When does exist matrices T and H such that HCE=TE? (all matrices are rectangular)
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could you please help me with this question; I want to find out the conditions (necessary and /or sufficient) for the existence of two matrices namely H and T such that the equality HCE=TE holds for given matrices C and E.
All matrices are rectangular. What about a simpler case would be when E is a column vector?
thanks a lot.
linear-algebra matrices matrix-rank
asked Nov 17 at 19:16
mahdy share pasand
93
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up vote
0
down vote
favorite
could you please help me with this question; I want to find out the conditions (necessary and /or sufficient) for the existence of two matrices namely H and T such that the equality HCE=TE holds for given matrices C and E.
All matrices are rectangular. What about a simpler case would be when E is a column vector?
thanks a lot.
linear-algebra matrices matrix-rank
asked Nov 17 at 19:16
mahdy share pasand
93
Are there constraints on the dimensions of $H$ and $T$? Otherwise, we could take $H$ to be the identity matrix and $T = C$.
– JimmyK4542
Nov 17 at 19:21
thanks a lot, but I needed a more general condition. In fact, H is a nxm matrix and T is a nxq matrix. where n>m,q. Therefore, this solution is not the solution I sought.
– mahdy share pasand
Nov 17 at 19:28
I am seeking for a solution like rank(CE)=rank(E) which guarantees that H and T exist.
– mahdy share pasand
Nov 17 at 19:29
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
could you please help me with this question; I want to find out the conditions (necessary and /or sufficient) for the existence of two matrices namely H and T such that the equality HCE=TE holds for given matrices C and E.
All matrices are rectangular. What about a simpler case would be when E is a column vector?
thanks a lot.
linear-algebra matrices matrix-rank
asked Nov 17 at 19:16
mahdy share pasand
93
could you please help me with this question; I want to find out the conditions (necessary and /or sufficient) for the existence of two matrices namely H and T such that the equality HCE=TE holds for given matrices C and E.
All matrices are rectangular. What about a simpler case would be when E is a column vector?
thanks a lot.
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
asked Nov 17 at 19:16
mahdy share pasand
93
asked Nov 17 at 19:16
mahdy share pasand
93
asked Nov 17 at 19:16
mahdy share pasand
93
asked Nov 17 at 19:16
mahdy share pasand
93
asked Nov 17 at 19:16
mahdy share pasand
93
93
Are there constraints on the dimensions of $H$ and $T$? Otherwise, we could take $H$ to be the identity matrix and $T = C$.
– JimmyK4542
Nov 17 at 19:21
thanks a lot, but I needed a more general condition. In fact, H is a nxm matrix and T is a nxq matrix. where n>m,q. Therefore, this solution is not the solution I sought.
– mahdy share pasand
Nov 17 at 19:28
I am seeking for a solution like rank(CE)=rank(E) which guarantees that H and T exist.
– mahdy share pasand
Nov 17 at 19:29
add a comment |
Are there constraints on the dimensions of $H$ and $T$? Otherwise, we could take $H$ to be the identity matrix and $T = C$.
– JimmyK4542
Nov 17 at 19:21
thanks a lot, but I needed a more general condition. In fact, H is a nxm matrix and T is a nxq matrix. where n>m,q. Therefore, this solution is not the solution I sought.
– mahdy share pasand
Nov 17 at 19:28
I am seeking for a solution like rank(CE)=rank(E) which guarantees that H and T exist.
– mahdy share pasand
Nov 17 at 19:29
Are there constraints on the dimensions of $H$ and $T$? Otherwise, we could take $H$ to be the identity matrix and $T = C$.
– JimmyK4542
Nov 17 at 19:21
Are there constraints on the dimensions of $H$ and $T$? Otherwise, we could take $H$ to be the identity matrix and $T = C$.
– JimmyK4542
Nov 17 at 19:21
thanks a lot, but I needed a more general condition. In fact, H is a nxm matrix and T is a nxq matrix. where n>m,q. Therefore, this solution is not the solution I sought.
– mahdy share pasand
Nov 17 at 19:28
thanks a lot, but I needed a more general condition. In fact, H is a nxm matrix and T is a nxq matrix. where n>m,q. Therefore, this solution is not the solution I sought.
– mahdy share pasand
Nov 17 at 19:28
I am seeking for a solution like rank(CE)=rank(E) which guarantees that H and T exist.
– mahdy share pasand
Nov 17 at 19:29
I am seeking for a solution like rank(CE)=rank(E) which guarantees that H and T exist.
– mahdy share pasand
Nov 17 at 19:29
add a comment |
1 Answer
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For the case where $E$ is a vector.
Well if $E$ is a vector then $CE$ is another vector. So your question is when there exists two matrices $T,H$ which sends one vector to another. The solution is always, unless one of the vectors is $0$.
If $E$ is the zero vector then any matrices $T,H$ would satisfy the equation.
If $E$ is non-zero but $CE$ is zero then we can take $T$ to be the zero matrix.
If non of them is zero then we can choose any matrix $T$, find a basis which consists $CE$ and then simply find a matrix which sends $CE$ to $TE$ and the rest of the vectors to zero.
