Intersection of row space and null space [closed]
I want to prove that
$N(A)$ $intersection$ $R(A)$={$0$} where $A$ be a $m×n$ matrix. & $N(A)$ is the null space of $A$ & $R(A)$ is the row space $A$.
Using their orthogonality we can say that my statement is true but I am trying another method.....
In faCt, I am trying to prove dimension of the intersection of the subspaces is 0.That's all.
linear-algebra vector-spaces
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
closed as off-topic by José Carlos Santos, Kavi Rama Murthy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I want to prove that
$N(A)$ $intersection$ $R(A)$={$0$} where $A$ be a $m×n$ matrix. & $N(A)$ is the null space of $A$ & $R(A)$ is the row space $A$.
Using their orthogonality we can say that my statement is true but I am trying another method.....
In faCt, I am trying to prove dimension of the intersection of the subspaces is 0.That's all.
linear-algebra vector-spaces
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
closed as off-topic by José Carlos Santos, Kavi Rama Murthy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
If $A$ is a square matrix, then in general $N(A) cap R(A) ne {0}$ ! Question: what are the properties of $A$ ?
– Fred
Nov 28 '18 at 12:11
1
@Fred please give me a counter example ...
– Supriyo Banerjee
Nov 28 '18 at 12:23
It might be of some use to clearly define the row-space. Is it a vector space consisting of the span of the rows of the matrix? If so, does it consist of row vectors, or are they all transposed to be column vectors? BTW, I suspect Fred's comment comes from the fact that $R(A)$ is sometimes used to denote the range of $A$ rather than the row-space; under that interpretation, the intersection may indeed be nonempty.
– John Hughes
Nov 28 '18 at 12:28
@JohnHughes I edited the body of the question please see that
– Supriyo Banerjee
Nov 28 '18 at 12:39
Yes, the row space is the vector space spanned by the rows of the matrix $A$
– Supriyo Banerjee
Nov 28 '18 at 12:40
add a comment |
I want to prove that
$N(A)$ $intersection$ $R(A)$={$0$} where $A$ be a $m×n$ matrix. & $N(A)$ is the null space of $A$ & $R(A)$ is the row space $A$.
Using their orthogonality we can say that my statement is true but I am trying another method.....
In faCt, I am trying to prove dimension of the intersection of the subspaces is 0.That's all.
linear-algebra vector-spaces
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
I want to prove that
$N(A)$ $intersection$ $R(A)$={$0$} where $A$ be a $m×n$ matrix. & $N(A)$ is the null space of $A$ & $R(A)$ is the row space $A$.
Using their orthogonality we can say that my statement is true but I am trying another method.....
In faCt, I am trying to prove dimension of the intersection of the subspaces is 0.That's all.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
edited Nov 28 '18 at 12:33
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
asked Nov 28 '18 at 12:04
Supriyo Banerjee
625
625
closed as off-topic by José Carlos Santos, Kavi Rama Murthy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by José Carlos Santos, Kavi Rama Murthy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:37
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
If $A$ is a square matrix, then in general $N(A) cap R(A) ne {0}$ ! Question: what are the properties of $A$ ?
– Fred
Nov 28 '18 at 12:11
1
@Fred please give me a counter example ...
– Supriyo Banerjee
Nov 28 '18 at 12:23
It might be of some use to clearly define the row-space. Is it a vector space consisting of the span of the rows of the matrix? If so, does it consist of row vectors, or are they all transposed to be column vectors? BTW, I suspect Fred's comment comes from the fact that $R(A)$ is sometimes used to denote the range of $A$ rather than the row-space; under that interpretation, the intersection may indeed be nonempty.
– John Hughes
Nov 28 '18 at 12:28
@JohnHughes I edited the body of the question please see that
– Supriyo Banerjee
Nov 28 '18 at 12:39
Yes, the row space is the vector space spanned by the rows of the matrix $A$
– Supriyo Banerjee
Nov 28 '18 at 12:40
add a comment |
If $A$ is a square matrix, then in general $N(A) cap R(A) ne {0}$ ! Question: what are the properties of $A$ ?
– Fred
Nov 28 '18 at 12:11
1
@Fred please give me a counter example ...
– Supriyo Banerjee
Nov 28 '18 at 12:23
It might be of some use to clearly define the row-space. Is it a vector space consisting of the span of the rows of the matrix? If so, does it consist of row vectors, or are they all transposed to be column vectors? BTW, I suspect Fred's comment comes from the fact that $R(A)$ is sometimes used to denote the range of $A$ rather than the row-space; under that interpretation, the intersection may indeed be nonempty.
– John Hughes
Nov 28 '18 at 12:28
@JohnHughes I edited the body of the question please see that
– Supriyo Banerjee
Nov 28 '18 at 12:39
Yes, the row space is the vector space spanned by the rows of the matrix $A$
– Supriyo Banerjee
Nov 28 '18 at 12:40
If $A$ is a square matrix, then in general $N(A) cap R(A) ne {0}$ ! Question: what are the properties of $A$ ?
