Mixing
Vectors are linearly independent if the rref of their matrix has columns of leading ones
$begingroup$
Say I choose the vectors (1,0,0),(0,1,0),(0,1,0), these are clearly not linearly independent, but the matrix the form is [1,0,0;0,1,1;0,0,0], which has leading ones in every column, or have I made a mistake somewhere?
Here is the proof:
linear-algebra
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
$endgroup$
add a comment |
$begingroup$
Say I choose the vectors (1,0,0),(0,1,0),(0,1,0), these are clearly not linearly independent, but the matrix the form is [1,0,0;0,1,1;0,0,0], which has leading ones in every column, or have I made a mistake somewhere?
Here is the proof:
linear-algebra
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
$endgroup$
add a comment |
$begingroup$
Say I choose the vectors (1,0,0),(0,1,0),(0,1,0), these are clearly not linearly independent, but the matrix the form is [1,0,0;0,1,1;0,0,0], which has leading ones in every column, or have I made a mistake somewhere?
Here is the proof:
linear-algebra
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
$endgroup$
Say I choose the vectors (1,0,0),(0,1,0),(0,1,0), these are clearly not linearly independent, but the matrix the form is [1,0,0;0,1,1;0,0,0], which has leading ones in every column, or have I made a mistake somewhere?
Here is the proof:
linear-algebra
linear-algebra
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
edited Feb 2 at 12:50
4M4D3U5 M0Z4RT
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
asked Feb 2 at 12:23
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
536
536
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have made a mistake (in interpretation). Your RREF doesn't have $3$ pivots: it only has $2$. The reason: you're vectors span a $2$-dimensional space.
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
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add a comment |
$begingroup$
There is no leading one (“pivot”) in the last column. The only nonzero entry has another nonzero entry to its left, so is quite clearly not the leading entry for that row.
answered Feb 3 at 0:34
amdamd
31.8k21053
$endgroup$
$begingroup$
Yes, but is it not a leading one in the column? As it is the first element of the column, and it is a one.
$endgroup$
– 4M4D3U5 M0Z4RT
Feb 3 at 21:16
$begingroup$
@4M4D3U5M0Z4RT “Leading” refers to the position in the row. Otherwise, one would conclude that $small{pmatrix{1&1\0&0}}$ has full rank, which is patently absurd. Review the definitions of these terms.
$endgroup$
– amd
Feb 3 at 21:29
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have made a mistake (in interpretation). Your RREF doesn't have $3$ pivots: it only has $2$. The reason: you're vectors span a $2$-dimensional space.
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
You have made a mistake (in interpretation). Your RREF doesn't have $3$ pivots: it only has $2$. The reason: you're vectors span a $2$-dimensional space.
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
$endgroup$
add a comment |
$begingroup$
You have made a mistake (in interpretation). Your RREF doesn't have $3$ pivots: it only has $2$. The reason: you're vectors span a $2$-dimensional space.
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
$endgroup$
You have made a mistake (in interpretation). Your RREF doesn't have $3$ pivots: it only has $2$. The reason: you're vectors span a $2$-dimensional space.
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
answered Feb 2 at 12:33
Chris CusterChris Custer
14.4k3827
14.4k3827
add a comment |
add a comment |
$begingroup$
There is no leading one (“pivot”) in the last column. The only nonzero entry has another nonzero entry to its left, so is quite clearly not the leading entry for that row.
answered Feb 3 at 0:34
amdamd
31.8k21053
$endgroup$
$begingroup$
Yes, but is it not a leading one in the column? As it is the first element of the column, and it is a one.
$endgroup$
– 4M4D3U5 M0Z4RT
Feb 3 at 21:16
$begingroup$
@4M4D3U5M0Z4RT “Leading” refers to the position in the row. Otherwise, one would conclude that $small{pmatrix{1&1\0&0}}$ has full rank, which is patently absurd. Review the definitions of these terms.
$endgroup$
– amd
Feb 3 at 21:29
add a comment |
$begingroup$
There is no leading one (“pivot”) in the last column. The only nonzero entry has another nonzero entry to its left, so is quite clearly not the leading entry for that row.
answered Feb 3 at 0:34
amdamd
31.8k21053
$endgroup$
$begingroup$
Yes, but is it not a leading one in the column? As it is the first element of the column, and it is a one.
$endgroup$
– 4M4D3U5 M0Z4RT
Feb 3 at 21:16
$begingroup$
@4M4D3U5M0Z4RT “Leading” refers to the position in the row. Otherwise, one would conclude that $small{pmatrix{1&1\0&0}}$ has full rank, which is patently absurd. Review the definitions of these terms.
$endgroup$
– amd
Feb 3 at 21:29
add a comment |
$begingroup$
There is no leading one (“pivot”) in the last column. The only nonzero entry has another nonzero entry to its left, so is quite clearly not the leading entry for that row.
answered Feb 3 at 0:34
amdamd
31.8k21053
$endgroup$
There is no leading one (“pivot”) in the last column. The only nonzero entry has another nonzero entry to its left, so is quite clearly not the leading entry for that row.
answered Feb 3 at 0:34
amdamd
31.8k21053
answered Feb 3 at 0:34
amdamd
31.8k21053
answered Feb 3 at 0:34
amdamd
31.8k21053
answered Feb 3 at 0:34
amdamd
31.8k21053
31.8k21053
$begingroup$
Yes, but is it not a leading one in the column? As it is the first element of the column, and it is a one.
$endgroup$
– 4M4D3U5 M0Z4RT
Feb 3 at 21:16
$begingroup$
@4M4D3U5M0Z4RT “Leading” refers to the position in the row. Otherwise, one would conclude that $small{pmatrix{1&1\0&0}}$ has full rank, which is patently absurd. Review the definitions of these terms.
$endgroup$
– amd
Feb 3 at 21:29
add a comment |
$begingroup$
Yes, but is it not a leading one in the column? As it is the first element of the column, and it is a one.
$endgroup$
– 4M4D3U5 M0Z4RT
Feb 3 at 21:16
$begingroup$
@4M4D3U5M0Z4RT “Leading” refers to the position in the row. Otherwise, one would conclude that $small{pmatrix{1&1\0&0}}$ has full rank, which is patently absurd. Review the definitions of these terms.
$endgroup$
– amd
Feb 3 at 21:29
$begingroup$
Yes, but is it not a leading one in the column? As it is the first element of the column, and it is a one.
$endgroup$
– 4M4D3U5 M0Z4RT
Feb 3 at 21:16
$begingroup$
Yes, but is it not a leading one in the column? As it is the first element of the column, and it is a one.
$endgroup$
– 4M4D3U5 M0Z4RT
Feb 3 at 21:16
$begingroup$
@4M4D3U5M0Z4RT “Leading” refers to the position in the row. Otherwise, one would conclude that $small{pmatrix{1&1\0&0}}$ has full rank, which is patently absurd. Review the definitions of these terms.
$endgroup$
– amd
Feb 3 at 21:29
$begingroup$
@4M4D3U5M0Z4RT “Leading” refers to the position in the row. Otherwise, one would conclude that $small{pmatrix{1&1\0&0}}$ has full rank, which is patently absurd. Review the definitions of these terms.
$endgroup$
– amd
Feb 3 at 21:29
add a comment |
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