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A set of $n$ vectors $A_1,dots, A_n$ in $n$-space is independent iff $d(A_1,dots, A_n) ne 0$.
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I found a proof of this theorem in the book Multivariate Calculus VOL 2 by T M Apostol. But in that proof I can't understand one assertion. I paste the proof here and also bold that line:
Theorem 3.6 (P-83). A set of $n$ vectors $A_1,dots, A_n$ in $n$-space is independent if and only if $d(A_1,dots, A_n) ne 0$.
Proof.(Only one direction) Assume that $A_1,dots, A_n$ are independent. Let $V_n$ denotes the linear space of $n$-tuples of scalars. Since $A_1,dots, A_n$ are $n$ independent elements in an $n$-dimensional space they form a basis for $V_n$. Therefore there is a linear transformation $T:V_n to V_n$ which maps these $n$ vectors onto the unit coordinate vectors, $$T(A_k)=I_k ~ textfor~ k=1,dots,n$$ Therefore there is an $ntimes n$ matrix $B$ such that $$bfA_kB=I_k ~ textfor ~k=1,dots, n$$....Now the proof continues.....
But I cannot guess how Such a matrix $B$ exists satisfying that condition...!!!
Please help me to clarify this existence of $B$. Thank you.
linear-algebra linear-transformations
asked Jul 24 at 4:17
Indrajit Ghosh
578415
 |Â
show 1 more comment
up vote
1
down vote
favorite
I found a proof of this theorem in the book Multivariate Calculus VOL 2 by T M Apostol. But in that proof I can't understand one assertion. I paste the proof here and also bold that line:
Theorem 3.6 (P-83). A set of $n$ vectors $A_1,dots, A_n$ in $n$-space is independent if and only if $d(A_1,dots, A_n) ne 0$.
Proof.(Only one direction) Assume that $A_1,dots, A_n$ are independent. Let $V_n$ denotes the linear space of $n$-tuples of scalars. Since $A_1,dots, A_n$ are $n$ independent elements in an $n$-dimensional space they form a basis for $V_n$. Therefore there is a linear transformation $T:V_n to V_n$ which maps these $n$ vectors onto the unit coordinate vectors, $$T(A_k)=I_k ~ textfor~ k=1,dots,n$$ Therefore there is an $ntimes n$ matrix $B$ such that $$bfA_kB=I_k ~ textfor ~k=1,dots, n$$....Now the proof continues.....
But I cannot guess how Such a matrix $B$ exists satisfying that condition...!!!
Please help me to clarify this existence of $B$. Thank you.
linear-algebra linear-transformations
asked Jul 24 at 4:17
Indrajit Ghosh
578415
en.wikipedia.org/wiki/Linear_map#Matrices
– Lorenzo
Jul 24 at 4:23
The existence of $B$ is the inverse $AA^-1 = A^-1A = I_k$
– RHowe
Jul 24 at 4:28
I know this... is B the matrix of T here? But I actually confused why $A_k$ is pre multiplied with B ti get $I_k$...
– Indrajit Ghosh
Jul 24 at 4:29
@Geronimo ..I can't get u..
– Indrajit Ghosh
Jul 24 at 4:33
@Geronimo The notation $I_k$ does not represent the identity matrix here.
– Dave
Jul 24 at 5:27
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I found a proof of this theorem in the book Multivariate Calculus VOL 2 by T M Apostol. But in that proof I can't understand one assertion. I paste the proof here and also bold that line:
Theorem 3.6 (P-83). A set of $n$ vectors $A_1,dots, A_n$ in $n$-space is independent if and only if $d(A_1,dots, A_n) ne 0$.
Proof.(Only one direction) Assume that $A_1,dots, A_n$ are independent. Let $V_n$ denotes the linear space of $n$-tuples of scalars. Since $A_1,dots, A_n$ are $n$ independent elements in an $n$-dimensional space they form a basis for $V_n$. Therefore there is a linear transformation $T:V_n to V_n$ which maps these $n$ vectors onto the unit coordinate vectors, $$T(A_k)=I_k ~ textfor~ k=1,dots,n$$ Therefore there is an $ntimes n$ matrix $B$ such that $$bfA_kB=I_k ~ textfor ~k=1,dots, n$$....Now the proof continues.....
But I cannot guess how Such a matrix $B$ exists satisfying that condition...!!!
