How is the column space of a matrix A orthogonal to its nullspace?
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How do you show that the column space of a matrix A is orthogonal to its nullspace?
matrices
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add a comment |
$begingroup$
How do you show that the column space of a matrix A is orthogonal to its nullspace?
matrices
$endgroup$
$begingroup$
You should make your question self-contained (such that it is understandable without referring to the title)...
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– Fabian
Mar 25 '11 at 22:54
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The following illustration might help: en.wikipedia.org/wiki/Kernel_%28linear_algebra%29#Illustration
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– Dor
Jan 28 '15 at 9:08
add a comment |
$begingroup$
How do you show that the column space of a matrix A is orthogonal to its nullspace?
matrices
$endgroup$
How do you show that the column space of a matrix A is orthogonal to its nullspace?
matrices
matrices
asked Mar 25 '11 at 22:52
PatiencePatience
3682716
3682716
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You should make your question self-contained (such that it is understandable without referring to the title)...
$endgroup$
– Fabian
Mar 25 '11 at 22:54
$begingroup$
The following illustration might help: en.wikipedia.org/wiki/Kernel_%28linear_algebra%29#Illustration
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– Dor
Jan 28 '15 at 9:08
add a comment |
$begingroup$
You should make your question self-contained (such that it is understandable without referring to the title)...
$endgroup$
– Fabian
Mar 25 '11 at 22:54
$begingroup$
The following illustration might help: en.wikipedia.org/wiki/Kernel_%28linear_algebra%29#Illustration
$endgroup$
– Dor
Jan 28 '15 at 9:08
$begingroup$
You should make your question self-contained (such that it is understandable without referring to the title)...
$endgroup$
– Fabian
Mar 25 '11 at 22:54
$begingroup$
You should make your question self-contained (such that it is understandable without referring to the title)...
$endgroup$
– Fabian
Mar 25 '11 at 22:54
$begingroup$
The following illustration might help: en.wikipedia.org/wiki/Kernel_%28linear_algebra%29#Illustration
$endgroup$
– Dor
Jan 28 '15 at 9:08
$begingroup$
The following illustration might help: en.wikipedia.org/wiki/Kernel_%28linear_algebra%29#Illustration
$endgroup$
– Dor
Jan 28 '15 at 9:08
add a comment |
4 Answers
4
active
oldest
votes
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What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix $A in mathbb{R}^{m times n}$. The right null space is defined as $$mathcal{N}(A) = {z in mathbb{R}^{n times 1} : Az = 0 }$$
Let $
A =
left[ {begin{array}{c}
a_1^T \
a_2^T \
ldots \
ldots \
a_m^T
end{array} } right]$. The row space of $A$ is defined as
$$mathcal{R}(A) = {y in mathbb{R}^{n times 1}: y = sum_{i=1}^m a_i x_i text{ , where }x_i in mathbb{R} text{ and }a_i in mathbb{R}^{n times 1} }$$
Now from the definition of right null space we have $a_i^T z = 0$.
So if we take a $y in mathcal{R}(A)$, then $y = displaystyle sum_{k=1}^m a_i x_i text{ , where }x_i in mathbb{R}$. Hence, $$y^Tz = (sum_{k=1}^m a_i x_i)^T z = (sum_{k=1}^m x_i a_i^T) z = sum_{k=1}^m x_i (a_i^T z) = 0$$
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of $A$ is orthogonal to the left null space of $A$.
Note: The left null space is defined as ${z in mathbb{R}^{m times 1}: z^TA = 0}$
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@Morgan Rodgers Can you explain the part on transpose? y transpose z to form 0. If i understand correctly y is a row vector and z is a column vector. If y were to transpose, it would become a column vector
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– aceminer
Jan 5 '17 at 14:52
2
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@aceminer I'm not the author of this answer, but I think I can say something. Frequently when we are using the convention that vectors are columns, we treat the row space as consisting of column vectors. This matches the definition given of the row space, where the $a_{i}$ are (column) vectors, and the $a_{i}^{T}$ are the rows of $A$; then $mathcal{R}(A)$ is the space spanned by the (column) vectors $a_{i}$. So $y$ and $z$ are both column vectors.
