linear algebra, dimensions problem

A

akleron

New member
Joined
Dec 28, 2019
Messages
42
Hey! would love some help with proving those two statements ! got no clue what to do here:
1577870020593.png
 
This is rather confusingly written! Usually the "given" is stated before what you are to prove. I think you are saying "Given that Rank(B)= m prove that Rank(AB)= Rank(A)". Since B is "m by m", saying that B has rank m means it has "full rank" so is invertible.
 
HallsofIvy said:
This is rather confusingly written! Usually the "given" is stated before what you are to prove. I think you are saying "Given that Rank(B)= m prove that Rank(AB)= Rank(A)". Since B is "m by m", saying that B has rank m means it has "full rank" so is invertible.
Click to expand...
you got the question right, my apology for the confusing way its written.
can you elaborate ? the fact the b is invertible means that ?
 
Hmm ... here's a thought that might help (or might not, I'm not sure exactly what you've learned).

I will use geometric vectors (arrows or directed lengths) by way of example. It's always good to remember that vectors are much more general than that.

In a 3D space, you might pick out three vectors that all stay in the same plane. In that case, those 3 vectors only span (by their linear combinations, I assume you understand span) that particular plane. That plane is a 2D space embedded in the 3D space, so we say that the span of those three vectors is only 2-dimensional. If you put those 3 vectors into a 3x3 matrix, they would turn out to be linearly dependent. If you pick any two of them, they may seem linearly independent, but the set of all 3 is not. One of the 3 will always be expressible as a linear combination of the other 2.

In that case, we say the matrix assembled that way is of rank 2. The 2 is the dimension of the space spanned by its vectors.

Now, if the three vectors are linearly independent, they will span the whole 3D space. They then have "full rank," the highest rank possible for three vectors in 3x3 matrix. A matrix like that will be invertible.
 
bZNyQ7C2 said:
Hmm ... here's a thought that might help (or might not, I'm not sure exactly what you've learned).

I will use geometric vectors (arrows or directed lengths) by way of example. It's always good to remember that vectors are much more general than that.

In a 3D space, you might pick out three vectors that all stay in the same plane. In that case, those 3 vectors only span (by their linear combinations, I assume you understand span) that particular plane. That plane is a 2D space embedded in the 3D space, so we say that the span of those three vectors is only 2-dimensional. If you put those 3 vectors into a 3x3 matrix, they would turn out to be linearly dependent. If you pick any two of them, they may seem linearly independent, but the set of all 3 is not. One of the 3 will always be expressible as a linear combination of the other 2.

In that case, we say the matrix assembled that way is of rank 2. The 2 is the dimension of the space spanned by its vectors.

Now, if the three vectors are linearly independent, they will span the whole 3D space. They then have "full rank," the highest rank possible for three vectors in 3x3 matrix. A matrix like that will be invertible.
Click to expand...

Hey, thank you for the explanation.
I think I do understand all the rules you described.
And I do understand why B is invertible.
My question is why the fact that B is invertible mean's that it doesn't change the rank of A by multiplication with B ? (Why rank(BA) stays as rank(A))
Is it because B is row identical to an mxm i matrix ?
 
You seem to want a rigorous, technical explanation, which is fine. I was offering a more intuitional approach. Let me think a bit.

It's like projecting a 2D object onto a line, or a 3D object onto a plane. After the projection is done and we have the result, we might pass it on to somebody else. This somebody else might then work out that a projection had been done, but s/he can't recover exactly what the situation was beforehand. Some information is lost.

If A and B are both full-rank matrices of the same dimensions, then they both have information about the whole space in which they live. When they're multiplied, the product preserves ... not all information, but some information about every location in the space.

If A is full rank and B is less than full rank, then the result destroys more information. Only a smaller space is preserved, say, one plane in a 3D space (the same kind of thing happens with higher dimensions, but that's harder to visualize).

If that helps you, great. If not, I hope you find some explanation that does.
 
Rank(B)=m implies that B is invertible.
If Ax=0, then BAx=B(Ax)=B0=0 and when BAx=0, then Ax=0 since B is invertible [ B'(BAx)= (B'B)(Ax)=Ax=0]. So the null(BA) = null(A). Hence rank(BA)=Rank(A)


Rank(C)=n implies that C is invertible.
Given any two matrices A and B such that AB makes sense we have Rank(AB)< Rank(A)
So Rank (A)=Rank(A*C*C')< Rank(AC) < Rank(A). This clearly implies that Rank(A) = Rank(AC)
 
bZNyQ7C2 said:
You seem to want a rigorous, technical explanation, which is fine. I was offering a more intuitional approach. Let me think a bit.

It's like projecting a 2D object onto a line, or a 3D object onto a plane. After the projection is done and we have the result, we might pass it on to somebody else. This somebody else might then work out that a projection had been done, but s/he can't recover exactly what the situation was beforehand. Some information is lost.

If A and B are both full-rank matrices of the same dimensions, then they both have information about the whole space in which they live. When they're multiplied, the product preserves ... not all information, but some information about every location in the space.

If A is full rank and B is less than full rank, then the result destroys more information. Only a smaller space is preserved, say, one plane in a 3D space (the same kind of thing happens with higher dimensions, but that's harder to visualize).

If that helps you, great. If not, I hope you find some explanation that does.
Click to expand...

Hey thanks alot for the help !
It did kinda help me imagine the idea of the question
but I need to prove it, so it's not enough though.
Hope another explanation will follow.
Thank you !
 
Jomo said:
Rank(B)=m implies that B is invertible.
If Ax=0, then BAx=B(Ax)=B0=0 and when BAx=0, then Ax=0 since B is invertible [ B'(BAx)= (B'B)(Ax)=Ax=0]. So the null(BA) = null(A). Hence rank(BA)=Rank(A)


Rank(C)=n implies that C is invertible.
Given any two matrices A and B such that AB makes sense we have Rank(AB)< Rank(A)
So Rank (A)=Rank(A*C*C')< Rank(AC) < Rank(A). This clearly implies that Rank(A) = Rank(AC)
Click to expand...

It seems correct !

Would you explain to me what null's means ? (language barrier here )
 
akleron said:
It seems correct !

Would you explain to me what null's means ? (language barrier here )
Click to expand...
Please fill in the blank. That will be what null is

____ (A) + rank(A) = dimension(A).
Does this help?
 
That is called the "nullity" of A (that's probably in your textbook). It is the dimension of the "null space" of A, the subspace of vectors that A maps to the 0 vector.
 
You must log in or register to reply here.
Top