Please help me to clarify this existence of $B$. Thank you.
linear-algebra linear-transformations
asked Jul 24 at 4:17
Indrajit Ghosh
578415
I found a proof of this theorem in the book Multivariate Calculus VOL 2 by T M Apostol. But in that proof I can't understand one assertion. I paste the proof here and also bold that line:
Theorem 3.6 (P-83). A set of $n$ vectors $A_1,dots, A_n$ in $n$-space is independent if and only if $d(A_1,dots, A_n) ne 0$.
Proof.(Only one direction) Assume that $A_1,dots, A_n$ are independent. Let $V_n$ denotes the linear space of $n$-tuples of scalars. Since $A_1,dots, A_n$ are $n$ independent elements in an $n$-dimensional space they form a basis for $V_n$. Therefore there is a linear transformation $T:V_n to V_n$ which maps these $n$ vectors onto the unit coordinate vectors, $$T(A_k)=I_k ~ textfor~ k=1,dots,n$$ Therefore there is an $ntimes n$ matrix $B$ such that $$bfA_kB=I_k ~ textfor ~k=1,dots, n$$....Now the proof continues.....
But I cannot guess how Such a matrix $B$ exists satisfying that condition...!!!
Please help me to clarify this existence of $B$. Thank you.
linear-algebra linear-transformations
asked Jul 24 at 4:17
Indrajit Ghosh
578415
asked Jul 24 at 4:17
Indrajit Ghosh
578415
asked Jul 24 at 4:17
Indrajit Ghosh
578415
asked Jul 24 at 4:17
Indrajit Ghosh
578415
578415
en.wikipedia.org/wiki/Linear_map#Matrices
– Lorenzo
Jul 24 at 4:23
The existence of $B$ is the inverse $AA^-1 = A^-1A = I_k$
– RHowe
Jul 24 at 4:28
I know this... is B the matrix of T here? But I actually confused why $A_k$ is pre multiplied with B ti get $I_k$...
– Indrajit Ghosh
Jul 24 at 4:29
@Geronimo ..I can't get u..
– Indrajit Ghosh
Jul 24 at 4:33
@Geronimo The notation $I_k$ does not represent the identity matrix here.
– Dave
Jul 24 at 5:27
 |Â
show 1 more comment
en.wikipedia.org/wiki/Linear_map#Matrices
– Lorenzo
Jul 24 at 4:23
The existence of $B$ is the inverse $AA^-1 = A^-1A = I_k$
– RHowe
Jul 24 at 4:28
I know this... is B the matrix of T here? But I actually confused why $A_k$ is pre multiplied with B ti get $I_k$...
– Indrajit Ghosh
Jul 24 at 4:29
@Geronimo ..I can't get u..
– Indrajit Ghosh
Jul 24 at 4:33
@Geronimo The notation $I_k$ does not represent the identity matrix here.
– Dave
Jul 24 at 5:27
en.wikipedia.org/wiki/Linear_map#Matrices
– Lorenzo
Jul 24 at 4:23
en.wikipedia.org/wiki/Linear_map#Matrices
– Lorenzo
Jul 24 at 4:23
The existence of $B$ is the inverse $AA^-1 = A^-1A = I_k$
– RHowe
Jul 24 at 4:28
The existence of $B$ is the inverse $AA^-1 = A^-1A = I_k$
– RHowe
Jul 24 at 4:28
I know this... is B the matrix of T here? But I actually confused why $A_k$ is pre multiplied with B ti get $I_k$...
– Indrajit Ghosh
Jul 24 at 4:29
I know this... is B the matrix of T here? But I actually confused why $A_k$ is pre multiplied with B ti get $I_k$...
– Indrajit Ghosh
Jul 24 at 4:29
@Geronimo ..I can't get u..
– Indrajit Ghosh
Jul 24 at 4:33
@Geronimo ..I can't get u..
– Indrajit Ghosh
Jul 24 at 4:33
@Geronimo The notation $I_k$ does not represent the identity matrix here.
– Dave
Jul 24 at 5:27
@Geronimo The notation $I_k$ does not represent the identity matrix here.
– Dave
Jul 24 at 5:27
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
I've had a look at this chapter of Apostol's book, and I have to agree that, unless I'm missing something, this assertion is poorly justified on the basis of the theory developed to that point.
Apostol states the correspondence between linear mappings and matrices but, as far as I can tell, fails to spell out that the effect of the linear mapping on coordinates is given by matrix multiplication. This fact should have been noted explicitly in the later section that defines matrix multiplication.
