Find rank of AB, given that A has linearly independent columns and B has rank 2
$begingroup$
I'm trying to prove to myself that given...
- Matrix A, which has linearly independent columns, and at least 2 columns...
- Matrix B, which has rank of 2
Their product, AB, will have rank of 2. I believe this is because...
- Matrix B has two linearly independent columns.
- Each column of AB will be a combination of the columns of A
- When multiplying matrix A by matrix B, each of the two independent columns of B will create a unique combination of the columns in A.
Is this true? If so, can this be made more rigorous? Thanks for the help in advance!
linear-algebra matrices matrix-rank
asked Jan 21 at 0:03
user2621707user2621707
184
$endgroup$
add a comment |
$begingroup$
I'm trying to prove to myself that given...
- Matrix A, which has linearly independent columns, and at least 2 columns...
- Matrix B, which has rank of 2
Their product, AB, will have rank of 2. I believe this is because...
- Matrix B has two linearly independent columns.
- Each column of AB will be a combination of the columns of A
- When multiplying matrix A by matrix B, each of the two independent columns of B will create a unique combination of the columns in A.
Is this true? If so, can this be made more rigorous? Thanks for the help in advance!
linear-algebra matrices matrix-rank
asked Jan 21 at 0:03
user2621707user2621707
184
$endgroup$
$begingroup$
Your conclusion is correct, but I don't quite follow your line of reasoning
$endgroup$
– Omnomnomnom
Jan 21 at 0:29
$begingroup$
@MattSamuel good point! Edited the question accordingly
$endgroup$
– user2621707
Jan 21 at 0:29
$begingroup$
@Omnomnomnom Let me try one more time: - Each column of AB is simply a combinations of the columns of A. This is because multiplying a matrix by a vector is equivalent to taking a linear combination of the columns of A. - There are two linearly independent columns in B. When we multiply A by these columns, we end up with two vectors which are linearly independent to each other. Does this make sense? This is the part I'm trying to prove to myself.
$endgroup$
– user2621707
Jan 21 at 0:36
add a comment |
$begingroup$
I'm trying to prove to myself that given...
- Matrix A, which has linearly independent columns, and at least 2 columns...
- Matrix B, which has rank of 2
Their product, AB, will have rank of 2. I believe this is because...
- Matrix B has two linearly independent columns.
- Each column of AB will be a combination of the columns of A
- When multiplying matrix A by matrix B, each of the two independent columns of B will create a unique combination of the columns in A.
Is this true? If so, can this be made more rigorous? Thanks for the help in advance!
linear-algebra matrices matrix-rank
asked Jan 21 at 0:03
user2621707user2621707
184
$endgroup$
I'm trying to prove to myself that given...
- Matrix A, which has linearly independent columns, and at least 2 columns...
- Matrix B, which has rank of 2
Their product, AB, will have rank of 2. I believe this is because...
- Matrix B has two linearly independent columns.
- Each column of AB will be a combination of the columns of A
- When multiplying matrix A by matrix B, each of the two independent columns of B will create a unique combination of the columns in A.
Is this true? If so, can this be made more rigorous? Thanks for the help in advance!
linear-algebra matrices matrix-rank
linear-algebra matrices matrix-rank
asked Jan 21 at 0:03
user2621707user2621707
184
asked Jan 21 at 0:03
user2621707user2621707
184
edited Jan 21 at 0:30
user2621707
asked Jan 21 at 0:03
user2621707user2621707
184
asked Jan 21 at 0:03
user2621707user2621707
184
asked Jan 21 at 0:03
user2621707user2621707
184
184
$begingroup$
Your conclusion is correct, but I don't quite follow your line of reasoning
$endgroup$
– Omnomnomnom
Jan 21 at 0:29
$begingroup$
@MattSamuel good point! Edited the question accordingly
$endgroup$
– user2621707
Jan 21 at 0:29
$begingroup$
@Omnomnomnom Let me try one more time: - Each column of AB is simply a combinations of the columns of A. This is because multiplying a matrix by a vector is equivalent to taking a linear combination of the columns of A. - There are two linearly independent columns in B. When we multiply A by these columns, we end up with two vectors which are linearly independent to each other. Does this make sense? This is the part I'm trying to prove to myself.
