Mixing
Rank, invertibility and solution spaces
$begingroup$
Let A and B be two matrices of the same order. Let's assume that ρ(B)=ρ(AB) (ρ=Rank).
Which of the following two statements are correct or incorrect?
- A is invertible
- The solution space of $ABmathbf x=mathbf0$ is equal to the solution space of $Bmathbf x=mathbf0$.
My answer is "incorrect" to 1): I see no theoretical reason to assume it's correct, and I have found a counterexample. Yet I believe 2). to be correct. If A were invertible it would be very easy to prove this, but as it is not I am looking for the proper evidence.
Can anyone help out?
Thank you!
linear-algebra matrices proof-verification matrix-rank
asked Feb 3 at 12:21
daltadalta
1578
$endgroup$
add a comment |
$begingroup$
Let A and B be two matrices of the same order. Let's assume that ρ(B)=ρ(AB) (ρ=Rank).
Which of the following two statements are correct or incorrect?
- A is invertible
- The solution space of $ABmathbf x=mathbf0$ is equal to the solution space of $Bmathbf x=mathbf0$.
My answer is "incorrect" to 1): I see no theoretical reason to assume it's correct, and I have found a counterexample. Yet I believe 2). to be correct. If A were invertible it would be very easy to prove this, but as it is not I am looking for the proper evidence.
Can anyone help out?
Thank you!
linear-algebra matrices proof-verification matrix-rank
asked Feb 3 at 12:21
daltadalta
1578
$endgroup$
$begingroup$
What does $rho$ stand for? Rank? Spectral radius?
$endgroup$
– PierreCarre
Feb 3 at 12:27
$begingroup$
Rank, let me edit...
$endgroup$
– dalta
Feb 3 at 12:30
add a comment |
$begingroup$
Let A and B be two matrices of the same order. Let's assume that ρ(B)=ρ(AB) (ρ=Rank).
Which of the following two statements are correct or incorrect?
- A is invertible
- The solution space of $ABmathbf x=mathbf0$ is equal to the solution space of $Bmathbf x=mathbf0$.
My answer is "incorrect" to 1): I see no theoretical reason to assume it's correct, and I have found a counterexample. Yet I believe 2). to be correct. If A were invertible it would be very easy to prove this, but as it is not I am looking for the proper evidence.
Can anyone help out?
Thank you!
linear-algebra matrices proof-verification matrix-rank
asked Feb 3 at 12:21
daltadalta
1578
$endgroup$
Let A and B be two matrices of the same order. Let's assume that ρ(B)=ρ(AB) (ρ=Rank).
Which of the following two statements are correct or incorrect?
- A is invertible
- The solution space of $ABmathbf x=mathbf0$ is equal to the solution space of $Bmathbf x=mathbf0$.
My answer is "incorrect" to 1): I see no theoretical reason to assume it's correct, and I have found a counterexample. Yet I believe 2). to be correct. If A were invertible it would be very easy to prove this, but as it is not I am looking for the proper evidence.
Can anyone help out?
Thank you!
linear-algebra matrices proof-verification matrix-rank
linear-algebra matrices proof-verification matrix-rank
asked Feb 3 at 12:21
daltadalta
1578
asked Feb 3 at 12:21
daltadalta
1578
edited Feb 3 at 12:30
dalta
asked Feb 3 at 12:21
daltadalta
1578
asked Feb 3 at 12:21
daltadalta
1578
asked Feb 3 at 12:21
daltadalta
1578
1578
$begingroup$
What does $rho$ stand for? Rank? Spectral radius?
$endgroup$
– PierreCarre
Feb 3 at 12:27
$begingroup$
Rank, let me edit...
$endgroup$
– dalta
Feb 3 at 12:30
add a comment |
$begingroup$
What does $rho$ stand for? Rank? Spectral radius?
$endgroup$
– PierreCarre
Feb 3 at 12:27
$begingroup$
Rank, let me edit...
$endgroup$
– dalta
Feb 3 at 12:30
$begingroup$
What does $rho$ stand for? Rank? Spectral radius?