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– Morgan Rodgers
Jan 5 '17 at 15:13
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Ah I see thanks
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– aceminer
Jan 5 '17 at 15:14
add a comment |
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The OP's question would have been well-phrased had he specified that the matrix $A$ is symmetric i.e. $A=A^top$, in which case ${rm colspan}(A)={rm rowspan}(A^top)={rm rowspan}(A)$. Now, consider the definition of ${rm null}(A)$ as the space of all vectors $mathbf{v}$ such that $Amathbf{v}=mathbf{0}$. Letting $mathbf{a}_1,ldots,mathbf{a}_n$ be the rows (and columns) of $A$, matrix multiplication tells us that $mathbf{a}_icdotmathbf{v}=0$ for each $i=1,ldots,dim(A)$. Thus any vector $mathbf{v}in{rm null}(A)$ is orthogonal to ${rm colspan}(A)$. It follows that ${rm null}(A)perp{rm colspan}(A)$.$square$
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add a comment |
$begingroup$
You can't show this: it is false. Take
$$begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix};$$
both its column space and null space are the subspace ${(x, 0) mid x in mathbb{R}}$ (the $x$-axis, so to speak). I think you need self-adjointness (symmetry in the real case) to get this.
EDIT: Sivaram anticipated me.
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add a comment |
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21
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix A∈Rm×n. The right null space is defined as
N(A)={z∈Rn×1:Az=0}
Let A=⎡⎣⎢⎢⎢⎢⎢⎢aT1aT2……aTm⎤⎦⎥⎥⎥⎥⎥⎥. The row space of A is defined as
R(A)={y∈Rn×1:y=∑i=1maixi , where xi∈R and ai∈Rn×1}
Now from the definition of right null space we have aTiz=0.
So if we take a y∈R(A), then y=∑k=1maixi , where xi∈R. Hence,
yTz=(∑k=1maixi)Tz=(∑k=1mxiaTi)z=∑k=1mxi(aTiz)=0
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of A is orthogonal to the left null space of A.
Note: The left null space is defined as {z∈Rm×1:zTA=0}
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For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
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– Kemono Chen
Dec 25 '18 at 7:38
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix $A in mathbb{R}^{m times n}$. The right null space is defined as $$mathcal{N}(A) = {z in mathbb{R}^{n times 1} : Az = 0 }$$
Let $
A =
left[ {begin{array}{c}
a_1^T \
a_2^T \
ldots \
ldots \
a_m^T
end{array} } right]$. The row space of $A$ is defined as
$$mathcal{R}(A) = {y in mathbb{R}^{n times 1}: y = sum_{i=1}^m a_i x_i text{ , where }x_i in mathbb{R} text{ and }a_i in mathbb{R}^{n times 1} }$$
Now from the definition of right null space we have $a_i^T z = 0$.
So if we take a $y in mathcal{R}(A)$, then $y = displaystyle sum_{k=1}^m a_i x_i text{ , where }x_i in mathbb{R}$. Hence, $$y^Tz = (sum_{k=1}^m a_i x_i)^T z = (sum_{k=1}^m x_i a_i^T) z = sum_{k=1}^m x_i (a_i^T z) = 0$$
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of $A$ is orthogonal to the left null space of $A$.
Note: The left null space is defined as ${z in mathbb{R}^{m times 1}: z^TA = 0}$
$endgroup$
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@Morgan Rodgers Can you explain the part on transpose? y transpose z to form 0. If i understand correctly y is a row vector and z is a column vector. If y were to transpose, it would become a column vector
$endgroup$
– aceminer
Jan 5 '17 at 14:52
2
$begingroup$
@aceminer I'm not the author of this answer, but I think I can say something. Frequently when we are using the convention that vectors are columns, we treat the row space as consisting of column vectors. This matches the definition given of the row space, where the $a_{i}$ are (column) vectors, and the $a_{i}^{T}$ are the rows of $A$; then $mathcal{R}(A)$ is the space spanned by the (column) vectors $a_{i}$. So $y$ and $z$ are both column vectors.
$endgroup$
– Morgan Rodgers
Jan 5 '17 at 15:13
$begingroup$
Ah I see thanks
$endgroup$
– aceminer
Jan 5 '17 at 15:14
add a comment |
$begingroup$
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix $A in mathbb{R}^{m times n}$. The right null space is defined as $$mathcal{N}(A) = {z in mathbb{R}^{n times 1} : Az = 0 }$$
Let $
A =
left[ {begin{array}{c}
a_1^T \
a_2^T \
ldots \
ldots \
a_m^T
end{array} } right]$. The row space of $A$ is defined as
$$mathcal{R}(A) = {y in mathbb{R}^{n times 1}: y = sum_{i=1}^m a_i x_i text{ , where }x_i in mathbb{R} text{ and }a_i in mathbb{R}^{n times 1} }$$
Now from the definition of right null space we have $a_i^T z = 0$.
