Understanding the Singular Form of ABA: Proving its Validity

[wrxue]

Homework Statement

A is MxN matrix.
B is NxN matrix.
If N<M

Relevant Equations

Prove ABA' is singular.(A' is transpose of A)

I don't know how to do.
Thanks in advance.

 

[WWGD]

And A' means A inverse?

 

[wrxue]

And A' means A inverse?

A' is transpose of A

 

[WWGD]

A' is transpose of A

Ah, dumb me, A is not square, so cannot have a 2-sided inverse.

 

[Delta2]

Is it also given that B is singular?. Cause if B is for example the identity matrix then I don't see how this can hold for any matrix A...

 

[wrxue]

Is it also given that B is singular?. Cause if B is for example the identity matrix then I don't see how this can hold for any matrix A...

No, B can be any matrix. No more conditions. Q_Q

 

[WWGD]

It is false then. Take, e.g., A=( 1 2) and B =$I_2$, the 2x2 id matrix. Then ABA'=(5), as a 1x1 matrix, which is invertible.

 

[wrxue]

It is false then. Take, e.g., A=( 1 2) and B =$I_2$, the 2x2 id matrix. Then ABA'=(5), as a 1x1 matrix, which is invertible.

But in this example, N(=2)>M(=1).

 

[WWGD]

But in this example, N(=2)>M(=1).

My bad, I jumped and been jumping the gun. Let me chill and get to it.

 

[StoneTemplePython]

But in this example, N(=2)>M(=1).

so $A'$ is short and fat... why does that imply there is some $\mathbf x \neq \mathbf 0$ such that $A'\mathbf x = \mathbf 0$?

also:
what tools/concepts do you have at your disposal? I'd like to do this in terms of rank but it can be done using other concepts e.g. discussing maximal sized square submatrix in $A$ that is invertible (nonzero determinant)

 

[wrxue]

so $A'$ is short and fat... why does that imply there is some $\mathbf x \neq \mathbf 0$ such that $A'\mathbf x = \mathbf 0$?

also:
what tools/concepts do you have at your disposal? I'd like to do this in terms of rank but it can be done using other concepts e.g. discussing maximal sized square submatrix in $A$ that is invertible (nonzero determinant)

My thought is
$$CS(ABA^T)\subseteq CS(AB)\subseteq CS(A)$$
so the dimensions of $CS(\cdot)$
$$dim(CS(ABA^T))\leq dim(CS(AB))\leq dim(CS(A))=Rank(A)\leq (min(N,M))$$
thus
$$dim(CS(ABA^T))\leq N$$
And
$$rank(ABA^T)=dim(CS(ABA^T))\leq N<M$$
$ABA^T$ must be singular
Just thought of it

 

[StoneTemplePython]

My thought is
$$CS(ABA^T)\subseteq CS(AB)\subseteq CS(A)$$
so the dimensions of $CS(\cdot)$
$$dim(CS(ABA^T))\leq dim(CS(AB))\leq dim(CS(A))=Rank(A)\leq (min(N,M))$$
thus
$$dim(CS(ABA^T))\leq N$$
And
$$rank(ABA^T)=dim(CS(ABA^T))\leq N<M$$
$ABA^T$ must be singular
Just thought of it

I think this works though I was trying to highlight that you don't need to consider all 3 matrices -- using associativity its sufficient to consider some $\mathbf x \neq \mathbf 0$

$\mathbf 0 = \mathbf {ABA}^T \mathbf x = \mathbf {AB}\big( \mathbf A^T\mathbf x\big) = \mathbf {AB} \mathbf 0$
because $\mathbf A^T$ has at most N linearly independent columns (as they each live in an N dimensional space), so there is at least one linearly independent vector in its nullspace (by say rank nullity or even more basic arguments since $M-N \geq 1 \gt 0$ there must be at least one column vector in $\mathbf A$ that flunks the definition of linearly independent and hence is linearly dependent...)

 

[WWGD]

EDIT I don't know if I am missing something but, using A an mxn ; m>n and A' nxm :
RankA= RankA'
and $Rank(AB) \leq RankA$,
we want to show $Rank(AB)<m$ . But, since $m>n$,
it follows
$RankA, RankB \leq n <m$ ,

And notice this sees to address the condition that m>n .
Doesn't it?

 

[FactChecker]

The proof will depend on what definitions and theorems you have to work with. Intuitively, the left-hand operator BA' maps a space of dimension m to a space of lower dimension n; then A maps that to a subspace of dimension n in a space of dimension m. So the entire ABA' operator must have a non-trivial kernal and be singular.