# Matrix Rank

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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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#2

So if we let and 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 is m. Now the matrix BA represents the linear map (composed maps, not multiplied). Now the image of this map is the image of with domain .

So for rank(AB)=rank(T_A T_B)=m we need with domain 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 for rank(AB)=rank(T_A T_B)=m we need with domain 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

So if we let and 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 is m. Now the matrix BA represents the linear map (composed maps, not multiplied). Now the image of this map is the image of with domain .

So for rank(AB)=rank(T_A T_B)=m we need with domain 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.

**B_9710**)So if we let and 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 is m. Now the matrix BA represents the linear map (composed maps, not multiplied). Now the image of this map is the image of with domain .

So for rank(AB)=rank(T_A T_B)=m we need with domain 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.

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 is invertible, where is the transpose of A. Here, has dimension 2 x n and so has dimension 2 x 2.

I know for sure that rank <= min (rank (, 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( = rank(A), so when is it that rank () = rank ( and so = rank (A)?

**Thoughts after I read your post:**Assuming it holds for AB and BA too, then since and A have the same rank, then their product must have the same rank too i.e. rank 2. So, if A had a rank of 1, then would have a rank of 1 too. But, since A has full column rank i.e. a rank of 2, then has a rank of 2 guaranteed (as has a rank of 2 as well) and so we know 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

**and the fact that if rank of A and rank of are the same, then the rank of must be equal to them**, then rank A implies full rank for and hence invertibility.

Last edited by Takeover Season; 1 month ago

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(Original post by

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.

I know for sure that rank <= min (rank (, 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( = rank(A), so when is it that rank () = rank ( and so = rank (A)?

**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 is invertible, where is the transpose of A. Here, has dimension 2 x n and so has dimension 2 x 2.I know for sure that rank <= min (rank (, 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( = rank(A), so when is it that rank () = rank ( and so = rank (A)?

**Thoughts after I read your post:**Assuming it holds for AB and BA too, then since and A have the same rank, then their product must have the same rank too i.e. rank 2. So, if A had a rank of 1, then would have a rank of 1 too. But, since A has full column rank i.e. a rank of 2, then has a rank of 2 guaranteed (as has a rank of 2 as well) and so we know 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**and the fact that if rank of A and rank of are the same, then the rank of must be equal to them**, then rank A implies full rank for and hence invertibility.
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#5

(Original post by

So basically if they both have rank m then AB will always have rank m too.

**B_9710**)So basically if they both have rank m then AB will always have rank m too.

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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(Original post by

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?

**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?

Last edited by Takeover Season; 1 month ago

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(Original post by

MAde a mistake when I said AB I meant BA, I think you understand this now

**B_9710**)MAde a mistake when I said AB I meant BA, I think you understand this now

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**B_9710**)

So if we let and 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 is m. Now the matrix BA represents the linear map (composed maps, not multiplied). Now the image of this map is the image of with domain .

So for rank(AB)=rank(T_A T_B)=m we need with domain 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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