So if we take a $y in mathcal{R}(A)$, then $y = displaystyle sum_{k=1}^m a_i x_i text{ , where }x_i in mathbb{R}$. Hence, $$y^Tz = (sum_{k=1}^m a_i x_i)^T z = (sum_{k=1}^m x_i a_i^T) z = sum_{k=1}^m x_i (a_i^T z) = 0$$
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of $A$ is orthogonal to the left null space of $A$.
Note: The left null space is defined as ${z in mathbb{R}^{m times 1}: z^TA = 0}$
$endgroup$
$begingroup$
@Morgan Rodgers Can you explain the part on transpose? y transpose z to form 0. If i understand correctly y is a row vector and z is a column vector. If y were to transpose, it would become a column vector
$endgroup$
– aceminer
Jan 5 '17 at 14:52
2
$begingroup$
@aceminer I'm not the author of this answer, but I think I can say something. Frequently when we are using the convention that vectors are columns, we treat the row space as consisting of column vectors. This matches the definition given of the row space, where the $a_{i}$ are (column) vectors, and the $a_{i}^{T}$ are the rows of $A$; then $mathcal{R}(A)$ is the space spanned by the (column) vectors $a_{i}$. So $y$ and $z$ are both column vectors.
$endgroup$
– Morgan Rodgers
Jan 5 '17 at 15:13
$begingroup$
Ah I see thanks
$endgroup$
– aceminer
Jan 5 '17 at 15:14
add a comment |
$begingroup$
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix $A in mathbb{R}^{m times n}$. The right null space is defined as $$mathcal{N}(A) = {z in mathbb{R}^{n times 1} : Az = 0 }$$
Let $
A =
left[ {begin{array}{c}
a_1^T \
a_2^T \
ldots \
ldots \
a_m^T
end{array} } right]$. The row space of $A$ is defined as
$$mathcal{R}(A) = {y in mathbb{R}^{n times 1}: y = sum_{i=1}^m a_i x_i text{ , where }x_i in mathbb{R} text{ and }a_i in mathbb{R}^{n times 1} }$$
Now from the definition of right null space we have $a_i^T z = 0$.
So if we take a $y in mathcal{R}(A)$, then $y = displaystyle sum_{k=1}^m a_i x_i text{ , where }x_i in mathbb{R}$. Hence, $$y^Tz = (sum_{k=1}^m a_i x_i)^T z = (sum_{k=1}^m x_i a_i^T) z = sum_{k=1}^m x_i (a_i^T z) = 0$$
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of $A$ is orthogonal to the left null space of $A$.
Note: The left null space is defined as ${z in mathbb{R}^{m times 1}: z^TA = 0}$
$endgroup$
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix $A in mathbb{R}^{m times n}$. The right null space is defined as $$mathcal{N}(A) = {z in mathbb{R}^{n times 1} : Az = 0 }$$
Let $
A =
left[ {begin{array}{c}
a_1^T \
a_2^T \
ldots \
ldots \
a_m^T
end{array} } right]$. The row space of $A$ is defined as
$$mathcal{R}(A) = {y in mathbb{R}^{n times 1}: y = sum_{i=1}^m a_i x_i text{ , where }x_i in mathbb{R} text{ and }a_i in mathbb{R}^{n times 1} }$$
Now from the definition of right null space we have $a_i^T z = 0$.
So if we take a $y in mathcal{R}(A)$, then $y = displaystyle sum_{k=1}^m a_i x_i text{ , where }x_i in mathbb{R}$. Hence, $$y^Tz = (sum_{k=1}^m a_i x_i)^T z = (sum_{k=1}^m x_i a_i^T) z = sum_{k=1}^m x_i (a_i^T z) = 0$$
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of $A$ is orthogonal to the left null space of $A$.
