How to know if $mathrmIm A = mathbb R^n$ given a matrix? The Next CEO of Stack Overflow$ker A capmathrmIm A = 0 Rightarrow ker A + mathrmIm A = R^n$.Given a linear map $T:Vto V$, is it true that $V=ker(T) oplus mathrmim(T)$?The space spanned by the rows of the Kernel matrix.Linear map and its matrixDetermining if $defrkoperatornamerankrk(A)$ is always equivalent to $rk(A^t*A)$Can we say something about the rows of the null space matrixlinear algebra: image, kernel and transpose$n - mathrmrank, A ge mathrmrank, A - mathrmrank, A^2$Find the farthest rotation matrix in $mathrmSO(3)$ from a given matrix.Existence of a subspace of the domain of a linear map with specific properties
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How to know if $mathrmIm A = mathbb R^n$ given a matrix?
The Next CEO of Stack Overflow$ker A capmathrmIm A = 0 Rightarrow ker A + mathrmIm A = R^n$.Given a linear map $T:Vto V$, is it true that $V=ker(T) oplus mathrmim(T)$?The space spanned by the rows of the Kernel matrix.Linear map and its matrixDetermining if $defrkoperatornamerankrk(A)$ is always equivalent to $rk(A^t*A)$Can we say something about the rows of the null space matrixlinear algebra: image, kernel and transpose$n - mathrmrank, A ge mathrmrank, A - mathrmrank, A^2$Find the farthest rotation matrix in $mathrmSO(3)$ from a given matrix.Existence of a subspace of the domain of a linear map with specific properties
$begingroup$
Knowing that I have $5times 6$ matrix $A$ with $dim mathrmIm, A = 4$ and $dim ker A= 2$, I am asked if $mathrmIm ,A = mathbb R^4$? The answer is that $mathrmIm, A$ is a subspace of $mathbb R^5$. Is it because we have 5 rows? If not why?
linear-algebra
edited Mar 28 at 1:07
Brian
1,208116
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Knowing that I have $5times 6$ matrix $A$ with $dim mathrmIm, A = 4$ and $dim ker A= 2$, I am asked if $mathrmIm ,A = mathbb R^4$? The answer is that $mathrmIm, A$ is a subspace of $mathbb R^5$. Is it because we have 5 rows? If not why?
linear-algebra
edited Mar 28 at 1:07
Brian
1,208116
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Yes, we have 5 rows and six columns. Therefore, $A$ as a linear mapping maps from $mathbb R^6$ to $mathbb R^5$. Hence, $Im(A)$ is a subspace of $mathbb R^5$. The space $mathbb R^4$ is another object and not a subspace of $mathbb R^5$. It can be embedded though, but the result will not be $mathbb R^4$ anymore.
$endgroup$
– amsmath
Mar 28 at 0:54
add a comment |
$begingroup$
Knowing that I have $5times 6$ matrix $A$ with $dim mathrmIm, A = 4$ and $dim ker A= 2$, I am asked if $mathrmIm ,A = mathbb R^4$? The answer is that $mathrmIm, A$ is a subspace of $mathbb R^5$. Is it because we have 5 rows? If not why?
linear-algebra
edited Mar 28 at 1:07
Brian
1,208116
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Knowing that I have $5times 6$ matrix $A$ with $dim mathrmIm, A = 4$ and $dim ker A= 2$, I am asked if $mathrmIm ,A = mathbb R^4$? The answer is that $mathrmIm, A$ is a subspace of $mathbb R^5$. Is it because we have 5 rows? If not why?
linear-algebra
linear-algebra
edited Mar 28 at 1:07
Brian
1,208116
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 1:07
Brian
1,208116
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 28 at 1:07
Brian
1,208116
edited Mar 28 at 1:07
Brian
1,208116
edited Mar 28 at 1:07
Brian
1,208116
1,208116
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
asked Mar 28 at 0:46
Dr.StoneDr.Stone
335
335
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dr.Stone is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Yes, we have 5 rows and six columns. Therefore, $A$ as a linear mapping maps from $mathbb R^6$ to $mathbb R^5$. Hence, $Im(A)$ is a subspace of $mathbb R^5$. The space $mathbb R^4$ is another object and not a subspace of $mathbb R^5$. It can be embedded though, but the result will not be $mathbb R^4$ anymore.
$endgroup$
– amsmath
Mar 28 at 0:54
add a comment |
1
$begingroup$
Yes, we have 5 rows and six columns. Therefore, $A$ as a linear mapping maps from $mathbb R^6$ to $mathbb R^5$. Hence, $Im(A)$ is a subspace of $mathbb R^5$. The space $mathbb R^4$ is another object and not a subspace of $mathbb R^5$. It can be embedded though, but the result will not be $mathbb R^4$ anymore.