$endgroup$
– PierreCarre
Feb 3 at 12:27
$begingroup$
What does $rho$ stand for? Rank? Spectral radius?
$endgroup$
– PierreCarre
Feb 3 at 12:27
$begingroup$
Rank, let me edit...
$endgroup$
– dalta
Feb 3 at 12:30
$begingroup$
Rank, let me edit...
$endgroup$
– dalta
Feb 3 at 12:30
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can make use of the following, which is not too hard to prove: $$ text{N}(B) subseteq text{N}(AB) $$ (null space of $B$ is a subset of the null space of $AB$) and combine it with the rank-nullity theorem.
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
$endgroup$
$begingroup$
Thank you! I've done what you suggested, with the following result: I am able to prove that $N(B) subseteq N(AB)$. So now I need to prove that the dimensions are equal, so that I can prove that they are equal. But I am wondering if my interpretation of the rank-nullity theorem is correct (I am confused because it's the first time I am applying it to a null space): dimP(ABx=0)=dimKer(ABx=0)+dimIm(ABx=0) vs. dimP(Bx=0)=dimKer(Bx=0)+dimIm(Bx=0). The dim of the images are 0, and the dimP(ABx=0)=dimP(Bx=0) because of them having the same rank?
$endgroup$
– dalta
Feb 4 at 10:17
$begingroup$
@dalta the solution space of $Ax=0$ is the null space of $A$ which is also the kernel of $A$ when $A$ is considered as a linear transformation...
$endgroup$
– Christiaan Hattingh
Feb 4 at 10:28
add a comment |
$begingroup$
Hint:
Try with simple examples. For $2 times 2$ matrices, you've already figured out that if the rank is $2$, then $A$ is invertible, and the claim is true. The same is true if the rank is zero. That leaves rank 1 matrices to investigate. So look at some examples.
For instance, here's a rank-1 $ 2 times 2$ matrix $B$:
$$
pmatrix{1 & 0 \ 0 & 0}
$$
If you pick $A = B$, then the ranks of $AB$ and $B$ are the same, and the solution space of $ABx = 0$ is the same as the solution space of $Bx = 0$, which suggests that maybe "2." is true. But that's not a lot of evidence. What other simple $2 times 2$ rank-1 matrices can you think of? Try a few.
Also: think about what $B$ does as a linear transformation -- I mean geometrically what it does. Then play some more.
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
Thank you for your answer, John! I will play around with it, as you suggested. Is there any algebraic evidence, though?
$endgroup$
– dalta
Feb 3 at 12:43
$begingroup$
I don't know what you mean by "algebraic evidence", I'm afraid. The key point of this exercise is for you to gain some intuition about what rank does and does not tell you, and having me give you the answer won't help, so my hint is where my help stops.
$endgroup$
– John Hughes
Feb 3 at 12:47
$begingroup$
OK, I understand… Thanks!
$endgroup$
– dalta
Feb 3 at 13:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
You can make use of the following, which is not too hard to prove: $$ text{N}(B) subseteq text{N}(AB) $$ (null space of $B$ is a subset of the null space of $AB$) and combine it with the rank-nullity theorem.
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
$endgroup$
$begingroup$
Thank you! I've done what you suggested, with the following result: I am able to prove that $N(B) subseteq N(AB)$. So now I need to prove that the dimensions are equal, so that I can prove that they are equal. But I am wondering if my interpretation of the rank-nullity theorem is correct (I am confused because it's the first time I am applying it to a null space): dimP(ABx=0)=dimKer(ABx=0)+dimIm(ABx=0) vs. dimP(Bx=0)=dimKer(Bx=0)+dimIm(Bx=0). The dim of the images are 0, and the dimP(ABx=0)=dimP(Bx=0) because of them having the same rank?
$endgroup$
– dalta
Feb 4 at 10:17
$begingroup$
@dalta the solution space of $Ax=0$ is the null space of $A$ which is also the kernel of $A$ when $A$ is considered as a linear transformation...
$endgroup$
– Christiaan Hattingh
Feb 4 at 10:28
add a comment |
$begingroup$
You can make use of the following, which is not too hard to prove: $$ text{N}(B) subseteq text{N}(AB) $$ (null space of $B$ is a subset of the null space of $AB$) and combine it with the rank-nullity theorem.
