Linear Algebra: Non-Singular Matrix and Zero Matrix

[Tim 1234]

Homework Statement

Suppose A is non-singular (nxn) matrix. Given that AB=(zero matrix), show that B=(zero matrix).

Hint: Express A as [A₁, A₂, ..., A_n]
and Express AB as [AB₁, AB₂, ..., AB_n]

Homework Equations

The Attempt at a Solution

[/B]
I argued that because A is non-singular, A=[A1, A2, ..., An] is linearly independent, thus Ax=(theta) has the unique solution x=(theta).

That is, for x1A1+x2A2+...+xnAn=(theta), x1=x2=...=xn=0.

Further, if A is a zero matrix, Ax=(theta) cannot have the unique solution x=(theta).

Because A cannot be a zero matrix and AB=(zero matrix), B=(zero matrix) by necessity.

Is my reasoning correct in that a non-singular matrix cannot be a zero matrix?

Thanks

 

[andrewkirk]

I don't understand your references to theta.

If x1A1+x2A2+...+xnAn = 0 and A1,...,An are linearly independent, what can we say about x1, ..., xn?

Then how can that help us make deductions about B?

 

[BvU]

thus Ax=(theta) has the unique solution x=(theta).

You mean $\vec{\vec A} \cdot \vec x = \vec \theta$ has the unique solution $\vec x = {\vec{\vec A}} ^{-1} \cdot \vec \theta\$ ?

And for

That is, for x1A1+x2A2+...+xnAn=(theta), x1=x2=...=xn=0

you mean

$\Sigma A_{ij} x_j = 0$ for all i $\ \Rightarrow x_j = 0 \$ for all $j$

--

 

[Tim 1234]

By theta, I mean the zero vector in R^n.

So x1,...,xn are all equal to 0 because A1...An are linearly independent.

Doesn't this imply that A is not a zero matrix?

If A was a zero matrix, x1, ..., xn would not need to equal 0 such that x1A1+...+xnAn = 0. x1, ..., xn could take on any value and the equation would still hold.

So a deduction we can make about B is because AB= the nxn zero matrix and A cannot be an nxn zero matrix, B has to be an nxn zero matrix?

 

[BvU]

By theta, I mean the zero vector in R^n

That helps ! Wasn't aware of that convention .
Nor of the convention that $A_1$ is the vector $A_{1j}\ \$. (although, upon second reading, the given hint wants you to do this)

But with that, your reasoning is a lot more valid. I would even say it's correct.

 

[andrewkirk]

That A is not the zero matrix follows immediately from it being non-singular, as the zero matrix is singular. You don't need to do the x1 ... xn thing to conclude that.

However, IIRC it is not the case that if the product of two matrices is zero, one of them must be zero.

Look at the second line of the hint they gave you. If we denote the elements of B1 by x1, ... , xn then what can you conclude from what you've done so far?

By the way, I assume that in this problem the definition of nonsingular you are using is that A has a nonzero determinant, and you may have a theorem that tells you that's equivalent to the columns being linearly independent and ditto for the rows. If we take nonsingular to mean that A is invertible (which I think I have seen used sometimes), or if you are allowed to use the theorem that nonsingular --> invertible then the problem is trivial.

 

[Tim 1234]

Denoting the elements of B1 by x1, ..., xn would render all elements of B1 equal to 0?

Then we could do this with every column of B, making B an nxn zero matrix?

Edit: This problem is from a section before invertibility and determinants.

 

[HallsofIvy]

One can also characterize B, the "zero- matrix", as the matrix such that Bx= 0 for every vector x. If AB= 0, then ABx= 0. Now, suppose B is not the zero matrix. Then there exist some x such that Bx is not 0. So we have ABx= A(Bx)= 0. Finish that.

 

[Tim 1234]

Assuming B is not the zero matrix, there is an x such that Bx is not 0. We know by virtue of A being non-singular that A is not the zero matrix.

Isn't it possible that there exists an x such that Bx is not 0, but A(Bx)=0?

 

[HallsofIvy]

No. If Av= 0 for any non-zero v, then A is singular. That was my point.

 

[Tim 1234]

If Av=0 for any non-zero vector v, then A is singular.

Under the assumption that B is not the zero matrix, there exists and x such that v=(Bx) is not 0.

If this is the case, and A(Bx)=Av=O, then A is singular.

But the question states A is non-singular.

So when we assume B not to be the zero matrix we arrive at a contradiction.

Therefore B must be the zero matrix.

Is my reasoning correct?

 

[Tim 1234]

Okay, all of the information I can deduce from the question:

B=[B₁, ..., B_n]
AB=[AB₁, ..., AB_n]

These were hints given in the question - I can't determine what their relevance is.

Also from the question, A is non-singular.

From this we know A is not the zero-vector (the zero-vector is singular).

I've read through the section a number of times, but I still have no idea where to go from here.

I can't see why its necessary for B to equal the zero-matrix to satisfy AB=zero-matrix...
Edit: I know that because A is non-singular, it is also invertible and because it is invertible, we can just multiply both sides of AB=zero-vector by A^-1. But invertibility and determinants are covered in the subsequent section, so I don't think we're meant to solve the problem this way.

 

[andrewkirk]

You are given that $AB=0$, ie that $\forall i:\ (A\times B)_i=A\times (B_i)=\vec{0}$.

You have worked out about $A_1x_1+...+A_nx_n=\vec{0}$ implies that all the $x_i$ are zero. Can you write that sum as the mutiplication of a matrix with a vector?

 

[Tim 1234]

Is the hint saying AB can be represented as a linear combination of the vector columns of B where the coefficients for each column are the matrix A.

And because A is non-zero and AB=0, the only way for the linear combination AB1+AB2+...+ABn=0 to be valid is if each column of B is composed of all zeros, thus making B a zero vector?

Please tell me this thought process is at least on track...