Equations and matrix...

[God]

Let A be a real matrix of M3 (3x3), if AX=(1,2,3) has only one for X, then is it true that AX=Y has only one solution for X, for every Y (3x1) ?
Let A be a real matrix of M3 (3x3), if AX=(1,2,3) has at least two solutions for X, then is it true that AX=Y has at least two solutions for X, for every Y (3x1) ?

I don't know how to answer these questions in that we don't know A. It can be any matrix. Showing that the linear function associated to the matrix A is or is not surjective/injective would have done the trick... but since I don't know A, I can't prove it.
Another strategy would have been finding a function whose matrix belongs to M3 ( like f(x,y,z)=(ax+by+cz,dx+ey+fz,gx+hy+kz) ) who's not injective or surjective but who has only one solution for f(x,y,z)=(1,2,3), but it doesn't seem too obvious to me since I can't figure one out...

Any idea on how I can proceed to prove/disprove these ?
Thanks

 

[daon2]

Let A be a real matrix of M3 (3x3), if AX=(1,2,3) has only one for X, then is it true that AX=Y has only one solution for X, for every Y (3x1) ?
Let A be a real matrix of M3 (3x3), if AX=(1,2,3) has at least two solutions for X, then is it true that AX=Y has at least two solutions for X, for every Y (3x1) ?

I don't know how to answer these questions in that we don't know A. It can be any matrix. Showing that the linear function associated to the matrix A is or is not surjective/injective would have done the trick... but since I don't know A, I can't prove it.
Another strategy would have been finding a function whose matrix belongs to M3 ( like f(x,y,z)=(ax+by+cz,dx+ey+fz,gx+hy+kz) ) who's not injective or surjective but who has only one solution for f(x,y,z)=(1,2,3), but it doesn't seem too obvious to me since I can't figure one out...

Any idea on how I can proceed to prove/disprove these ?
Thanks

If A is not invertible, then there is a vector v=/=0 such that Av=0. Suppose there is only one x such that Ax=(1,2,3). What then of the vector (v+x)?

 

[God]

If A is not invertible, then there is a vector v=/=0 such that Av=0. Suppose there is only one x such that Ax=(1,2,3). What then of the vector (v+x)?

Then v+x is also a solution. Since it is supposed there is only one solution, A must be invertible.
So ker(A)=0 and rank(A)=3
Since this is a matrix of a function from R^3 to R^3, this function is bijection, which means AX=Y has only one solution for X, and that for every Y, so this is true.

I think I got it, thanks.

For the second question, it must also be true for every invertible matrix.
However, let's suppose A is NOT invertible.
ker(A)=/=0 => the function associated is not surjective so there is a Y such as AX=Y has no solution
so it's false.

Am I right ?
Thanks again !

 

[daon2]

Then v+x is also a solution. Since it is supposed there is only one solution, A must be invertible.
So ker(A)=0 and rank(A)=3
Since this is a matrix of a function from R^3 to R^3, this function is bijection, which means AX=Y has only one solution for X, and that for every Y, so this is true.

I think I got it, thanks.

YUP!

For the second question, it must also be true for every invertible matrix.
However, let's suppose A is NOT invertible.
ker(A)=/=0 => the function associated is not surjective so there is a Y such as AX=Y has no solution
so it's false.

Am I right ?
Thanks again !

Keep in mind "At least two solutions" and "infinitely many solutions" are synonymous for linear systems. Now, A is certainly not invertible because that implies a linear bijection. Thus there being more than one solution for Ax=(1,2,3) implies there are more than one solution for every Y in the column space of A. But dim(ColA)=rank(A) is at most two-dimensional (or as you put it ker(A)=/={0}), so there is some vector y in R^3-Col(A) for which there is NO solution to Ax=y.

edit: rereading what you had written, It looks like you have the correct idea, but there is no reason to suppose it is not invertible. It is a certainty from your hypothesis.

 

[God]

Nevermind, indeed it cannot be invertible for the second question.
Thanks