When does exist matrices T and H such that HCE=TE? (all matrices are rectangular)

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.

• 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 '18 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 '18 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 '18 at 19:29

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 $$T,H$$).

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