$endgroup$
– amsmath
Mar 28 at 0:54
1
1
$begingroup$
Yes, we have 5 rows and six columns. Therefore, $A$ as a linear mapping maps from $mathbb R^6$ to $mathbb R^5$. Hence, $Im(A)$ is a subspace of $mathbb R^5$. The space $mathbb R^4$ is another object and not a subspace of $mathbb R^5$. It can be embedded though, but the result will not be $mathbb R^4$ anymore.
$endgroup$
– amsmath
Mar 28 at 0:54
$begingroup$
Yes, we have 5 rows and six columns. Therefore, $A$ as a linear mapping maps from $mathbb R^6$ to $mathbb R^5$. Hence, $Im(A)$ is a subspace of $mathbb R^5$. The space $mathbb R^4$ is another object and not a subspace of $mathbb R^5$. It can be embedded though, but the result will not be $mathbb R^4$ anymore.
$endgroup$
– amsmath
Mar 28 at 0:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any $mtimes n$ matrix can be thought of as describing a linear transformation from $mathbb R^n$ to $mathbb R^m$. The fact that $A$ is a $5 times 6$ matrix tells us that its column space (or, image) will be a subspace of $mathbb R^5$. Since $dim mathrmIm, A = 4$, we know that this subspace has dimension four, but this is not the same as $mathbb R^4$.
Consider the the following subspace of $mathbb R^3$:
$$
V=mathrmspan, left beginbmatrix1\1\0endbmatrix, beginbmatrix0\1\1endbmatrixright
$$
It is easy to see that the subspace $V$ is a plane situated in $mathbb R^3$. However, this plane is clearly not the same as $mathbb R^2$; it is a separate object altogether. The same is true for the column space of your matrix. Though it has four dimensions, it is not the "four-dimensional space," but rather a subsection of "five-dimensional space."
answered Mar 28 at 0:57
BrianBrian
1,208116
$endgroup$
$begingroup$
"Though it is four dimensional, it is not the whole of "four-dimensional space"" -- I think this is misleading...
$endgroup$
– amsmath
Mar 28 at 1:31
$begingroup$
@amsmath How so? Is there a way that this could be better expressed?
$endgroup$
– Brian
Mar 28 at 1:35
$begingroup$
"Not a whole of" somehow indicates that it is a part of four-dimensional space, which is of course not true. Rather the contrary. It is a "whole" four-dimensional space. But just not $mathbb R^4$. ;o)
$endgroup$
– amsmath
Mar 28 at 1:38
$begingroup$
@amsmath I see your point, though I feel that my phrasing adequately conveys that sentiment, especially with the surrounding context. Nonetheless, I have edited my post to remove any possible ambiguity for future readers. Thanks for your close review!
$endgroup$
– Brian
Mar 28 at 1:49
$begingroup$
The thing is that you can very easily confuse people that are not really safe with maths because they are often insecured.
$endgroup$
– amsmath
Mar 28 at 1:55
add a comment |
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1 Answer
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1 Answer
1
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oldest
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oldest
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active
oldest
votes
$begingroup$
Any $mtimes n$ matrix can be thought of as describing a linear transformation from $mathbb R^n$ to $mathbb R^m$. The fact that $A$ is a $5 times 6$ matrix tells us that its column space (or, image) will be a subspace of $mathbb R^5$. Since $dim mathrmIm, A = 4$, we know that this subspace has dimension four, but this is not the same as $mathbb R^4$.
Consider the the following subspace of $mathbb R^3$:
$$
V=mathrmspan, left beginbmatrix1\1\0endbmatrix, beginbmatrix0\1\1endbmatrixright
$$
It is easy to see that the subspace $V$ is a plane situated in $mathbb R^3$. However, this plane is clearly not the same as $mathbb R^2$; it is a separate object altogether. The same is true for the column space of your matrix. Though it has four dimensions, it is not the "four-dimensional space," but rather a subsection of "five-dimensional space."
answered Mar 28 at 0:57
BrianBrian
1,208116
$endgroup$
$begingroup$
"Though it is four dimensional, it is not the whole of "four-dimensional space"" -- I think this is misleading...
$endgroup$
– amsmath
Mar 28 at 1:31
$begingroup$
@amsmath How so? Is there a way that this could be better expressed?