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
$endgroup$
$begingroup$
Thank you! I've done what you suggested, with the following result: I am able to prove that $N(B) subseteq N(AB)$. So now I need to prove that the dimensions are equal, so that I can prove that they are equal. But I am wondering if my interpretation of the rank-nullity theorem is correct (I am confused because it's the first time I am applying it to a null space): dimP(ABx=0)=dimKer(ABx=0)+dimIm(ABx=0) vs. dimP(Bx=0)=dimKer(Bx=0)+dimIm(Bx=0). The dim of the images are 0, and the dimP(ABx=0)=dimP(Bx=0) because of them having the same rank?
$endgroup$
– dalta
Feb 4 at 10:17
$begingroup$
@dalta the solution space of $Ax=0$ is the null space of $A$ which is also the kernel of $A$ when $A$ is considered as a linear transformation...
$endgroup$
– Christiaan Hattingh
Feb 4 at 10:28
add a comment |
$begingroup$
You can make use of the following, which is not too hard to prove: $$ text{N}(B) subseteq text{N}(AB) $$ (null space of $B$ is a subset of the null space of $AB$) and combine it with the rank-nullity theorem.
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
$endgroup$
You can make use of the following, which is not too hard to prove: $$ text{N}(B) subseteq text{N}(AB) $$ (null space of $B$ is a subset of the null space of $AB$) and combine it with the rank-nullity theorem.
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
answered Feb 3 at 18:56
Christiaan HattinghChristiaan Hattingh
3,887922
3,887922
$begingroup$
Thank you! I've done what you suggested, with the following result: I am able to prove that $N(B) subseteq N(AB)$. So now I need to prove that the dimensions are equal, so that I can prove that they are equal. But I am wondering if my interpretation of the rank-nullity theorem is correct (I am confused because it's the first time I am applying it to a null space): dimP(ABx=0)=dimKer(ABx=0)+dimIm(ABx=0) vs. dimP(Bx=0)=dimKer(Bx=0)+dimIm(Bx=0). The dim of the images are 0, and the dimP(ABx=0)=dimP(Bx=0) because of them having the same rank?
$endgroup$
– dalta
Feb 4 at 10:17
$begingroup$
@dalta the solution space of $Ax=0$ is the null space of $A$ which is also the kernel of $A$ when $A$ is considered as a linear transformation...
$endgroup$
– Christiaan Hattingh
Feb 4 at 10:28
add a comment |
$begingroup$
Thank you! I've done what you suggested, with the following result: I am able to prove that $N(B) subseteq N(AB)$. So now I need to prove that the dimensions are equal, so that I can prove that they are equal. But I am wondering if my interpretation of the rank-nullity theorem is correct (I am confused because it's the first time I am applying it to a null space): dimP(ABx=0)=dimKer(ABx=0)+dimIm(ABx=0) vs. dimP(Bx=0)=dimKer(Bx=0)+dimIm(Bx=0). The dim of the images are 0, and the dimP(ABx=0)=dimP(Bx=0) because of them having the same rank?
$endgroup$
– dalta
Feb 4 at 10:17
$begingroup$
@dalta the solution space of $Ax=0$ is the null space of $A$ which is also the kernel of $A$ when $A$ is considered as a linear transformation...
$endgroup$
– Christiaan Hattingh
Feb 4 at 10:28
$begingroup$
Thank you! I've done what you suggested, with the following result: I am able to prove that $N(B) subseteq N(AB)$. So now I need to prove that the dimensions are equal, so that I can prove that they are equal. But I am wondering if my interpretation of the rank-nullity theorem is correct (I am confused because it's the first time I am applying it to a null space): dimP(ABx=0)=dimKer(ABx=0)+dimIm(ABx=0) vs. dimP(Bx=0)=dimKer(Bx=0)+dimIm(Bx=0). The dim of the images are 0, and the dimP(ABx=0)=dimP(Bx=0) because of them having the same rank?