There are two essential points to understand to justify this part of the proof of Theorem 3.6.
First, in the statement of Theorem 2.13, if we let $X = (x_1, dots, x_n)$ and $Y = (y_1, dots, y_m)$ be the coordinate representations, as column vectors, of $x$ and $y$ in the given bases, then equation $(2.13)$ can be written in the matrix form $Y = CX$, where $C = (t_ik)$ is the matrix of $T$ relative to the given bases.
If we instead consider $X$ and $Y$ as row vectors, then we must write $Y = X C^t$. Here, $C^t$ denotes the transpose of $C$, a concept defined later on page 91. Since Apostol doesn't develop the properties of transposes, it's best to check the equivalence of this matrix equality with equation $(2.13)$ directly.
Second, according to Theorem 2.13, there is a matrix C associated with the linear mapping $T$, if the standard basis $(I_1,dots,I_n)$ is selected in each of the two copies of $V_n$.
Then by Theorem 2.13, and remembering that $A_k$ and $I_k$ are considered row vectors in this chapter, we have $I_k = A_k C^t$ for each index $k$. This assertion relies on the fact that in the basis $(I_1,dots,I_n)$, the coordinates of the vector $A_k$ are in fact the elements of $A_k$.
So we can take $B = C^t$.
answered Jul 24 at 5:12
Dave
912
Oh... I see... Thank you so much....!!
– Indrajit Ghosh
Jul 24 at 5:16
I don't have access to the first edition of the book, but I suspect this is the kind of error that might have been introduced in the process of producing a new edition.
– Dave
Jul 24 at 5:19
add a comment |Â
up vote
0
down vote
If you look in your book in chapter 2.19
There is a theorem for inverses of square matrices.Now when you read the proof.
The reasoning is somewhat circular however the idea actually follows from this.
If
$$ A = U Lambda U^T $$
then
$$ det(A) = det(U Lambda U^T) = det(U)det(Lambda)det(U^T) $$
now the matrices $U,U^T$ are orthogonal have determinant one
$$det(A) = det(Lambda) = prod_i=1^n lambda_i $$
if the matrix is not linearly independent then one of the eigenvalues is zero. Then the product of the eigenvalues is zero.
answered Jul 24 at 5:45
RHowe
1,010815
How does this relate to the OP's question?
– copper.hat
Jul 24 at 14:22
It is the proof in the book he is referring to
– RHowe
Jul 24 at 14:23
He didnt even read the page before hand.
– RHowe
Jul 24 at 14:24
How do you know? And how does it relate to the OP's question?
– copper.hat
Jul 24 at 14:56
does this help...the proof was in chapter 2
– RHowe
Jul 24 at 15:06
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I've had a look at this chapter of Apostol's book, and I have to agree that, unless I'm missing something, this assertion is poorly justified on the basis of the theory developed to that point.
Apostol states the correspondence between linear mappings and matrices but, as far as I can tell, fails to spell out that the effect of the linear mapping on coordinates is given by matrix multiplication. This fact should have been noted explicitly in the later section that defines matrix multiplication.
There are two essential points to understand to justify this part of the proof of Theorem 3.6.
First, in the statement of Theorem 2.13, if we let $X = (x_1, dots, x_n)$ and $Y = (y_1, dots, y_m)$ be the coordinate representations, as column vectors, of $x$ and $y$ in the given bases, then equation $(2.13)$ can be written in the matrix form $Y = CX$, where $C = (t_ik)$ is the matrix of $T$ relative to the given bases.
If we instead consider $X$ and $Y$ as row vectors, then we must write $Y = X C^t$. Here, $C^t$ denotes the transpose of $C$, a concept defined later on page 91. Since Apostol doesn't develop the properties of transposes, it's best to check the equivalence of this matrix equality with equation $(2.13)$ directly.
Second, according to Theorem 2.13, there is a matrix C associated with the linear mapping $T$, if the standard basis $(I_1,dots,I_n)$ is selected in each of the two copies of $V_n$.
Then by Theorem 2.13, and remembering that $A_k$ and $I_k$ are considered row vectors in this chapter, we have $I_k = A_k C^t$ for each index $k$. This assertion relies on the fact that in the basis $(I_1,dots,I_n)$, the coordinates of the vector $A_k$ are in fact the elements of $A_k$.