$endgroup$
– Brian
Mar 28 at 1:35
$begingroup$
"Not a whole of" somehow indicates that it is a part of four-dimensional space, which is of course not true. Rather the contrary. It is a "whole" four-dimensional space. But just not $mathbb R^4$. ;o)
$endgroup$
– amsmath
Mar 28 at 1:38
$begingroup$
@amsmath I see your point, though I feel that my phrasing adequately conveys that sentiment, especially with the surrounding context. Nonetheless, I have edited my post to remove any possible ambiguity for future readers. Thanks for your close review!
$endgroup$
– Brian
Mar 28 at 1:49
$begingroup$
The thing is that you can very easily confuse people that are not really safe with maths because they are often insecured.
$endgroup$
– amsmath
Mar 28 at 1:55
add a comment |
$begingroup$
Any $mtimes n$ matrix can be thought of as describing a linear transformation from $mathbb R^n$ to $mathbb R^m$. The fact that $A$ is a $5 times 6$ matrix tells us that its column space (or, image) will be a subspace of $mathbb R^5$. Since $dim mathrmIm, A = 4$, we know that this subspace has dimension four, but this is not the same as $mathbb R^4$.
Consider the the following subspace of $mathbb R^3$:
$$
V=mathrmspan, left beginbmatrix1\1\0endbmatrix, beginbmatrix0\1\1endbmatrixright
$$
It is easy to see that the subspace $V$ is a plane situated in $mathbb R^3$. However, this plane is clearly not the same as $mathbb R^2$; it is a separate object altogether. The same is true for the column space of your matrix. Though it has four dimensions, it is not the "four-dimensional space," but rather a subsection of "five-dimensional space."
answered Mar 28 at 0:57
BrianBrian
1,208116
$endgroup$
$begingroup$
"Though it is four dimensional, it is not the whole of "four-dimensional space"" -- I think this is misleading...
$endgroup$
– amsmath
Mar 28 at 1:31
$begingroup$
@amsmath How so? Is there a way that this could be better expressed?
$endgroup$
– Brian
Mar 28 at 1:35
$begingroup$
"Not a whole of" somehow indicates that it is a part of four-dimensional space, which is of course not true. Rather the contrary. It is a "whole" four-dimensional space. But just not $mathbb R^4$. ;o)
$endgroup$
– amsmath
Mar 28 at 1:38
$begingroup$
@amsmath I see your point, though I feel that my phrasing adequately conveys that sentiment, especially with the surrounding context. Nonetheless, I have edited my post to remove any possible ambiguity for future readers. Thanks for your close review!
$endgroup$
– Brian
Mar 28 at 1:49
$begingroup$
The thing is that you can very easily confuse people that are not really safe with maths because they are often insecured.
$endgroup$
– amsmath
Mar 28 at 1:55
add a comment |
$begingroup$
Any $mtimes n$ matrix can be thought of as describing a linear transformation from $mathbb R^n$ to $mathbb R^m$. The fact that $A$ is a $5 times 6$ matrix tells us that its column space (or, image) will be a subspace of $mathbb R^5$. Since $dim mathrmIm, A = 4$, we know that this subspace has dimension four, but this is not the same as $mathbb R^4$.
Consider the the following subspace of $mathbb R^3$:
$$
V=mathrmspan, left beginbmatrix1\1\0endbmatrix, beginbmatrix0\1\1endbmatrixright
$$
It is easy to see that the subspace $V$ is a plane situated in $mathbb R^3$. However, this plane is clearly not the same as $mathbb R^2$; it is a separate object altogether. The same is true for the column space of your matrix. Though it has four dimensions, it is not the "four-dimensional space," but rather a subsection of "five-dimensional space."
answered Mar 28 at 0:57
BrianBrian
1,208116
$endgroup$
Any $mtimes n$ matrix can be thought of as describing a linear transformation from $mathbb R^n$ to $mathbb R^m$. The fact that $A$ is a $5 times 6$ matrix tells us that its column space (or, image) will be a subspace of $mathbb R^5$. Since $dim mathrmIm, A = 4$, we know that this subspace has dimension four, but this is not the same as $mathbb R^4$.
Consider the the following subspace of $mathbb R^3$:
$$
V=mathrmspan, left beginbmatrix1\1\0endbmatrix, beginbmatrix0\1\1endbmatrixright
$$
It is easy to see that the subspace $V$ is a plane situated in $mathbb R^3$. However, this plane is clearly not the same as $mathbb R^2$; it is a separate object altogether. The same is true for the column space of your matrix. Though it has four dimensions, it is not the "four-dimensional space," but rather a subsection of "five-dimensional space."
answered Mar 28 at 0:57
BrianBrian
1,208116
edited Mar 28 at 1:49
answered Mar 28 at 0:57
BrianBrian
1,208116
answered Mar 28 at 0:57
BrianBrian
1,208116
answered Mar 28 at 0:57
BrianBrian
1,208116
1,208116
$begingroup$
"Though it is four dimensional, it is not the whole of "four-dimensional space"" -- I think this is misleading...
