 \ 1 endarray ight> +etaleft< eginarrayr 3 \ 5 \ 2 endarray ight>, quad alpha = -3, eta = 2.$$


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answered Mar 13 "17 at 19:49
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dantopadantopa
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By finding the rref of $A$ you’ve determined that the column space is two-dimensional and also the the initially and third columns of $A$ for a basis for this space. The 2 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 expectancy the column room. You can have the ability to spot that the given vectors are linear combinations of some of the columns of $A$, which mirrors that they’re aspects of the column area. Taken along with the other facts you have, that’s enough to present that they’re a basis. There’s an extra systematic method to go around this, though.

In his answer, dantopa suggests trying to expush these 2 vectors in the basis that you’ve discovered. If so, then they span the exact same room and you’re done. I indicate turning that roughly and also rather seeing if you deserve to expush the basis that you’ve discovered in terms of the 2 offered vectors. That is, solving the system $$eginalignapmatrix1\4\3+bpmatrix3\4\1&=pmatrix1\2\1 \ cpmatrix1\4\3+dpmatrix3\4\1&=pmatrix3\5\2endalign$$ which have the right to be composed as the single equation $$pmatrix1&3\4&4\3&1pmatrixa&c\b&d=pmatrix1&3\2&5\1&2.$$ This deserve to be resolved by creating an augmented matrix and row-reducing, simply as you’re used to doing. If the system has actually a solution, you’ll finish up via pivots in the first 2 columns and also the compelled coefficients in the other 2.

Observe, though that you might have done this via the matrix $A$ in the initially place and also skipped the intermediate action of finding an additional basis for its column room. That is, prepend the 2 vectors to $A$ and also row-alleviate as usual. If those 2 vectors develop a basis for the column area, then you’ll have pivots in the first 2 columns and nowbelow else. For the problem at hand, you’d produce the augmented matrix $$left(eginarraycc1&3 & 1&2&3&2&1&0 \ 4&4 & 2&4&5&3&3&1 \ 3&1 & 1&2&2&1&2&1 endarray ight)$$ and also row-minimize it as usual.

As for finding a “nice” basis for the column area, that really depends on what “nice” indicates. That sassist, I find that the basis that you acquire by taking the columns that correspond to pivot columns in the rref doesn’t commonly 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 favor row-reduction, except that you run on the columns of the matrix rather of its rows. If you like, you have the right to think of it as row-reducing the transpose and then transposing the outcome. The non-zero columns of the matrix created by this process are a basis for the column room. You have the right to watch why this functions if you remember that the non-zero rows of the rref of a matrix create a basis for its row room, and also that the row area of a matrix is equal to the column space of its transpose.

For the matrix in this trouble, we finish up via $$left(eginarrayrc1&0 & 0&cdots&0 \ 0&1 & 0&cdots&0 \ -1&1 & 0&cdots&0 endarray ight)$$ so a “nice” basis for the column room can be $(1,0,-1)^T$ and $(0,1,1)^T$. In general, the vectors for a basis computed this method will be sporadic, i.e., they will have $r-1$ zeros as components, wright here $r=operatornamerankA$, and one more of the components of each vector will certainly be $1$.