Note: The left null space is defined as ${z in mathbb{R}^{m times 1}: z^TA = 0}$
edited Feb 10 '16 at 13:47
Morgan Rodgers
9,78821440
9,78821440
answered Mar 25 '11 at 22:56
user17762
$begingroup$
@Morgan Rodgers Can you explain the part on transpose? y transpose z to form 0. If i understand correctly y is a row vector and z is a column vector. If y were to transpose, it would become a column vector
$endgroup$
– aceminer
Jan 5 '17 at 14:52
2
$begingroup$
@aceminer I'm not the author of this answer, but I think I can say something. Frequently when we are using the convention that vectors are columns, we treat the row space as consisting of column vectors. This matches the definition given of the row space, where the $a_{i}$ are (column) vectors, and the $a_{i}^{T}$ are the rows of $A$; then $mathcal{R}(A)$ is the space spanned by the (column) vectors $a_{i}$. So $y$ and $z$ are both column vectors.
$endgroup$
– Morgan Rodgers
Jan 5 '17 at 15:13
$begingroup$
Ah I see thanks
$endgroup$
– aceminer
Jan 5 '17 at 15:14
add a comment |
$begingroup$
@Morgan Rodgers Can you explain the part on transpose? y transpose z to form 0. If i understand correctly y is a row vector and z is a column vector. If y were to transpose, it would become a column vector
$endgroup$
– aceminer
Jan 5 '17 at 14:52
2
$begingroup$
@aceminer I'm not the author of this answer, but I think I can say something. Frequently when we are using the convention that vectors are columns, we treat the row space as consisting of column vectors. This matches the definition given of the row space, where the $a_{i}$ are (column) vectors, and the $a_{i}^{T}$ are the rows of $A$; then $mathcal{R}(A)$ is the space spanned by the (column) vectors $a_{i}$. So $y$ and $z$ are both column vectors.
$endgroup$
– Morgan Rodgers
Jan 5 '17 at 15:13
$begingroup$
Ah I see thanks
$endgroup$
– aceminer
Jan 5 '17 at 15:14
$begingroup$
@Morgan Rodgers Can you explain the part on transpose? y transpose z to form 0. If i understand correctly y is a row vector and z is a column vector. If y were to transpose, it would become a column vector
$endgroup$
– aceminer
Jan 5 '17 at 14:52
$begingroup$
@Morgan Rodgers Can you explain the part on transpose? y transpose z to form 0. If i understand correctly y is a row vector and z is a column vector. If y were to transpose, it would become a column vector
$endgroup$
– aceminer
Jan 5 '17 at 14:52
2
2
$begingroup$
@aceminer I'm not the author of this answer, but I think I can say something. Frequently when we are using the convention that vectors are columns, we treat the row space as consisting of column vectors. This matches the definition given of the row space, where the $a_{i}$ are (column) vectors, and the $a_{i}^{T}$ are the rows of $A$; then $mathcal{R}(A)$ is the space spanned by the (column) vectors $a_{i}$. So $y$ and $z$ are both column vectors.
$endgroup$
– Morgan Rodgers
Jan 5 '17 at 15:13
$begingroup$
@aceminer I'm not the author of this answer, but I think I can say something. Frequently when we are using the convention that vectors are columns, we treat the row space as consisting of column vectors. This matches the definition given of the row space, where the $a_{i}$ are (column) vectors, and the $a_{i}^{T}$ are the rows of $A$; then $mathcal{R}(A)$ is the space spanned by the (column) vectors $a_{i}$. So $y$ and $z$ are both column vectors.