– Fred
Nov 28 '18 at 12:11
If $A$ is a square matrix, then in general $N(A) cap R(A) ne {0}$ ! Question: what are the properties of $A$ ?
– Fred
Nov 28 '18 at 12:11
1
1
@Fred please give me a counter example ...
– Supriyo Banerjee
Nov 28 '18 at 12:23
@Fred please give me a counter example ...
– Supriyo Banerjee
Nov 28 '18 at 12:23
It might be of some use to clearly define the row-space. Is it a vector space consisting of the span of the rows of the matrix? If so, does it consist of row vectors, or are they all transposed to be column vectors? BTW, I suspect Fred's comment comes from the fact that $R(A)$ is sometimes used to denote the range of $A$ rather than the row-space; under that interpretation, the intersection may indeed be nonempty.
– John Hughes
Nov 28 '18 at 12:28
It might be of some use to clearly define the row-space. Is it a vector space consisting of the span of the rows of the matrix? If so, does it consist of row vectors, or are they all transposed to be column vectors? BTW, I suspect Fred's comment comes from the fact that $R(A)$ is sometimes used to denote the range of $A$ rather than the row-space; under that interpretation, the intersection may indeed be nonempty.
– John Hughes
Nov 28 '18 at 12:28
@JohnHughes I edited the body of the question please see that
– Supriyo Banerjee
Nov 28 '18 at 12:39
@JohnHughes I edited the body of the question please see that
– Supriyo Banerjee
Nov 28 '18 at 12:39
Yes, the row space is the vector space spanned by the rows of the matrix $A$
– Supriyo Banerjee
Nov 28 '18 at 12:40
Yes, the row space is the vector space spanned by the rows of the matrix $A$
– Supriyo Banerjee
Nov 28 '18 at 12:40
add a comment |
2 Answers
2
active
oldest
votes
If you want to consider the intersection of $N(A)$ and $R(A)$ (row space), you have to define the latter as the column space of $A^T$, because the span of the rows of $A$ lives in a different vector space than $N(A)$.
With that interpretation, let $vin N(A)cap R(A)$. Then there exists $x$ such that $v=A^Tx$. Also $Av=0$, so $AA^Tx=0$; in particular
$$
0=x^TAA^Tx=(A^Tx)^T(A^Tx)=v^Tv
$$
implying $v=0$.
Here $X^T$ denotes the transpose matrix.
answered Nov 28 '18 at 12:41
egreg
178k1484201
Thank you so much......
– Supriyo Banerjee
Nov 28 '18 at 12:53
add a comment |
Orthogonality is how you prove this, since it's a fundamental definition of the null space of a matrix: the null space is the space that is orthogonal to the row space. We can make some progress with dimensions, but we'll have to return to the orthogonality argument eventually:
If by some reason the two spaces had non trivial intersection, then some nonzero vectors in the null space could be represented as linear combinations of those in the row space. If that were the case, then some sub-space of the row space would be the basis for a sub-space of the null space, and therefore the some nonzero vectors of one are linearly dependent on nonzero vectors in the other. However, there is a theorem that states that orthogonal sets of vectors are always linearly independent. This is a contradiction unless the only vector in the intersection is the zero vector.
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you want to consider the intersection of $N(A)$ and $R(A)$ (row space), you have to define the latter as the column space of $A^T$, because the span of the rows of $A$ lives in a different vector space than $N(A)$.
With that interpretation, let $vin N(A)cap R(A)$. Then there exists $x$ such that $v=A^Tx$. Also $Av=0$, so $AA^Tx=0$; in particular
$$
0=x^TAA^Tx=(A^Tx)^T(A^Tx)=v^Tv
$$
implying $v=0$.
Here $X^T$ denotes the transpose matrix.
answered Nov 28 '18 at 12:41
egreg
178k1484201
Thank you so much......
– Supriyo Banerjee
Nov 28 '18 at 12:53
add a comment |
If you want to consider the intersection of $N(A)$ and $R(A)$ (row space), you have to define the latter as the column space of $A^T$, because the span of the rows of $A$ lives in a different vector space than $N(A)$.
With that interpretation, let $vin N(A)cap R(A)$. Then there exists $x$ such that $v=A^Tx$. Also $Av=0$, so $AA^Tx=0$; in particular
$$
0=x^TAA^Tx=(A^Tx)^T(A^Tx)=v^Tv
$$
implying $v=0$.
Here $X^T$ denotes the transpose matrix.
answered Nov 28 '18 at 12:41
egreg
178k1484201
Thank you so much......
– Supriyo Banerjee
Nov 28 '18 at 12:53
add a comment |
If you want to consider the intersection of $N(A)$ and $R(A)$ (row space), you have to define the latter as the column space of $A^T$, because the span of the rows of $A$ lives in a different vector space than $N(A)$.
With that interpretation, let $vin N(A)cap R(A)$. Then there exists $x$ such that $v=A^Tx$. Also $Av=0$, so $AA^Tx=0$; in particular
$$
0=x^TAA^Tx=(A^Tx)^T(A^Tx)=v^Tv
$$
implying $v=0$.