$endgroup$
– dalta
Feb 4 at 10:17
$begingroup$
Thank you! I've done what you suggested, with the following result: I am able to prove that $N(B) subseteq N(AB)$. So now I need to prove that the dimensions are equal, so that I can prove that they are equal. But I am wondering if my interpretation of the rank-nullity theorem is correct (I am confused because it's the first time I am applying it to a null space): dimP(ABx=0)=dimKer(ABx=0)+dimIm(ABx=0) vs. dimP(Bx=0)=dimKer(Bx=0)+dimIm(Bx=0). The dim of the images are 0, and the dimP(ABx=0)=dimP(Bx=0) because of them having the same rank?
$endgroup$
– dalta
Feb 4 at 10:17
$begingroup$
@dalta the solution space of $Ax=0$ is the null space of $A$ which is also the kernel of $A$ when $A$ is considered as a linear transformation...
$endgroup$
– Christiaan Hattingh
Feb 4 at 10:28
$begingroup$
@dalta the solution space of $Ax=0$ is the null space of $A$ which is also the kernel of $A$ when $A$ is considered as a linear transformation...
$endgroup$
– Christiaan Hattingh
Feb 4 at 10:28
add a comment |
$begingroup$
Hint:
Try with simple examples. For $2 times 2$ matrices, you've already figured out that if the rank is $2$, then $A$ is invertible, and the claim is true. The same is true if the rank is zero. That leaves rank 1 matrices to investigate. So look at some examples.
For instance, here's a rank-1 $ 2 times 2$ matrix $B$:
$$
pmatrix{1 & 0 \ 0 & 0}
$$
If you pick $A = B$, then the ranks of $AB$ and $B$ are the same, and the solution space of $ABx = 0$ is the same as the solution space of $Bx = 0$, which suggests that maybe "2." is true. But that's not a lot of evidence. What other simple $2 times 2$ rank-1 matrices can you think of? Try a few.
Also: think about what $B$ does as a linear transformation -- I mean geometrically what it does. Then play some more.
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
Thank you for your answer, John! I will play around with it, as you suggested. Is there any algebraic evidence, though?
$endgroup$
– dalta
Feb 3 at 12:43
$begingroup$
I don't know what you mean by "algebraic evidence", I'm afraid. The key point of this exercise is for you to gain some intuition about what rank does and does not tell you, and having me give you the answer won't help, so my hint is where my help stops.
$endgroup$
– John Hughes
Feb 3 at 12:47
$begingroup$
OK, I understand… Thanks!
$endgroup$
– dalta
Feb 3 at 13:02
add a comment |
$begingroup$
Hint:
Try with simple examples. For $2 times 2$ matrices, you've already figured out that if the rank is $2$, then $A$ is invertible, and the claim is true. The same is true if the rank is zero. That leaves rank 1 matrices to investigate. So look at some examples.
For instance, here's a rank-1 $ 2 times 2$ matrix $B$:
$$
pmatrix{1 & 0 \ 0 & 0}
$$
If you pick $A = B$, then the ranks of $AB$ and $B$ are the same, and the solution space of $ABx = 0$ is the same as the solution space of $Bx = 0$, which suggests that maybe "2." is true. But that's not a lot of evidence. What other simple $2 times 2$ rank-1 matrices can you think of? Try a few.
Also: think about what $B$ does as a linear transformation -- I mean geometrically what it does. Then play some more.
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
$endgroup$
$begingroup$
Thank you for your answer, John! I will play around with it, as you suggested. Is there any algebraic evidence, though?
$endgroup$
– dalta
Feb 3 at 12:43
$begingroup$
I don't know what you mean by "algebraic evidence", I'm afraid. The key point of this exercise is for you to gain some intuition about what rank does and does not tell you, and having me give you the answer won't help, so my hint is where my help stops.
$endgroup$
– John Hughes
Feb 3 at 12:47
$begingroup$
OK, I understand… Thanks!
$endgroup$
– dalta
Feb 3 at 13:02
add a comment |
$begingroup$
Hint:
Try with simple examples. For $2 times 2$ matrices, you've already figured out that if the rank is $2$, then $A$ is invertible, and the claim is true. The same is true if the rank is zero. That leaves rank 1 matrices to investigate. So look at some examples.