$endgroup$
– Morgan Rodgers
Jan 5 '17 at 15:13
$begingroup$
Ah I see thanks
$endgroup$
– aceminer
Jan 5 '17 at 15:14
$begingroup$
Ah I see thanks
$endgroup$
– aceminer
Jan 5 '17 at 15:14
add a comment |
$begingroup$
The OP's question would have been well-phrased had he specified that the matrix $A$ is symmetric i.e. $A=A^top$, in which case ${rm colspan}(A)={rm rowspan}(A^top)={rm rowspan}(A)$. Now, consider the definition of ${rm null}(A)$ as the space of all vectors $mathbf{v}$ such that $Amathbf{v}=mathbf{0}$. Letting $mathbf{a}_1,ldots,mathbf{a}_n$ be the rows (and columns) of $A$, matrix multiplication tells us that $mathbf{a}_icdotmathbf{v}=0$ for each $i=1,ldots,dim(A)$. Thus any vector $mathbf{v}in{rm null}(A)$ is orthogonal to ${rm colspan}(A)$. It follows that ${rm null}(A)perp{rm colspan}(A)$.$square$
$endgroup$
add a comment |
$begingroup$
The OP's question would have been well-phrased had he specified that the matrix $A$ is symmetric i.e. $A=A^top$, in which case ${rm colspan}(A)={rm rowspan}(A^top)={rm rowspan}(A)$. Now, consider the definition of ${rm null}(A)$ as the space of all vectors $mathbf{v}$ such that $Amathbf{v}=mathbf{0}$. Letting $mathbf{a}_1,ldots,mathbf{a}_n$ be the rows (and columns) of $A$, matrix multiplication tells us that $mathbf{a}_icdotmathbf{v}=0$ for each $i=1,ldots,dim(A)$. Thus any vector $mathbf{v}in{rm null}(A)$ is orthogonal to ${rm colspan}(A)$. It follows that ${rm null}(A)perp{rm colspan}(A)$.$square$
$endgroup$
add a comment |
$begingroup$
The OP's question would have been well-phrased had he specified that the matrix $A$ is symmetric i.e. $A=A^top$, in which case ${rm colspan}(A)={rm rowspan}(A^top)={rm rowspan}(A)$. Now, consider the definition of ${rm null}(A)$ as the space of all vectors $mathbf{v}$ such that $Amathbf{v}=mathbf{0}$. Letting $mathbf{a}_1,ldots,mathbf{a}_n$ be the rows (and columns) of $A$, matrix multiplication tells us that $mathbf{a}_icdotmathbf{v}=0$ for each $i=1,ldots,dim(A)$. Thus any vector $mathbf{v}in{rm null}(A)$ is orthogonal to ${rm colspan}(A)$. It follows that ${rm null}(A)perp{rm colspan}(A)$.$square$
$endgroup$
The OP's question would have been well-phrased had he specified that the matrix $A$ is symmetric i.e. $A=A^top$, in which case ${rm colspan}(A)={rm rowspan}(A^top)={rm rowspan}(A)$. Now, consider the definition of ${rm null}(A)$ as the space of all vectors $mathbf{v}$ such that $Amathbf{v}=mathbf{0}$. Letting $mathbf{a}_1,ldots,mathbf{a}_n$ be the rows (and columns) of $A$, matrix multiplication tells us that $mathbf{a}_icdotmathbf{v}=0$ for each $i=1,ldots,dim(A)$. Thus any vector $mathbf{v}in{rm null}(A)$ is orthogonal to ${rm colspan}(A)$. It follows that ${rm null}(A)perp{rm colspan}(A)$.$square$
answered Sep 16 '14 at 5:20
PikalaxALTPikalaxALT
8113
8113
add a comment |
add a comment |
$begingroup$
You can't show this: it is false. Take
$$begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix};$$
both its column space and null space are the subspace ${(x, 0) mid x in mathbb{R}}$ (the $x$-axis, so to speak). I think you need self-adjointness (symmetry in the real case) to get this.
EDIT: Sivaram anticipated me.
$endgroup$
add a comment |
$begingroup$
You can't show this: it is false. Take
$$begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix};$$
both its column space and null space are the subspace ${(x, 0) mid x in mathbb{R}}$ (the $x$-axis, so to speak). I think you need self-adjointness (symmetry in the real case) to get this.
EDIT: Sivaram anticipated me.
$endgroup$
add a comment |
$begingroup$
You can't show this: it is false. Take
$$begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix};$$
both its column space and null space are the subspace ${(x, 0) mid x in mathbb{R}}$ (the $x$-axis, so to speak). I think you need self-adjointness (symmetry in the real case) to get this.
EDIT: Sivaram anticipated me.
$endgroup$
You can't show this: it is false. Take
$$begin{bmatrix} 0 & 1 \ 0 & 0 end{bmatrix};$$
both its column space and null space are the subspace ${(x, 0) mid x in mathbb{R}}$ (the $x$-axis, so to speak). I think you need self-adjointness (symmetry in the real case) to get this.