Here $X^T$ denotes the transpose matrix.
answered Nov 28 '18 at 12:41
egreg
178k1484201
If you want to consider the intersection of $N(A)$ and $R(A)$ (row space), you have to define the latter as the column space of $A^T$, because the span of the rows of $A$ lives in a different vector space than $N(A)$.
With that interpretation, let $vin N(A)cap R(A)$. Then there exists $x$ such that $v=A^Tx$. Also $Av=0$, so $AA^Tx=0$; in particular
$$
0=x^TAA^Tx=(A^Tx)^T(A^Tx)=v^Tv
$$
implying $v=0$.
Here $X^T$ denotes the transpose matrix.
answered Nov 28 '18 at 12:41
egreg
178k1484201
answered Nov 28 '18 at 12:41
egreg
178k1484201
answered Nov 28 '18 at 12:41
egreg
178k1484201
answered Nov 28 '18 at 12:41
egreg
178k1484201
178k1484201
Thank you so much......
– Supriyo Banerjee
Nov 28 '18 at 12:53
add a comment |
Thank you so much......
– Supriyo Banerjee
Nov 28 '18 at 12:53
Thank you so much......
– Supriyo Banerjee
Nov 28 '18 at 12:53
Thank you so much......
– Supriyo Banerjee
Nov 28 '18 at 12:53
add a comment |
Orthogonality is how you prove this, since it's a fundamental definition of the null space of a matrix: the null space is the space that is orthogonal to the row space. We can make some progress with dimensions, but we'll have to return to the orthogonality argument eventually:
If by some reason the two spaces had non trivial intersection, then some nonzero vectors in the null space could be represented as linear combinations of those in the row space. If that were the case, then some sub-space of the row space would be the basis for a sub-space of the null space, and therefore the some nonzero vectors of one are linearly dependent on nonzero vectors in the other. However, there is a theorem that states that orthogonal sets of vectors are always linearly independent. This is a contradiction unless the only vector in the intersection is the zero vector.
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
add a comment |
Orthogonality is how you prove this, since it's a fundamental definition of the null space of a matrix: the null space is the space that is orthogonal to the row space. We can make some progress with dimensions, but we'll have to return to the orthogonality argument eventually:
If by some reason the two spaces had non trivial intersection, then some nonzero vectors in the null space could be represented as linear combinations of those in the row space. If that were the case, then some sub-space of the row space would be the basis for a sub-space of the null space, and therefore the some nonzero vectors of one are linearly dependent on nonzero vectors in the other. However, there is a theorem that states that orthogonal sets of vectors are always linearly independent. This is a contradiction unless the only vector in the intersection is the zero vector.
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
add a comment |
Orthogonality is how you prove this, since it's a fundamental definition of the null space of a matrix: the null space is the space that is orthogonal to the row space. We can make some progress with dimensions, but we'll have to return to the orthogonality argument eventually:
If by some reason the two spaces had non trivial intersection, then some nonzero vectors in the null space could be represented as linear combinations of those in the row space. If that were the case, then some sub-space of the row space would be the basis for a sub-space of the null space, and therefore the some nonzero vectors of one are linearly dependent on nonzero vectors in the other. However, there is a theorem that states that orthogonal sets of vectors are always linearly independent. This is a contradiction unless the only vector in the intersection is the zero vector.
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
Orthogonality is how you prove this, since it's a fundamental definition of the null space of a matrix: the null space is the space that is orthogonal to the row space. We can make some progress with dimensions, but we'll have to return to the orthogonality argument eventually:
If by some reason the two spaces had non trivial intersection, then some nonzero vectors in the null space could be represented as linear combinations of those in the row space. If that were the case, then some sub-space of the row space would be the basis for a sub-space of the null space, and therefore the some nonzero vectors of one are linearly dependent on nonzero vectors in the other. However, there is a theorem that states that orthogonal sets of vectors are always linearly independent. This is a contradiction unless the only vector in the intersection is the zero vector.
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
answered Nov 28 '18 at 12:43
Michael Stachowsky
1,260417
1,260417
add a comment |
add a comment |
If $A$ is a square matrix, then in general $N(A) cap R(A) ne {0}$ ! Question: what are the properties of $A$ ?
– Fred
Nov 28 '18 at 12:11
1
@Fred please give me a counter example ...
– Supriyo Banerjee
Nov 28 '18 at 12:23
It might be of some use to clearly define the row-space. Is it a vector space consisting of the span of the rows of the matrix? If so, does it consist of row vectors, or are they all transposed to be column vectors? BTW, I suspect Fred's comment comes from the fact that $R(A)$ is sometimes used to denote the range of $A$ rather than the row-space; under that interpretation, the intersection may indeed be nonempty.
– John Hughes
Nov 28 '18 at 12:28
@JohnHughes I edited the body of the question please see that
– Supriyo Banerjee
Nov 28 '18 at 12:39
Yes, the row space is the vector space spanned by the rows of the matrix $A$
– Supriyo Banerjee
Nov 28 '18 at 12:40