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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.

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  • $\begingroup$ Are there constraints on the dimensions of $H$ and $T$? Otherwise, we could take $H$ to be the identity matrix and $T = C$. $\endgroup$ – JimmyK4542 Nov 17 '18 at 19:21
  • $\begingroup$ 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. $\endgroup$ – mahdy share pasand Nov 17 '18 at 19:28
  • $\begingroup$ I am seeking for a solution like rank(CE)=rank(E) which guarantees that H and T exist. $\endgroup$ – mahdy share pasand Nov 17 '18 at 19:29
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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 $T,H$).

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  • $\begingroup$ 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)? $\endgroup$ – mahdy share pasand Nov 17 '18 at 19:45
  • $\begingroup$ @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 $E\in\ker T$ $\endgroup$ – Yanko Nov 17 '18 at 19:48
  • $\begingroup$ thank you very much @Yanko $\endgroup$ – mahdy share pasand Nov 17 '18 at 19:53

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