$endgroup$
– user2621707
Jan 21 at 0:36
add a comment |
$begingroup$
Your conclusion is correct, but I don't quite follow your line of reasoning
$endgroup$
– Omnomnomnom
Jan 21 at 0:29
$begingroup$
@MattSamuel good point! Edited the question accordingly
$endgroup$
– user2621707
Jan 21 at 0:29
$begingroup$
@Omnomnomnom Let me try one more time: - Each column of AB is simply a combinations of the columns of A. This is because multiplying a matrix by a vector is equivalent to taking a linear combination of the columns of A. - There are two linearly independent columns in B. When we multiply A by these columns, we end up with two vectors which are linearly independent to each other. Does this make sense? This is the part I'm trying to prove to myself.
$endgroup$
– user2621707
Jan 21 at 0:36
$begingroup$
Your conclusion is correct, but I don't quite follow your line of reasoning
$endgroup$
– Omnomnomnom
Jan 21 at 0:29
$begingroup$
Your conclusion is correct, but I don't quite follow your line of reasoning
$endgroup$
– Omnomnomnom
Jan 21 at 0:29
$begingroup$
@MattSamuel good point! Edited the question accordingly
$endgroup$
– user2621707
Jan 21 at 0:29
$begingroup$
@MattSamuel good point! Edited the question accordingly
$endgroup$
– user2621707
Jan 21 at 0:29
$begingroup$
@Omnomnomnom Let me try one more time: - Each column of AB is simply a combinations of the columns of A. This is because multiplying a matrix by a vector is equivalent to taking a linear combination of the columns of A. - There are two linearly independent columns in B. When we multiply A by these columns, we end up with two vectors which are linearly independent to each other. Does this make sense? This is the part I'm trying to prove to myself.
$endgroup$
– user2621707
Jan 21 at 0:36
$begingroup$
@Omnomnomnom Let me try one more time: - Each column of AB is simply a combinations of the columns of A. This is because multiplying a matrix by a vector is equivalent to taking a linear combination of the columns of A. - There are two linearly independent columns in B. When we multiply A by these columns, we end up with two vectors which are linearly independent to each other. Does this make sense? This is the part I'm trying to prove to myself.
$endgroup$
– user2621707
Jan 21 at 0:36
add a comment |
1 Answer
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Yes, it can be made more rigorous/clear, here's how I would do it. Note that it does require thinking about the linear transformations represented by the matrices. I will use abuse notation slightly and use $A$ and $B$ to refer to both the matrices and the linear transformations they represent.
- $newcommandim{operatorname{im}}dim im B = 2$
- $im AB = A(im B)$
$A$ is injective (since the columns are linearly independent), so $$newcommandrk{operatorname{rank}}rk(AB)=dim im AB = dim A(im B) = dimim B = rk(B) = 2$$
Note that this says more generally that if $A$ has linearly independent columns, then $rk(AB)=rk(B)$. Also, I tried to keep the proof as close to what you wrote as possible, to make it clear how it rigorizes what you wrote, but it's probably not the most clear formulation of the proof.
answered Jan 21 at 0:39
jgonjgon
14.6k22042
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1 Answer
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$begingroup$
Yes, it can be made more rigorous/clear, here's how I would do it. Note that it does require thinking about the linear transformations represented by the matrices. I will use abuse notation slightly and use $A$ and $B$ to refer to both the matrices and the linear transformations they represent.
