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#1
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What conditions need to hold in order the rank of BA to be the same as the rank of A and B? For example, suppose A is a n x m matrix of rank m, and B is a m x n matrix of rank m, then BA is a m x m matrix of rank m. When is this necessarily true?
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B_9710
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So if we let  T_A and  T_B be the linear transformations that are represented by matrices A and B respectively. Both these have rank m. So the dimension of the image of  T_A is m. Now the matrix BA represents the linear map  T_B T_A (composed maps, not multiplied). Now the image of this map is the image of  T_B with domain  \text{image} (T_A) .
So for rank(AB)=rank(T_A T_B)=m we need  T_B with domain  \text{image} (T_A) to map to a m-dimensional vector space - which is the whole of the image of T_B since T_B has rank m.
So basically if they both have rank m then AB will always have rank m too.
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(Original post by B_9710)
So if we let  T_A and  T_B be the linear transformations that are represented by matrices A and B respectively. Both these have rank m. So the dimension of the image of  T_A is m. Now the matrix BA represents the linear map  T_B T_A (composed maps, not multiplied). Now the image of this map is the image of  T_B with domain  \text{image} (T_A) .
So for rank(AB)=rank(T_A T_B)=m we need  T_B with domain  \text{image} (T_A) to map to a m-dimensional vector space - which is the whole of the image of T_B since T_B has rank m.
So basically if they both have rank m then AB will always have rank m too.
So basically if both A and B have rank m then AB will always have rank m too?
Oh ok, so just clarifying, does this hold for BA too, because you talked about BA then went onto the rank of AB.

I was just wondering regarding a similar problem that I had.
So, let's say A is a n x 2 matrix, with rank 2. So, A has full column rank. This means that the system of linear equations Ax = b will have a unique solution x, just saying randomly.

Question before I read your post: My question is, if A has that following information, then why does it (if it does) mean that A^TA is invertible, where A^T is the transpose of A. Here, A^T has dimension 2 x n and so A^TA has dimension 2 x 2.

I know for sure that rank A^TA <= min (rank (A^T, rank A) because in general a matrix product AB will always have the rows of A and the columns of B and so its rank will be the min of those rows and columns, which also dictate the minimum of the individual matrices A and B. In this case, we have that rank(A^T = rank(A), so when is it that rank (A^TA) = rank (A^T and so = rank (A)?

Thoughts after I read your post: Assuming it holds for AB and BA too, then since A^T and A have the same rank, then their product A^TA must have the same rank too i.e. rank 2. So, if A had a rank of 1, then A^TA would have a rank of 1 too. But, since A has full column rank i.e. a rank of 2, then A^TA has a rank of 2 guaranteed (as A^T has a rank of 2 as well) and so we know A^TA has full rank i.e. a rank of 2, and hence is invertible. So, the fact that rank of A = number of rows/columns of A^TA and the fact that if rank of A and rank of A^T are the same, then the rank of A^TA must be equal to them, then rank A implies full rank for A^TA and hence invertibility.
Last edited by Takeover Season; 1 month ago
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B_9710
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(Original post by Takeover Season)
So basically if both A and B have rank m then AB will always have rank m too?
Oh ok, so just clarifying, does this hold for BA too, because you talked about BA then went onto the rank of AB.

I was just wondering regarding a similar problem that I had.
So, let's say A is a n x 2 matrix, with rank 2. So, A has full column rank. This means that the system of linear equations Ax = b will have a unique solution x, just saying randomly.

Question before I read your post: My question is, if A has that following information, then why does it (if it does) mean that A^TA is invertible, where A^T is the transpose of A. Here, A^T has dimension 2 x n and so A^TA has dimension 2 x 2.

I know for sure that rank A^TA <= min (rank (A^T, rank A) because in general a matrix product AB will always have the rows of A and the columns of B and so its rank will be the min of those rows and columns, which also dictate the minimum of the individual matrices A and B. In this case, we have that rank(A^T = rank(A), so when is it that rank (A^TA) = rank (A^T and so = rank (A)?

Thoughts after I read your post: Assuming it holds for AB and BA too, then since A^T and A have the same rank, then their product A^TA must have the same rank too i.e. rank 2. So, if A had a rank of 1, then A^TA would have a rank of 1 too. But, since A has full column rank i.e. a rank of 2, then A^TA has a rank of 2 guaranteed (as A^T has a rank of 2 as well) and so we know A^TA has full rank i.e. a rank of 2, and hence is invertible. So, the fact that rank of A = number of rows/columns of A^TA and the fact that if rank of A and rank of A^T are the same, then the rank of A^TA must be equal to them, then rank A implies full rank for A^TA and hence invertibility.
MAde a mistake when I said AB I meant BA, I think you understand this now
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RichE
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#5
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(Original post by B_9710)
So basically if they both have rank m then AB will always have rank m too.
Unless I'm missing some hypothesis, this won't always be true. For example

T_A (x,y) = (x,y,0)
T_B (x,y,z) = (y,z)
both have rank 2

but T_B T_A (x,y) = (y,0)
has rank 1.

Or have I missed something?
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#6
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(Original post by RichE)
Unless I'm missing some hypothesis, this won't always be true. For example

T_A (x,y) = (x,y,0)
T_B (x,y,z) = (y,z)
both have rank 2

but T_B T_A (x,y) = (y,0)
has rank 1.

Or have I missed something?
I think A has to be a vector into the domain of B. In my case, I just have two matrices of opposite dimensions like tranposes 2 x n and n x 2 and in this case it is always meant to hold.
Last edited by Takeover Season; 1 month ago
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#7
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(Original post by B_9710)
MAde a mistake when I said AB I meant BA, I think you understand this now
Thank you so much!!
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#8
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(Original post by B_9710)
So if we let  T_A and  T_B be the linear transformations that are represented by matrices A and B respectively. Both these have rank m. So the dimension of the image of  T_A is m. Now the matrix BA represents the linear map  T_B T_A (composed maps, not multiplied). Now the image of this map is the image of  T_B with domain  \text{image} (T_A) .
So for rank(AB)=rank(T_A T_B)=m we need  T_B with domain  \text{image} (T_A) to map to a m-dimensional vector space - which is the whole of the image of T_B since T_B has rank m.
So basically if they both have rank m then AB will always have rank m too.
Just wondering, if you swap the B's and the A's, does the same argument also hold that if both A and B have rank m, then BA will always have rank m too?
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