Solbing equation A(u)=B(v) for square matrices A and B

[ivl]

Dear all,

this is perhaps a trivial question, so I apologise in advance. Any help is greatly appreciated nonetheless.

==The Equation==
The equation under consideration is:

A(u)=B(v)

where A and B are n times n matrices, while u and v are n-dimensional vectors.

==The Question==
From the above equation, I would like to determine u as a function of v.
Question: is the only way of determining u as a function of v to require that A is invertible?
In other words, is it correct to say that the only way of solving the above equation for u is u=A^{-1}B(v) ?

Thanks a lot,
IVL

 

[DonAntonio]

Dear all,

this is perhaps a trivial question, so I apologise in advance. Any help is greatly appreciated nonetheless.

==The Equation==
The equation under consideration is:

A(u)=B(v)

where A and B are n times n matrices, while u and v are n-dimensional vectors.

==The Question==
From the above equation, I would like to determine u as a function of v.
Question: is the only way of determining u as a function of v to require that A is invertible?
In other words, is it correct to say that the only way of solving the above equation for u is u=A^{-1}B(v) ?

Thanks a lot,
IVL

Well, more than "saying that the only way...", I'd say that requiring A to be invertible is the only way to guarantee that there's a solution to your problem, otherwise we'd have to look at each case in particular.

DonAntonio

 

[ivl]

Thanks for your reply, DonAntonio.

You are right, my question was a bit too vague. But you understood what I meant.

Perhaps a more meaningful way to put the question would be:

-given two n times n matrices, A and B
-let u and v be two n-dimensional vectors

A(u)=B(v)

For each choice of v, the above equation has a UNIQUE solution u if and only if det(A) is non-zero.

Correct?

Thanks
IVL

 

[HallsofIvy]

There is a unique solution if and only if A is invertible and in that case u= A^-1Bv.

 

[ivl]

Thanks everybody,

glad to see things got clearer in my head. Problem solved.

Cheers
IVL