Conclusion: if $E$ is a column vector the existence of $T,H$ is guaranteed with no additional conditions.
Some notes on the general case: If $E$ is a matrix, then denote by $E_1,E_2,...,E_n$ it's colum vectors. Then the column of $TE$ are $TE_1,TE_2,...,TE_n$. So the question above is basically whether we can find $T,H$ for multiple vectors $E_1,...,E_n$ simultaneously and so I believe it also works in the general case (i.e you can always find such $T,H$).
answered Nov 17 at 19:38
Yanko
5,023722
thanks a lot. In the second case; where E is non-zero but CE is zero, could you suggest a non-zero solution? (other than T=0)?
– mahdy share pasand
Nov 17 at 19:45
@mahdysharepasand You know that such $T$ exists abstractly (if the dimension of the space is more than 1). Because if $E$ is non-zero you can find a basis which consist $E$ and then choose a matrix that sends $E$ to zero and the other basis elements to themselves (or any other non-zero vectors). In other words you only need that $Einker T$
– Yanko
Nov 17 at 19:48
thank you very much @Yanko
– mahdy share pasand
Nov 17 at 19:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
For the case where $E$ is a vector.
Well if $E$ is a vector then $CE$ is another vector. So your question is when there exists two matrices $T,H$ which sends one vector to another. The solution is always, unless one of the vectors is $0$.
If $E$ is the zero vector then any matrices $T,H$ would satisfy the equation.
If $E$ is non-zero but $CE$ is zero then we can take $T$ to be the zero matrix.
If non of them is zero then we can choose any matrix $T$, find a basis which consists $CE$ and then simply find a matrix which sends $CE$ to $TE$ and the rest of the vectors to zero.
Conclusion: if $E$ is a column vector the existence of $T,H$ is guaranteed with no additional conditions.
Some notes on the general case: If $E$ is a matrix, then denote by $E_1,E_2,...,E_n$ it's colum vectors. Then the column of $TE$ are $TE_1,TE_2,...,TE_n$. So the question above is basically whether we can find $T,H$ for multiple vectors $E_1,...,E_n$ simultaneously and so I believe it also works in the general case (i.e you can always find such $T,H$).
answered Nov 17 at 19:38
Yanko
5,023722
thanks a lot. In the second case; where E is non-zero but CE is zero, could you suggest a non-zero solution? (other than T=0)?
– mahdy share pasand
Nov 17 at 19:45
@mahdysharepasand You know that such $T$ exists abstractly (if the dimension of the space is more than 1). Because if $E$ is non-zero you can find a basis which consist $E$ and then choose a matrix that sends $E$ to zero and the other basis elements to themselves (or any other non-zero vectors). In other words you only need that $Einker T$
– Yanko
Nov 17 at 19:48
thank you very much @Yanko
– mahdy share pasand
Nov 17 at 19:53
add a comment |
up vote
0
down vote
For the case where $E$ is a vector.
Well if $E$ is a vector then $CE$ is another vector. So your question is when there exists two matrices $T,H$ which sends one vector to another. The solution is always, unless one of the vectors is $0$.
If $E$ is the zero vector then any matrices $T,H$ would satisfy the equation.
If $E$ is non-zero but $CE$ is zero then we can take $T$ to be the zero matrix.
If non of them is zero then we can choose any matrix $T$, find a basis which consists $CE$ and then simply find a matrix which sends $CE$ to $TE$ and the rest of the vectors to zero.
Conclusion: if $E$ is a column vector the existence of $T,H$ is guaranteed with no additional conditions.
Some notes on the general case: If $E$ is a matrix, then denote by $E_1,E_2,...,E_n$ it's colum vectors. Then the column of $TE$ are $TE_1,TE_2,...,TE_n$. So the question above is basically whether we can find $T,H$ for multiple vectors $E_1,...,E_n$ simultaneously and so I believe it also works in the general case (i.e you can always find such $T,H$).
answered Nov 17 at 19:38
Yanko
5,023722
thanks a lot. In the second case; where E is non-zero but CE is zero, could you suggest a non-zero solution? (other than T=0)?
– mahdy share pasand
Nov 17 at 19:45
@mahdysharepasand You know that such $T$ exists abstractly (if the dimension of the space is more than 1). Because if $E$ is non-zero you can find a basis which consist $E$ and then choose a matrix that sends $E$ to zero and the other basis elements to themselves (or any other non-zero vectors). In other words you only need that $Einker T$
– Yanko
Nov 17 at 19:48
thank you very much @Yanko
– mahdy share pasand
Nov 17 at 19:53
add a comment |
up vote
0
down vote
up vote
0
down vote
For the case where $E$ is a vector.
Well if $E$ is a vector then $CE$ is another vector. So your question is when there exists two matrices $T,H$ which sends one vector to another. The solution is always, unless one of the vectors is $0$.
If $E$ is the zero vector then any matrices $T,H$ would satisfy the equation.
If $E$ is non-zero but $CE$ is zero then we can take $T$ to be the zero matrix.
