Does changing rows in matrix changes column space order?
$begingroup$
For example in matrix:
$begin{bmatrix} 0 & -2 & 3\
4 & 0 & 11end{bmatrix}$
Column Spaces are asked.
The answer is
{(4,0), (0,-2)}
Shouldn't we take the original matrix's columns which should be
{(0,4), (-2,0)}
Why did we change the order? Or the books answer is wrong?
linear-algebra matrices matrix-equations
edited Dec 25 '18 at 0:27
user376343
3,9234829
asked Dec 24 '18 at 23:58
NakurodNakurod
132
$endgroup$
add a comment |
$begingroup$
For example in matrix:
$begin{bmatrix} 0 & -2 & 3\
4 & 0 & 11end{bmatrix}$
Column Spaces are asked.
The answer is
{(4,0), (0,-2)}
Shouldn't we take the original matrix's columns which should be
{(0,4), (-2,0)}
Why did we change the order? Or the books answer is wrong?
linear-algebra matrices matrix-equations
edited Dec 25 '18 at 0:27
user376343
3,9234829
asked Dec 24 '18 at 23:58
NakurodNakurod
132
$endgroup$
add a comment |
$begingroup$
For example in matrix:
$begin{bmatrix} 0 & -2 & 3\
4 & 0 & 11end{bmatrix}$
Column Spaces are asked.
The answer is
{(4,0), (0,-2)}
Shouldn't we take the original matrix's columns which should be
{(0,4), (-2,0)}
Why did we change the order? Or the books answer is wrong?
linear-algebra matrices matrix-equations
edited Dec 25 '18 at 0:27
user376343
3,9234829
asked Dec 24 '18 at 23:58
NakurodNakurod
132
$endgroup$
For example in matrix:
$begin{bmatrix} 0 & -2 & 3\
4 & 0 & 11end{bmatrix}$
Column Spaces are asked.
The answer is
{(4,0), (0,-2)}
Shouldn't we take the original matrix's columns which should be
{(0,4), (-2,0)}
Why did we change the order? Or the books answer is wrong?
linear-algebra matrices matrix-equations
linear-algebra matrices matrix-equations
edited Dec 25 '18 at 0:27
user376343
3,9234829
asked Dec 24 '18 at 23:58
NakurodNakurod
132
edited Dec 25 '18 at 0:27
user376343
3,9234829
asked Dec 24 '18 at 23:58
NakurodNakurod
132
edited Dec 25 '18 at 0:27
user376343
3,9234829
edited Dec 25 '18 at 0:27
user376343
3,9234829
edited Dec 25 '18 at 0:27
user376343
3,9234829
3,9234829
asked Dec 24 '18 at 23:58
NakurodNakurod
132
asked Dec 24 '18 at 23:58
NakurodNakurod
132
asked Dec 24 '18 at 23:58
NakurodNakurod
132
132
add a comment |
add a comment |
1 Answer
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$begingroup$
While in this particular case, both the sets ${(0,4),(-2,0)}$ and ${(4,0),(0,-2)}$ span $Bbb R^2$, which is the column space, in general, you cannot change the top-to-bottom order of the column vectors. For example, take $A=begin{bmatrix}1&0\0&0end{bmatrix}$. The column space is the span of $(1,0)$, not $(0,1).$
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thanks for the answer, it really helped me.
$endgroup$
– Nakurod
Dec 25 '18 at 0:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
While in this particular case, both the sets ${(0,4),(-2,0)}$ and ${(4,0),(0,-2)}$ span $Bbb R^2$, which is the column space, in general, you cannot change the top-to-bottom order of the column vectors. For example, take $A=begin{bmatrix}1&0\0&0end{bmatrix}$. The column space is the span of $(1,0)$, not $(0,1).$
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thanks for the answer, it really helped me.
$endgroup$
– Nakurod
Dec 25 '18 at 0:29
add a comment |
$begingroup$
While in this particular case, both the sets ${(0,4),(-2,0)}$ and ${(4,0),(0,-2)}$ span $Bbb R^2$, which is the column space, in general, you cannot change the top-to-bottom order of the column vectors. For example, take $A=begin{bmatrix}1&0\0&0end{bmatrix}$. The column space is the span of $(1,0)$, not $(0,1).$
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thanks for the answer, it really helped me.
$endgroup$
– Nakurod
Dec 25 '18 at 0:29
add a comment |
$begingroup$
While in this particular case, both the sets ${(0,4),(-2,0)}$ and ${(4,0),(0,-2)}$ span $Bbb R^2$, which is the column space, in general, you cannot change the top-to-bottom order of the column vectors. For example, take $A=begin{bmatrix}1&0\0&0end{bmatrix}$. The column space is the span of $(1,0)$, not $(0,1).$
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
$endgroup$
While in this particular case, both the sets ${(0,4),(-2,0)}$ and ${(4,0),(0,-2)}$ span $Bbb R^2$, which is the column space, in general, you cannot change the top-to-bottom order of the column vectors. For example, take $A=begin{bmatrix}1&0\0&0end{bmatrix}$. The column space is the span of $(1,0)$, not $(0,1).$
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 0:19
Shubham JohriShubham Johri
5,204718
5,204718
$begingroup$
Thanks for the answer, it really helped me.
$endgroup$
– Nakurod
Dec 25 '18 at 0:29
add a comment |
$begingroup$
Thanks for the answer, it really helped me.
$endgroup$
– Nakurod
Dec 25 '18 at 0:29
$begingroup$
Thanks for the answer, it really helped me.
$endgroup$
– Nakurod
Dec 25 '18 at 0:29
$begingroup$
Thanks for the answer, it really helped me.
$endgroup$
– Nakurod
Dec 25 '18 at 0:29
add a comment |
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