Exploring the Solvability of a Matrix Equation

[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 [itex]A_2x_1\in im(A_1)[/tex].
But we know that $im(A_1)=ker(A_1^T)^\bot$.
So the system has a solution if and only if $A_2x_1\in ker(A_1^T)^\bot=ker(A_1)^\bot$.

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 [itex]A_2x_1\in im(A_1)[/tex].
But we know that $im(A_1)=ker(A_1^T)^\bot$.
So the system has a solution if and only if $A_2x_1\in ker(A_1^T)^\bot=ker(A_1)^\bot$.

So it must hold that $x_1^TA_1x_1=0$. I fear that this is not a sufficient condition in general...

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??