- $newcommandim{operatorname{im}}dim im B = 2$
- $im AB = A(im B)$
$A$ is injective (since the columns are linearly independent), so $$newcommandrk{operatorname{rank}}rk(AB)=dim im AB = dim A(im B) = dimim B = rk(B) = 2$$
Note that this says more generally that if $A$ has linearly independent columns, then $rk(AB)=rk(B)$. Also, I tried to keep the proof as close to what you wrote as possible, to make it clear how it rigorizes what you wrote, but it's probably not the most clear formulation of the proof.
answered Jan 21 at 0:39
jgonjgon
14.6k22042
$endgroup$
add a comment |
$begingroup$
Yes, it can be made more rigorous/clear, here's how I would do it. Note that it does require thinking about the linear transformations represented by the matrices. I will use abuse notation slightly and use $A$ and $B$ to refer to both the matrices and the linear transformations they represent.
- $newcommandim{operatorname{im}}dim im B = 2$
- $im AB = A(im B)$
$A$ is injective (since the columns are linearly independent), so $$newcommandrk{operatorname{rank}}rk(AB)=dim im AB = dim A(im B) = dimim B = rk(B) = 2$$
Note that this says more generally that if $A$ has linearly independent columns, then $rk(AB)=rk(B)$. Also, I tried to keep the proof as close to what you wrote as possible, to make it clear how it rigorizes what you wrote, but it's probably not the most clear formulation of the proof.
answered Jan 21 at 0:39
jgonjgon
14.6k22042
$endgroup$
add a comment |
$begingroup$
Yes, it can be made more rigorous/clear, here's how I would do it. Note that it does require thinking about the linear transformations represented by the matrices. I will use abuse notation slightly and use $A$ and $B$ to refer to both the matrices and the linear transformations they represent.
- $newcommandim{operatorname{im}}dim im B = 2$
- $im AB = A(im B)$
$A$ is injective (since the columns are linearly independent), so $$newcommandrk{operatorname{rank}}rk(AB)=dim im AB = dim A(im B) = dimim B = rk(B) = 2$$
Note that this says more generally that if $A$ has linearly independent columns, then $rk(AB)=rk(B)$. Also, I tried to keep the proof as close to what you wrote as possible, to make it clear how it rigorizes what you wrote, but it's probably not the most clear formulation of the proof.
answered Jan 21 at 0:39
jgonjgon
14.6k22042
$endgroup$
Yes, it can be made more rigorous/clear, here's how I would do it. Note that it does require thinking about the linear transformations represented by the matrices. I will use abuse notation slightly and use $A$ and $B$ to refer to both the matrices and the linear transformations they represent.
- $newcommandim{operatorname{im}}dim im B = 2$
- $im AB = A(im B)$
$A$ is injective (since the columns are linearly independent), so $$newcommandrk{operatorname{rank}}rk(AB)=dim im AB = dim A(im B) = dimim B = rk(B) = 2$$
Note that this says more generally that if $A$ has linearly independent columns, then $rk(AB)=rk(B)$. Also, I tried to keep the proof as close to what you wrote as possible, to make it clear how it rigorizes what you wrote, but it's probably not the most clear formulation of the proof.
answered Jan 21 at 0:39
jgonjgon
14.6k22042
answered Jan 21 at 0:39
jgonjgon
14.6k22042
answered Jan 21 at 0:39
jgonjgon
14.6k22042
answered Jan 21 at 0:39
jgonjgon
14.6k22042
14.6k22042
add a comment |
add a comment |
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$begingroup$
Your conclusion is correct, but I don't quite follow your line of reasoning
$endgroup$
– Omnomnomnom
Jan 21 at 0:29
$begingroup$
@MattSamuel good point! Edited the question accordingly
$endgroup$
– user2621707
Jan 21 at 0:29
$begingroup$
@Omnomnomnom Let me try one more time: - Each column of AB is simply a combinations of the columns of A. This is because multiplying a matrix by a vector is equivalent to taking a linear combination of the columns of A. - There are two linearly independent columns in B. When we multiply A by these columns, we end up with two vectors which are linearly independent to each other. Does this make sense? This is the part I'm trying to prove to myself.
$endgroup$
– user2621707
Jan 21 at 0:36