So we can take $B = C^t$.
answered Jul 24 at 5:12
Dave
912
Oh... I see... Thank you so much....!!
– Indrajit Ghosh
Jul 24 at 5:16
I don't have access to the first edition of the book, but I suspect this is the kind of error that might have been introduced in the process of producing a new edition.
– Dave
Jul 24 at 5:19
add a comment |Â
up vote
1
down vote
accepted
I've had a look at this chapter of Apostol's book, and I have to agree that, unless I'm missing something, this assertion is poorly justified on the basis of the theory developed to that point.
Apostol states the correspondence between linear mappings and matrices but, as far as I can tell, fails to spell out that the effect of the linear mapping on coordinates is given by matrix multiplication. This fact should have been noted explicitly in the later section that defines matrix multiplication.
There are two essential points to understand to justify this part of the proof of Theorem 3.6.
First, in the statement of Theorem 2.13, if we let $X = (x_1, dots, x_n)$ and $Y = (y_1, dots, y_m)$ be the coordinate representations, as column vectors, of $x$ and $y$ in the given bases, then equation $(2.13)$ can be written in the matrix form $Y = CX$, where $C = (t_ik)$ is the matrix of $T$ relative to the given bases.
If we instead consider $X$ and $Y$ as row vectors, then we must write $Y = X C^t$. Here, $C^t$ denotes the transpose of $C$, a concept defined later on page 91. Since Apostol doesn't develop the properties of transposes, it's best to check the equivalence of this matrix equality with equation $(2.13)$ directly.
Second, according to Theorem 2.13, there is a matrix C associated with the linear mapping $T$, if the standard basis $(I_1,dots,I_n)$ is selected in each of the two copies of $V_n$.
Then by Theorem 2.13, and remembering that $A_k$ and $I_k$ are considered row vectors in this chapter, we have $I_k = A_k C^t$ for each index $k$. This assertion relies on the fact that in the basis $(I_1,dots,I_n)$, the coordinates of the vector $A_k$ are in fact the elements of $A_k$.
So we can take $B = C^t$.
answered Jul 24 at 5:12
Dave
912
Oh... I see... Thank you so much....!!
– Indrajit Ghosh
Jul 24 at 5:16
I don't have access to the first edition of the book, but I suspect this is the kind of error that might have been introduced in the process of producing a new edition.
– Dave
Jul 24 at 5:19
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I've had a look at this chapter of Apostol's book, and I have to agree that, unless I'm missing something, this assertion is poorly justified on the basis of the theory developed to that point.
Apostol states the correspondence between linear mappings and matrices but, as far as I can tell, fails to spell out that the effect of the linear mapping on coordinates is given by matrix multiplication. This fact should have been noted explicitly in the later section that defines matrix multiplication.
There are two essential points to understand to justify this part of the proof of Theorem 3.6.
First, in the statement of Theorem 2.13, if we let $X = (x_1, dots, x_n)$ and $Y = (y_1, dots, y_m)$ be the coordinate representations, as column vectors, of $x$ and $y$ in the given bases, then equation $(2.13)$ can be written in the matrix form $Y = CX$, where $C = (t_ik)$ is the matrix of $T$ relative to the given bases.
If we instead consider $X$ and $Y$ as row vectors, then we must write $Y = X C^t$. Here, $C^t$ denotes the transpose of $C$, a concept defined later on page 91. Since Apostol doesn't develop the properties of transposes, it's best to check the equivalence of this matrix equality with equation $(2.13)$ directly.
Second, according to Theorem 2.13, there is a matrix C associated with the linear mapping $T$, if the standard basis $(I_1,dots,I_n)$ is selected in each of the two copies of $V_n$.
Then by Theorem 2.13, and remembering that $A_k$ and $I_k$ are considered row vectors in this chapter, we have $I_k = A_k C^t$ for each index $k$. This assertion relies on the fact that in the basis $(I_1,dots,I_n)$, the coordinates of the vector $A_k$ are in fact the elements of $A_k$.
So we can take $B = C^t$.
answered Jul 24 at 5:12
Dave
912
I've had a look at this chapter of Apostol's book, and I have to agree that, unless I'm missing something, this assertion is poorly justified on the basis of the theory developed to that point.
Apostol states the correspondence between linear mappings and matrices but, as far as I can tell, fails to spell out that the effect of the linear mapping on coordinates is given by matrix multiplication. This fact should have been noted explicitly in the later section that defines matrix multiplication.
