How to prove that orthogonal complement of null space is range of transpose of A [duplicate]
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The range of $T^*$ is the orthogonal complement of $Ker(T)$
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Let A be a linear map from $R^n$ to $R^m$, Then
Prove that $null(A)^bot = range(A^T)$.
So far, I can easily show that $range(A^T) subset null(A)^bot$. See below.
if $x in range(A^T)$, then $(x,n)=(A^Tu,n)=(u^T,An)=(u^T,0)=0$ for all $n in null(A)^bot$.
But I cannot proof the opposite direction.
How can I prove that if $x in null(A)^bot,$ then $ x in range(A^T)$?
linear-algebra vector-spaces
asked Aug 22 at 9:17
DongukJu
725
marked as duplicate by Arnaud D., uniquesolution, KittyL, Claudius, Adrian Keister Aug 23 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
0
down vote
favorite
This question already has an answer here:
The range of $T^*$ is the orthogonal complement of $Ker(T)$
2 answers
Let A be a linear map from $R^n$ to $R^m$, Then
Prove that $null(A)^bot = range(A^T)$.
So far, I can easily show that $range(A^T) subset null(A)^bot$. See below.
if $x in range(A^T)$, then $(x,n)=(A^Tu,n)=(u^T,An)=(u^T,0)=0$ for all $n in null(A)^bot$.
But I cannot proof the opposite direction.
How can I prove that if $x in null(A)^bot,$ then $ x in range(A^T)$?
linear-algebra vector-spaces
asked Aug 22 at 9:17
DongukJu
725
marked as duplicate by Arnaud D., uniquesolution, KittyL, Claudius, Adrian Keister Aug 23 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
No, it's different. Without any verification we cannot complement the orthogonal complement.
– DongukJu
Aug 22 at 9:28
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up vote
0
down vote
favorite
This question already has an answer here:
The range of $T^*$ is the orthogonal complement of $Ker(T)$
2 answers
Let A be a linear map from $R^n$ to $R^m$, Then
Prove that $null(A)^bot = range(A^T)$.
So far, I can easily show that $range(A^T) subset null(A)^bot$. See below.
if $x in range(A^T)$, then $(x,n)=(A^Tu,n)=(u^T,An)=(u^T,0)=0$ for all $n in null(A)^bot$.
But I cannot proof the opposite direction.
How can I prove that if $x in null(A)^bot,$ then $ x in range(A^T)$?
linear-algebra vector-spaces
asked Aug 22 at 9:17
DongukJu
725
This question already has an answer here:
The range of $T^*$ is the orthogonal complement of $Ker(T)$
2 answers
Let A be a linear map from $R^n$ to $R^m$, Then
Prove that $null(A)^bot = range(A^T)$.
So far, I can easily show that $range(A^T) subset null(A)^bot$. See below.
if $x in range(A^T)$, then $(x,n)=(A^Tu,n)=(u^T,An)=(u^T,0)=0$ for all $n in null(A)^bot$.
But I cannot proof the opposite direction.
How can I prove that if $x in null(A)^bot,$ then $ x in range(A^T)$?
This question already has an answer here:
The range of $T^*$ is the orthogonal complement of $Ker(T)$
2 answers
linear-algebra vector-spaces
asked Aug 22 at 9:17
DongukJu
725
asked Aug 22 at 9:17
DongukJu
725
asked Aug 22 at 9:17
DongukJu
725
asked Aug 22 at 9:17
DongukJu
725
725
marked as duplicate by Arnaud D., uniquesolution, KittyL, Claudius, Adrian Keister Aug 23 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arnaud D., uniquesolution, KittyL, Claudius, Adrian Keister Aug 23 at 13:03
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
No, it's different. Without any verification we cannot complement the orthogonal complement.
– DongukJu
Aug 22 at 9:28
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No, it's different. Without any verification we cannot complement the orthogonal complement.
– DongukJu
Aug 22 at 9:28
No, it's different. Without any verification we cannot complement the orthogonal complement.
– DongukJu
Aug 22 at 9:28
No, it's different. Without any verification we cannot complement the orthogonal complement.
– DongukJu
Aug 22 at 9:28
add a comment |Â
1 Answer
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Suppose $x^T in null(A)^perp$, we want to show that $x^T in range(A^T)$, in other words, $x^T$ is in the row space of $A$.
Let $d$ be the dimension of the row space of $A$ and $ x_1^T, ldots, x_d^T$ be a basis of the row space of $A$.
Then the dimension of $null(A)$ is $n-d$. Construct the matrix $B=beginbmatrix y_1 ldots y_n-dendbmatrix$ where each columns are independent and they are elements of nullspace of $A$. By definition, we have $AB=0$which is just $B^TA^T=0$.