$endgroup$
– amsmath
Mar 28 at 1:31
$begingroup$
@amsmath How so? Is there a way that this could be better expressed?
$endgroup$
– Brian
Mar 28 at 1:35
$begingroup$
"Not a whole of" somehow indicates that it is a part of four-dimensional space, which is of course not true. Rather the contrary. It is a "whole" four-dimensional space. But just not $mathbb R^4$. ;o)
$endgroup$
– amsmath
Mar 28 at 1:38
$begingroup$
@amsmath I see your point, though I feel that my phrasing adequately conveys that sentiment, especially with the surrounding context. Nonetheless, I have edited my post to remove any possible ambiguity for future readers. Thanks for your close review!
$endgroup$
– Brian
Mar 28 at 1:49
$begingroup$
The thing is that you can very easily confuse people that are not really safe with maths because they are often insecured.
$endgroup$
– amsmath
Mar 28 at 1:55
add a comment |
$begingroup$
"Though it is four dimensional, it is not the whole of "four-dimensional space"" -- I think this is misleading...
$endgroup$
– amsmath
Mar 28 at 1:31
$begingroup$
@amsmath How so? Is there a way that this could be better expressed?
$endgroup$
– Brian
Mar 28 at 1:35
$begingroup$
"Not a whole of" somehow indicates that it is a part of four-dimensional space, which is of course not true. Rather the contrary. It is a "whole" four-dimensional space. But just not $mathbb R^4$. ;o)
$endgroup$
– amsmath
Mar 28 at 1:38
$begingroup$
@amsmath I see your point, though I feel that my phrasing adequately conveys that sentiment, especially with the surrounding context. Nonetheless, I have edited my post to remove any possible ambiguity for future readers. Thanks for your close review!
$endgroup$
– Brian
Mar 28 at 1:49
$begingroup$
The thing is that you can very easily confuse people that are not really safe with maths because they are often insecured.
$endgroup$
– amsmath
Mar 28 at 1:55
$begingroup$
"Though it is four dimensional, it is not the whole of "four-dimensional space"" -- I think this is misleading...
$endgroup$
– amsmath
Mar 28 at 1:31
$begingroup$
"Though it is four dimensional, it is not the whole of "four-dimensional space"" -- I think this is misleading...
$endgroup$
– amsmath
Mar 28 at 1:31
$begingroup$
@amsmath How so? Is there a way that this could be better expressed?
$endgroup$
– Brian
Mar 28 at 1:35
$begingroup$
@amsmath How so? Is there a way that this could be better expressed?
$endgroup$
– Brian
Mar 28 at 1:35
$begingroup$
"Not a whole of" somehow indicates that it is a part of four-dimensional space, which is of course not true. Rather the contrary. It is a "whole" four-dimensional space. But just not $mathbb R^4$. ;o)
$endgroup$
– amsmath
Mar 28 at 1:38
$begingroup$
"Not a whole of" somehow indicates that it is a part of four-dimensional space, which is of course not true. Rather the contrary. It is a "whole" four-dimensional space. But just not $mathbb R^4$. ;o)
$endgroup$
– amsmath
Mar 28 at 1:38
$begingroup$
@amsmath I see your point, though I feel that my phrasing adequately conveys that sentiment, especially with the surrounding context. Nonetheless, I have edited my post to remove any possible ambiguity for future readers. Thanks for your close review!
$endgroup$
– Brian
Mar 28 at 1:49
$begingroup$
@amsmath I see your point, though I feel that my phrasing adequately conveys that sentiment, especially with the surrounding context. Nonetheless, I have edited my post to remove any possible ambiguity for future readers. Thanks for your close review!
$endgroup$
– Brian
Mar 28 at 1:49
$begingroup$
The thing is that you can very easily confuse people that are not really safe with maths because they are often insecured.
$endgroup$
– amsmath
Mar 28 at 1:55
$begingroup$
The thing is that you can very easily confuse people that are not really safe with maths because they are often insecured.
$endgroup$
– amsmath
Mar 28 at 1:55
add a comment |
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
Dr.Stone is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Yes, we have 5 rows and six columns. Therefore, $A$ as a linear mapping maps from $mathbb R^6$ to $mathbb R^5$. Hence, $Im(A)$ is a subspace of $mathbb R^5$. The space $mathbb R^4$ is another object and not a subspace of $mathbb R^5$. It can be embedded though, but the result will not be $mathbb R^4$ anymore.
$endgroup$
– amsmath
Mar 28 at 0:54