EDIT: Sivaram anticipated me.
answered Mar 25 '11 at 22:57
Giuseppe NegroGiuseppe Negro
17.4k332126
17.4k332126
add a comment |
add a comment |
$begingroup$
21
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix A∈Rm×n. The right null space is defined as
N(A)={z∈Rn×1:Az=0}
Let A=⎡⎣⎢⎢⎢⎢⎢⎢aT1aT2……aTm⎤⎦⎥⎥⎥⎥⎥⎥. The row space of A is defined as
R(A)={y∈Rn×1:y=∑i=1maixi , where xi∈R and ai∈Rn×1}
Now from the definition of right null space we have aTiz=0.
So if we take a y∈R(A), then y=∑k=1maixi , where xi∈R. Hence,
yTz=(∑k=1maixi)Tz=(∑k=1mxiaTi)z=∑k=1mxi(aTiz)=0
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of A is orthogonal to the left null space of A.
Note: The left null space is defined as {z∈Rm×1:zTA=0}
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For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
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– Kemono Chen
Dec 25 '18 at 7:38
add a comment |
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21
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix A∈Rm×n. The right null space is defined as
N(A)={z∈Rn×1:Az=0}
Let A=⎡⎣⎢⎢⎢⎢⎢⎢aT1aT2……aTm⎤⎦⎥⎥⎥⎥⎥⎥. The row space of A is defined as
R(A)={y∈Rn×1:y=∑i=1maixi , where xi∈R and ai∈Rn×1}
Now from the definition of right null space we have aTiz=0.
So if we take a y∈R(A), then y=∑k=1maixi , where xi∈R. Hence,
yTz=(∑k=1maixi)Tz=(∑k=1mxiaTi)z=∑k=1mxi(aTiz)=0
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of A is orthogonal to the left null space of A.
Note: The left null space is defined as {z∈Rm×1:zTA=0}
$endgroup$
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Kemono Chen
Dec 25 '18 at 7:38
add a comment |
$begingroup$
21
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix A∈Rm×n. The right null space is defined as
N(A)={z∈Rn×1:Az=0}
Let A=⎡⎣⎢⎢⎢⎢⎢⎢aT1aT2……aTm⎤⎦⎥⎥⎥⎥⎥⎥. The row space of A is defined as
R(A)={y∈Rn×1:y=∑i=1maixi , where xi∈R and ai∈Rn×1}
Now from the definition of right null space we have aTiz=0.
So if we take a y∈R(A), then y=∑k=1maixi , where xi∈R. Hence,
yTz=(∑k=1maixi)Tz=(∑k=1mxiaTi)z=∑k=1mxi(aTiz)=0
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of A is orthogonal to the left null space of A.
Note: The left null space is defined as {z∈Rm×1:zTA=0}
$endgroup$
21
What you have written is only correct if you are referring to the left nullspace (it is more standard to use the term "nullspace" to refer to the right nullspace).
The row space (not the column space) is orthogonal to the right null space.
Showing that row space is orthogonal to the right null space follows directly from the definition of right null space.
Let the matrix A∈Rm×n. The right null space is defined as
N(A)={z∈Rn×1:Az=0}
Let A=⎡⎣⎢⎢⎢⎢⎢⎢aT1aT2……aTm⎤⎦⎥⎥⎥⎥⎥⎥. The row space of A is defined as
R(A)={y∈Rn×1:y=∑i=1maixi , where xi∈R and ai∈Rn×1}
Now from the definition of right null space we have aTiz=0.
So if we take a y∈R(A), then y=∑k=1maixi , where xi∈R. Hence,
yTz=(∑k=1maixi)Tz=(∑k=1mxiaTi)z=∑k=1mxi(aTiz)=0
This proves that row space is orthogonal to the right null space. A similar analysis proves that column space of A is orthogonal to the left null space of A.
Note: The left null space is defined as {z∈Rm×1:zTA=0}
answered Dec 25 '18 at 7:32
user629195user629195
1
1
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For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Kemono Chen
Dec 25 '18 at 7:38
add a comment |
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Kemono Chen
Dec 25 '18 at 7:38
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Kemono Chen
Dec 25 '18 at 7:38
$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Kemono Chen
Dec 25 '18 at 7:38
add a comment |
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You should make your question self-contained (such that it is understandable without referring to the title)...
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– Fabian
Mar 25 '11 at 22:54
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The following illustration might help: en.wikipedia.org/wiki/Kernel_%28linear_algebra%29#Illustration
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– Dor
Jan 28 '15 at 9:08