For instance, here's a rank-1 $ 2 times 2$ matrix $B$:
$$
pmatrix{1 & 0 \ 0 & 0}
$$
If you pick $A = B$, then the ranks of $AB$ and $B$ are the same, and the solution space of $ABx = 0$ is the same as the solution space of $Bx = 0$, which suggests that maybe "2." is true. But that's not a lot of evidence. What other simple $2 times 2$ rank-1 matrices can you think of? Try a few.
Also: think about what $B$ does as a linear transformation -- I mean geometrically what it does. Then play some more.
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
$endgroup$
Hint:
Try with simple examples. For $2 times 2$ matrices, you've already figured out that if the rank is $2$, then $A$ is invertible, and the claim is true. The same is true if the rank is zero. That leaves rank 1 matrices to investigate. So look at some examples.
For instance, here's a rank-1 $ 2 times 2$ matrix $B$:
$$
pmatrix{1 & 0 \ 0 & 0}
$$
If you pick $A = B$, then the ranks of $AB$ and $B$ are the same, and the solution space of $ABx = 0$ is the same as the solution space of $Bx = 0$, which suggests that maybe "2." is true. But that's not a lot of evidence. What other simple $2 times 2$ rank-1 matrices can you think of? Try a few.
Also: think about what $B$ does as a linear transformation -- I mean geometrically what it does. Then play some more.
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
answered Feb 3 at 12:34
John HughesJohn Hughes
65.5k24293
65.5k24293
$begingroup$
Thank you for your answer, John! I will play around with it, as you suggested. Is there any algebraic evidence, though?
$endgroup$
– dalta
Feb 3 at 12:43
$begingroup$
I don't know what you mean by "algebraic evidence", I'm afraid. The key point of this exercise is for you to gain some intuition about what rank does and does not tell you, and having me give you the answer won't help, so my hint is where my help stops.
$endgroup$
– John Hughes
Feb 3 at 12:47
$begingroup$
OK, I understand… Thanks!
$endgroup$
– dalta
Feb 3 at 13:02
add a comment |
$begingroup$
Thank you for your answer, John! I will play around with it, as you suggested. Is there any algebraic evidence, though?
$endgroup$
– dalta
Feb 3 at 12:43
$begingroup$
I don't know what you mean by "algebraic evidence", I'm afraid. The key point of this exercise is for you to gain some intuition about what rank does and does not tell you, and having me give you the answer won't help, so my hint is where my help stops.
$endgroup$
– John Hughes
Feb 3 at 12:47
$begingroup$
OK, I understand… Thanks!
$endgroup$
– dalta
Feb 3 at 13:02
$begingroup$
Thank you for your answer, John! I will play around with it, as you suggested. Is there any algebraic evidence, though?
$endgroup$
– dalta
Feb 3 at 12:43
$begingroup$
Thank you for your answer, John! I will play around with it, as you suggested. Is there any algebraic evidence, though?
$endgroup$
– dalta
Feb 3 at 12:43
$begingroup$
I don't know what you mean by "algebraic evidence", I'm afraid. The key point of this exercise is for you to gain some intuition about what rank does and does not tell you, and having me give you the answer won't help, so my hint is where my help stops.
$endgroup$
– John Hughes
Feb 3 at 12:47
$begingroup$
I don't know what you mean by "algebraic evidence", I'm afraid. The key point of this exercise is for you to gain some intuition about what rank does and does not tell you, and having me give you the answer won't help, so my hint is where my help stops.
$endgroup$
– John Hughes
Feb 3 at 12:47
$begingroup$
OK, I understand… Thanks!
$endgroup$
– dalta
Feb 3 at 13:02
$begingroup$
OK, I understand… Thanks!
$endgroup$
– dalta
Feb 3 at 13:02
add a comment |
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$begingroup$
What does $rho$ stand for? Rank? Spectral radius?
$endgroup$
– PierreCarre
Feb 3 at 12:27
$begingroup$
Rank, let me edit...
$endgroup$
– dalta
Feb 3 at 12:30