Rank of $B^T$ is $n-d$, hence $null(B^T)=n-(n-d)=d$, hence $ x_1, ldots, x_d$ is a basis of $null(B^T)$.
Since $x^T in null(A)^perp$, we must have $B^Tx=0$, that is $x in null(B^T)$, which means $x$ can be expressed as $$x=sum_i=1^d c_i x_i$$
$$x^T=sum_i=1^d c_i x_i^T$$
That is $x^T$ is in the row space of $A$.
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Suppose $x^T in null(A)^perp$, we want to show that $x^T in range(A^T)$, in other words, $x^T$ is in the row space of $A$.
Let $d$ be the dimension of the row space of $A$ and $ x_1^T, ldots, x_d^T$ be a basis of the row space of $A$.
Then the dimension of $null(A)$ is $n-d$. Construct the matrix $B=beginbmatrix y_1 ldots y_n-dendbmatrix$ where each columns are independent and they are elements of nullspace of $A$. By definition, we have $AB=0$which is just $B^TA^T=0$.
Rank of $B^T$ is $n-d$, hence $null(B^T)=n-(n-d)=d$, hence $ x_1, ldots, x_d$ is a basis of $null(B^T)$.
Since $x^T in null(A)^perp$, we must have $B^Tx=0$, that is $x in null(B^T)$, which means $x$ can be expressed as $$x=sum_i=1^d c_i x_i$$
$$x^T=sum_i=1^d c_i x_i^T$$
That is $x^T$ is in the row space of $A$.
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
add a comment |Â
up vote
0
down vote
Suppose $x^T in null(A)^perp$, we want to show that $x^T in range(A^T)$, in other words, $x^T$ is in the row space of $A$.
Let $d$ be the dimension of the row space of $A$ and $ x_1^T, ldots, x_d^T$ be a basis of the row space of $A$.
Then the dimension of $null(A)$ is $n-d$. Construct the matrix $B=beginbmatrix y_1 ldots y_n-dendbmatrix$ where each columns are independent and they are elements of nullspace of $A$. By definition, we have $AB=0$which is just $B^TA^T=0$.
Rank of $B^T$ is $n-d$, hence $null(B^T)=n-(n-d)=d$, hence $ x_1, ldots, x_d$ is a basis of $null(B^T)$.
Since $x^T in null(A)^perp$, we must have $B^Tx=0$, that is $x in null(B^T)$, which means $x$ can be expressed as $$x=sum_i=1^d c_i x_i$$
$$x^T=sum_i=1^d c_i x_i^T$$
That is $x^T$ is in the row space of $A$.
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Suppose $x^T in null(A)^perp$, we want to show that $x^T in range(A^T)$, in other words, $x^T$ is in the row space of $A$.
Let $d$ be the dimension of the row space of $A$ and $ x_1^T, ldots, x_d^T$ be a basis of the row space of $A$.
Then the dimension of $null(A)$ is $n-d$. Construct the matrix $B=beginbmatrix y_1 ldots y_n-dendbmatrix$ where each columns are independent and they are elements of nullspace of $A$. By definition, we have $AB=0$which is just $B^TA^T=0$.
Rank of $B^T$ is $n-d$, hence $null(B^T)=n-(n-d)=d$, hence $ x_1, ldots, x_d$ is a basis of $null(B^T)$.
Since $x^T in null(A)^perp$, we must have $B^Tx=0$, that is $x in null(B^T)$, which means $x$ can be expressed as $$x=sum_i=1^d c_i x_i$$
$$x^T=sum_i=1^d c_i x_i^T$$
That is $x^T$ is in the row space of $A$.
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
Suppose $x^T in null(A)^perp$, we want to show that $x^T in range(A^T)$, in other words, $x^T$ is in the row space of $A$.
Let $d$ be the dimension of the row space of $A$ and $ x_1^T, ldots, x_d^T$ be a basis of the row space of $A$.
Then the dimension of $null(A)$ is $n-d$. Construct the matrix $B=beginbmatrix y_1 ldots y_n-dendbmatrix$ where each columns are independent and they are elements of nullspace of $A$. By definition, we have $AB=0$which is just $B^TA^T=0$.
Rank of $B^T$ is $n-d$, hence $null(B^T)=n-(n-d)=d$, hence $ x_1, ldots, x_d$ is a basis of $null(B^T)$.
Since $x^T in null(A)^perp$, we must have $B^Tx=0$, that is $x in null(B^T)$, which means $x$ can be expressed as $$x=sum_i=1^d c_i x_i$$
$$x^T=sum_i=1^d c_i x_i^T$$
That is $x^T$ is in the row space of $A$.
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
answered Aug 22 at 18:44
Siong Thye Goh
80.5k1453101
80.5k1453101
add a comment |Â
add a comment |Â
No, it's different. Without any verification we cannot complement the orthogonal complement.
– DongukJu
Aug 22 at 9:28