There are two essential points to understand to justify this part of the proof of Theorem 3.6.
First, in the statement of Theorem 2.13, if we let $X = (x_1, dots, x_n)$ and $Y = (y_1, dots, y_m)$ be the coordinate representations, as column vectors, of $x$ and $y$ in the given bases, then equation $(2.13)$ can be written in the matrix form $Y = CX$, where $C = (t_ik)$ is the matrix of $T$ relative to the given bases.
If we instead consider $X$ and $Y$ as row vectors, then we must write $Y = X C^t$. Here, $C^t$ denotes the transpose of $C$, a concept defined later on page 91. Since Apostol doesn't develop the properties of transposes, it's best to check the equivalence of this matrix equality with equation $(2.13)$ directly.
Second, according to Theorem 2.13, there is a matrix C associated with the linear mapping $T$, if the standard basis $(I_1,dots,I_n)$ is selected in each of the two copies of $V_n$.
Then by Theorem 2.13, and remembering that $A_k$ and $I_k$ are considered row vectors in this chapter, we have $I_k = A_k C^t$ for each index $k$. This assertion relies on the fact that in the basis $(I_1,dots,I_n)$, the coordinates of the vector $A_k$ are in fact the elements of $A_k$.
So we can take $B = C^t$.
answered Jul 24 at 5:12
Dave
912
edited Jul 24 at 5:17
answered Jul 24 at 5:12
Dave
912
answered Jul 24 at 5:12
Dave
912
answered Jul 24 at 5:12
Dave
912
912
Oh... I see... Thank you so much....!!
– Indrajit Ghosh
Jul 24 at 5:16
I don't have access to the first edition of the book, but I suspect this is the kind of error that might have been introduced in the process of producing a new edition.
– Dave
Jul 24 at 5:19
add a comment |Â
Oh... I see... Thank you so much....!!
– Indrajit Ghosh
Jul 24 at 5:16
I don't have access to the first edition of the book, but I suspect this is the kind of error that might have been introduced in the process of producing a new edition.
– Dave
Jul 24 at 5:19
Oh... I see... Thank you so much....!!
– Indrajit Ghosh
Jul 24 at 5:16
Oh... I see... Thank you so much....!!
– Indrajit Ghosh
Jul 24 at 5:16
I don't have access to the first edition of the book, but I suspect this is the kind of error that might have been introduced in the process of producing a new edition.
– Dave
Jul 24 at 5:19
I don't have access to the first edition of the book, but I suspect this is the kind of error that might have been introduced in the process of producing a new edition.
– Dave
Jul 24 at 5:19
add a comment |Â
up vote
0
down vote
If you look in your book in chapter 2.19
There is a theorem for inverses of square matrices.Now when you read the proof.
The reasoning is somewhat circular however the idea actually follows from this.
If
$$ A = U Lambda U^T $$
then
$$ det(A) = det(U Lambda U^T) = det(U)det(Lambda)det(U^T) $$
now the matrices $U,U^T$ are orthogonal have determinant one
$$det(A) = det(Lambda) = prod_i=1^n lambda_i $$
if the matrix is not linearly independent then one of the eigenvalues is zero. Then the product of the eigenvalues is zero.
answered Jul 24 at 5:45
RHowe
1,010815
How does this relate to the OP's question?
– copper.hat
Jul 24 at 14:22
It is the proof in the book he is referring to
– RHowe
Jul 24 at 14:23
He didnt even read the page before hand.
– RHowe
Jul 24 at 14:24
How do you know? And how does it relate to the OP's question?
– copper.hat
Jul 24 at 14:56
does this help...the proof was in chapter 2
– RHowe
Jul 24 at 15:06
 |Â
show 2 more comments
up vote
0
down vote
If you look in your book in chapter 2.19
There is a theorem for inverses of square matrices.Now when you read the proof.
The reasoning is somewhat circular however the idea actually follows from this.
If
$$ A = U Lambda U^T $$
then
$$ det(A) = det(U Lambda U^T) = det(U)det(Lambda)det(U^T) $$
now the matrices $U,U^T$ are orthogonal have determinant one
$$det(A) = det(Lambda) = prod_i=1^n lambda_i $$
if the matrix is not linearly independent then one of the eigenvalues is zero. Then the product of the eigenvalues is zero.
answered Jul 24 at 5:45
RHowe
1,010815
How does this relate to the OP's question?
