What is the solvability condition for the matrix equation A_1x_2 = A_2x_1?

[Hootenanny]

I have a quick question regarding matrix equations. Usually, I would look this up but unfortunately I'm away from the office and library and it can't wait until I get back.

Let A_1 and A_2 be n\times n square matrices with real elements and let \boldsymbol{x}_1\;,\boldsymbol{x}_2\in\mathbb{R}^n. Further, let A_1 \boldsymbol{x}_1 = \boldsymbol{0}. What is the solvability condition for the following system?

A_1\boldsymbol{x}_2 = A_2\boldsymbol{x}_1

The result would suggest \boldsymbol{x}_1^\text{T}A_2\boldsymbol{x}_1 = 0, but I'm clearly missing something. I fairly certain its something minor that I just can't see.

Any help would be very much appreciated.

 

[kdbnlin78]

Can any of the x_{i} or A_{i} be inverted? (i.e., do you know anything about their determinants?)

 

[Hootenanny]

Can any of the x_{i} or A_{i} be inverted? (i.e., do you know anything about their determinants?)

\boldsymbol{x}_i are vectors in \mathbb{R}^n, and A_i are singular in general.

 

[kdbnlin78]

Apologies, can now see the x_{i} are vectors. I'm at work and scanning articles when no-one is looking.

I your reasoning has lead to to conclude that /boldsymbol{x_{1}^T}A_{1} = /boldsymbol{0} - How do you know this?

 

[Hootenanny]

Apologies, can now see the x_{i} are vectors. I'm at work and scanning articles when no-one is looking.

I your reasoning has lead to to conclude that /boldsymbol{x_{1}^T}A_{1} = /boldsymbol{0} - How do you know this?

No problem :)

I'm tracing back a result and I've found that the result would only hold if the above relation is true.

I only asked because I assumed that solvability condition for an equation of the forum that I posted in my original post would be fairly well known, or at least established.

 

[micromass]

Is your matrix symmetric by any chance?? In that case we have that

The thing is that

A_1x_2=A_2x_1

has a solution if and only if A_2x_1\in im(A_1)[/tex].<br /> But we know that im(A_1)=ker(A_1^T)^\bot.<br /> So the system has a solution if and only if A_2x_1\in ker(A_1^T)^\bot=ker(A_1)^\bot.<br /> <br /> So it must hold that x_1^TA_1x_1=0. I fear that this is not a sufficient condition in general...

 

[Hootenanny]

Is your matrix symmetric by any chance?? In that case we have that

The thing is that

A_1x_2=A_2x_1

has a solution if and only if A_2x_1\in im(A_1)[/tex].<br /> But we know that im(A_1)=ker(A_1^T)^\bot.<br /> So the system has a solution if and only if A_2x_1\in ker(A_1^T)^\bot=ker(A_1)^\bot.<br /> <br /> So it must hold that x_1^TA_1x_1=0. I fear that this is not a sufficient condition in general...

<br /> That was my first thought as well. Alas, there are no symmetry conditions on the matrices A_i.

 

[Hootenanny]

Nevermind - I've figure it out :D

 

[micromass]

Nevermind - I've figure it out :D

Can you tell us the solution??