If non of them is zero then we can choose any matrix $T$, find a basis which consists $CE$ and then simply find a matrix which sends $CE$ to $TE$ and the rest of the vectors to zero.
Conclusion: if $E$ is a column vector the existence of $T,H$ is guaranteed with no additional conditions.
Some notes on the general case: If $E$ is a matrix, then denote by $E_1,E_2,...,E_n$ it's colum vectors. Then the column of $TE$ are $TE_1,TE_2,...,TE_n$. So the question above is basically whether we can find $T,H$ for multiple vectors $E_1,...,E_n$ simultaneously and so I believe it also works in the general case (i.e you can always find such $T,H$).
answered Nov 17 at 19:38
Yanko
5,023722
For the case where $E$ is a vector.
Well if $E$ is a vector then $CE$ is another vector. So your question is when there exists two matrices $T,H$ which sends one vector to another. The solution is always, unless one of the vectors is $0$.
If $E$ is the zero vector then any matrices $T,H$ would satisfy the equation.
If $E$ is non-zero but $CE$ is zero then we can take $T$ to be the zero matrix.
If non of them is zero then we can choose any matrix $T$, find a basis which consists $CE$ and then simply find a matrix which sends $CE$ to $TE$ and the rest of the vectors to zero.
Conclusion: if $E$ is a column vector the existence of $T,H$ is guaranteed with no additional conditions.
Some notes on the general case: If $E$ is a matrix, then denote by $E_1,E_2,...,E_n$ it's colum vectors. Then the column of $TE$ are $TE_1,TE_2,...,TE_n$. So the question above is basically whether we can find $T,H$ for multiple vectors $E_1,...,E_n$ simultaneously and so I believe it also works in the general case (i.e you can always find such $T,H$).
answered Nov 17 at 19:38
Yanko
5,023722
edited Nov 17 at 19:46
answered Nov 17 at 19:38
Yanko
5,023722
answered Nov 17 at 19:38
Yanko
5,023722
answered Nov 17 at 19:38
Yanko
5,023722
5,023722
thanks a lot. In the second case; where E is non-zero but CE is zero, could you suggest a non-zero solution? (other than T=0)?
– mahdy share pasand
Nov 17 at 19:45
@mahdysharepasand You know that such $T$ exists abstractly (if the dimension of the space is more than 1). Because if $E$ is non-zero you can find a basis which consist $E$ and then choose a matrix that sends $E$ to zero and the other basis elements to themselves (or any other non-zero vectors). In other words you only need that $Einker T$
– Yanko
Nov 17 at 19:48
thank you very much @Yanko
– mahdy share pasand
Nov 17 at 19:53
add a comment |
thanks a lot. In the second case; where E is non-zero but CE is zero, could you suggest a non-zero solution? (other than T=0)?
– mahdy share pasand
Nov 17 at 19:45
@mahdysharepasand You know that such $T$ exists abstractly (if the dimension of the space is more than 1). Because if $E$ is non-zero you can find a basis which consist $E$ and then choose a matrix that sends $E$ to zero and the other basis elements to themselves (or any other non-zero vectors). In other words you only need that $Einker T$
– Yanko
Nov 17 at 19:48
thank you very much @Yanko
– mahdy share pasand
Nov 17 at 19:53
thanks a lot. In the second case; where E is non-zero but CE is zero, could you suggest a non-zero solution? (other than T=0)?
– mahdy share pasand
Nov 17 at 19:45
thanks a lot. In the second case; where E is non-zero but CE is zero, could you suggest a non-zero solution? (other than T=0)?
– mahdy share pasand
Nov 17 at 19:45
@mahdysharepasand You know that such $T$ exists abstractly (if the dimension of the space is more than 1). Because if $E$ is non-zero you can find a basis which consist $E$ and then choose a matrix that sends $E$ to zero and the other basis elements to themselves (or any other non-zero vectors). In other words you only need that $Einker T$
– Yanko
Nov 17 at 19:48
@mahdysharepasand You know that such $T$ exists abstractly (if the dimension of the space is more than 1). Because if $E$ is non-zero you can find a basis which consist $E$ and then choose a matrix that sends $E$ to zero and the other basis elements to themselves (or any other non-zero vectors). In other words you only need that $Einker T$
– Yanko
Nov 17 at 19:48
thank you very much @Yanko
– mahdy share pasand
Nov 17 at 19:53
thank you very much @Yanko
– mahdy share pasand
Nov 17 at 19:53
add a comment |
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Are there constraints on the dimensions of $H$ and $T$? Otherwise, we could take $H$ to be the identity matrix and $T = C$.
– JimmyK4542
Nov 17 at 19:21
thanks a lot, but I needed a more general condition. In fact, H is a nxm matrix and T is a nxq matrix. where n>m,q. Therefore, this solution is not the solution I sought.
– mahdy share pasand
Nov 17 at 19:28
I am seeking for a solution like rank(CE)=rank(E) which guarantees that H and T exist.
– mahdy share pasand
Nov 17 at 19:29