– copper.hat
Jul 24 at 14:22
It is the proof in the book he is referring to
– RHowe
Jul 24 at 14:23
He didnt even read the page before hand.
– RHowe
Jul 24 at 14:24
How do you know? And how does it relate to the OP's question?
– copper.hat
Jul 24 at 14:56
does this help...the proof was in chapter 2
– RHowe
Jul 24 at 15:06
 |Â
show 2 more comments
up vote
0
down vote
up vote
0
down vote
If you look in your book in chapter 2.19
There is a theorem for inverses of square matrices.Now when you read the proof.
The reasoning is somewhat circular however the idea actually follows from this.
If
$$ A = U Lambda U^T $$
then
$$ det(A) = det(U Lambda U^T) = det(U)det(Lambda)det(U^T) $$
now the matrices $U,U^T$ are orthogonal have determinant one
$$det(A) = det(Lambda) = prod_i=1^n lambda_i $$
if the matrix is not linearly independent then one of the eigenvalues is zero. Then the product of the eigenvalues is zero.
answered Jul 24 at 5:45
RHowe
1,010815
If you look in your book in chapter 2.19
There is a theorem for inverses of square matrices.Now when you read the proof.
The reasoning is somewhat circular however the idea actually follows from this.
If
$$ A = U Lambda U^T $$
then
$$ det(A) = det(U Lambda U^T) = det(U)det(Lambda)det(U^T) $$
now the matrices $U,U^T$ are orthogonal have determinant one
$$det(A) = det(Lambda) = prod_i=1^n lambda_i $$
if the matrix is not linearly independent then one of the eigenvalues is zero. Then the product of the eigenvalues is zero.
answered Jul 24 at 5:45
RHowe
1,010815
edited Jul 24 at 15:11
answered Jul 24 at 5:45
RHowe
1,010815
answered Jul 24 at 5:45
RHowe
1,010815
answered Jul 24 at 5:45
RHowe
1,010815
1,010815
How does this relate to the OP's question?
– copper.hat
Jul 24 at 14:22
It is the proof in the book he is referring to
– RHowe
Jul 24 at 14:23
He didnt even read the page before hand.
– RHowe
Jul 24 at 14:24
How do you know? And how does it relate to the OP's question?
– copper.hat
Jul 24 at 14:56
does this help...the proof was in chapter 2
– RHowe
Jul 24 at 15:06
 |Â
show 2 more comments
How does this relate to the OP's question?
– copper.hat
Jul 24 at 14:22
It is the proof in the book he is referring to
– RHowe
Jul 24 at 14:23
He didnt even read the page before hand.
– RHowe
Jul 24 at 14:24
How do you know? And how does it relate to the OP's question?
– copper.hat
Jul 24 at 14:56
does this help...the proof was in chapter 2
– RHowe
Jul 24 at 15:06
How does this relate to the OP's question?
– copper.hat
Jul 24 at 14:22
How does this relate to the OP's question?
– copper.hat
Jul 24 at 14:22
It is the proof in the book he is referring to
– RHowe
Jul 24 at 14:23
It is the proof in the book he is referring to
– RHowe
Jul 24 at 14:23
He didnt even read the page before hand.
– RHowe
Jul 24 at 14:24
He didnt even read the page before hand.
– RHowe
Jul 24 at 14:24
How do you know? And how does it relate to the OP's question?
– copper.hat
Jul 24 at 14:56
How do you know? And how does it relate to the OP's question?
– copper.hat
Jul 24 at 14:56
does this help...the proof was in chapter 2
– RHowe
Jul 24 at 15:06
does this help...the proof was in chapter 2
– RHowe
Jul 24 at 15:06
 |Â
show 2 more comments
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en.wikipedia.org/wiki/Linear_map#Matrices
– Lorenzo
Jul 24 at 4:23
The existence of $B$ is the inverse $AA^-1 = A^-1A = I_k$
– RHowe
Jul 24 at 4:28
I know this... is B the matrix of T here? But I actually confused why $A_k$ is pre multiplied with B ti get $I_k$...
– Indrajit Ghosh
Jul 24 at 4:29
@Geronimo ..I can't get u..
– Indrajit Ghosh
Jul 24 at 4:33
@Geronimo The notation $I_k$ does not represent the identity matrix here.
– Dave
Jul 24 at 5:27