$B = A^T A$ can only be one of the following. Which one?
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I know that the answer is (b) but I have no idea on how to arrive there. What would be a good approach to arrive at the answer?
linear-algebra matrices
edited Aug 13 at 11:03
A.Γ.
20.3k22353
asked Aug 13 at 8:55
Abdul Malek Altawekji
317115
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up vote
1
down vote
favorite
I know that the answer is (b) but I have no idea on how to arrive there. What would be a good approach to arrive at the answer?
linear-algebra matrices
edited Aug 13 at 11:03
A.Γ.
20.3k22353
asked Aug 13 at 8:55
Abdul Malek Altawekji
317115
What have you tried? Did you use the given information that the matrix is nonsingular? And what are the properties of $A^T A$? This might get you started :)
– Jan
Aug 13 at 9:03
1
The title is a little misleading. Maybe consider re-wording it?
– awllower
Aug 13 at 9:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that the answer is (b) but I have no idea on how to arrive there. What would be a good approach to arrive at the answer?
linear-algebra matrices
edited Aug 13 at 11:03
A.Γ.
20.3k22353
asked Aug 13 at 8:55
Abdul Malek Altawekji
317115
I know that the answer is (b) but I have no idea on how to arrive there. What would be a good approach to arrive at the answer?
linear-algebra matrices
edited Aug 13 at 11:03
A.Γ.
20.3k22353
asked Aug 13 at 8:55
Abdul Malek Altawekji
317115
edited Aug 13 at 11:03
A.Γ.
20.3k22353
edited Aug 13 at 11:03
A.Γ.
20.3k22353
edited Aug 13 at 11:03
A.Γ.
20.3k22353
20.3k22353
asked Aug 13 at 8:55
Abdul Malek Altawekji
317115
asked Aug 13 at 8:55
Abdul Malek Altawekji
317115
asked Aug 13 at 8:55
Abdul Malek Altawekji
317115
317115
What have you tried? Did you use the given information that the matrix is nonsingular? And what are the properties of $A^T A$? This might get you started :)
– Jan
Aug 13 at 9:03
1
The title is a little misleading. Maybe consider re-wording it?
– awllower
Aug 13 at 9:04
add a comment |Â
What have you tried? Did you use the given information that the matrix is nonsingular? And what are the properties of $A^T A$? This might get you started :)
– Jan
Aug 13 at 9:03
1
The title is a little misleading. Maybe consider re-wording it?
– awllower
Aug 13 at 9:04
What have you tried? Did you use the given information that the matrix is nonsingular? And what are the properties of $A^T A$? This might get you started :)
– Jan
Aug 13 at 9:03
What have you tried? Did you use the given information that the matrix is nonsingular? And what are the properties of $A^T A$? This might get you started :)
– Jan
Aug 13 at 9:03
1
1
The title is a little misleading. Maybe consider re-wording it?
– awllower
Aug 13 at 9:04
The title is a little misleading. Maybe consider re-wording it?
– awllower
Aug 13 at 9:04
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
I would check it in this order:
- $B$ needs to be symmetric, which rules out (c).
- $B$ needs to be nonsingular (as $A$ was nonsingular) which rules out (d).
- The diagonal of $B$ carries norms and hence cannot contain negative values (which rules out (e)).
- $B$ needs to be positive definite which rules out (a) (negative determinant).
In the particular example you show, the determinant condition actually rules out all options except (b), as awllower mentioned. This does not always need to be the case. For example, $B = -I_2$ has determinant +1 and yet is not positive definite.
answered Aug 13 at 10:51
Florian
1,2741617
I think $B$ in $(d)$ is symmetric.
– awllower
Aug 13 at 12:38
1
Looks like you’ve swapped (c) and (d) in your list.
– amd
Aug 13 at 18:27
Yes I did, thanks for noticing! Corrected.
– Florian
Aug 14 at 21:54
add a comment |Â
up vote
5
down vote
The determinant of $A^TA$ is the product $det Atimesdet A^T=(det A)^2$, so must be $>0$.
In the five options there is only one satisfying this criterion.
Hope this helps
answered Aug 13 at 9:01
awllower
9,85132470
add a comment |Â
up vote
0
down vote
This can be deduced as follows. Note that the product of A with it's transpose is symmetric. Thus one can eliminate C. Also if A is nonsingular, then B is also non singular. Thus, the determinant cannot be zero which eliminates D. Finally remember the determinant of transpose(A) is equal to that of A. This implies that the determinant of B is positive since it is equal to (det(A))^2. Therefore B is the correct answer. Hope this helps.
answered Aug 15 at 5:23
Rohpa
383
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I would check it in this order:
- $B$ needs to be symmetric, which rules out (c).
- $B$ needs to be nonsingular (as $A$ was nonsingular) which rules out (d).
- The diagonal of $B$ carries norms and hence cannot contain negative values (which rules out (e)).
- $B$ needs to be positive definite which rules out (a) (negative determinant).
In the particular example you show, the determinant condition actually rules out all options except (b), as awllower mentioned. This does not always need to be the case. For example, $B = -I_2$ has determinant +1 and yet is not positive definite.
answered Aug 13 at 10:51
Florian
1,2741617
I think $B$ in $(d)$ is symmetric.
– awllower
Aug 13 at 12:38
1
Looks like you’ve swapped (c) and (d) in your list.
– amd
Aug 13 at 18:27
Yes I did, thanks for noticing! Corrected.
– Florian
Aug 14 at 21:54
add a comment |Â
up vote
1
down vote
accepted
I would check it in this order:
- $B$ needs to be symmetric, which rules out (c).
- $B$ needs to be nonsingular (as $A$ was nonsingular) which rules out (d).
- The diagonal of $B$ carries norms and hence cannot contain negative values (which rules out (e)).
- $B$ needs to be positive definite which rules out (a) (negative determinant).
In the particular example you show, the determinant condition actually rules out all options except (b), as awllower mentioned. This does not always need to be the case. For example, $B = -I_2$ has determinant +1 and yet is not positive definite.
answered Aug 13 at 10:51
Florian
1,2741617
I think $B$ in $(d)$ is symmetric.
– awllower
Aug 13 at 12:38
1
Looks like you’ve swapped (c) and (d) in your list.
– amd
Aug 13 at 18:27
Yes I did, thanks for noticing! Corrected.
– Florian
Aug 14 at 21:54
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I would check it in this order:
- $B$ needs to be symmetric, which rules out (c).
- $B$ needs to be nonsingular (as $A$ was nonsingular) which rules out (d).
- The diagonal of $B$ carries norms and hence cannot contain negative values (which rules out (e)).
- $B$ needs to be positive definite which rules out (a) (negative determinant).
In the particular example you show, the determinant condition actually rules out all options except (b), as awllower mentioned. This does not always need to be the case. For example, $B = -I_2$ has determinant +1 and yet is not positive definite.
answered Aug 13 at 10:51
Florian
1,2741617
I would check it in this order:
- $B$ needs to be symmetric, which rules out (c).
- $B$ needs to be nonsingular (as $A$ was nonsingular) which rules out (d).
- The diagonal of $B$ carries norms and hence cannot contain negative values (which rules out (e)).
- $B$ needs to be positive definite which rules out (a) (negative determinant).
In the particular example you show, the determinant condition actually rules out all options except (b), as awllower mentioned. This does not always need to be the case. For example, $B = -I_2$ has determinant +1 and yet is not positive definite.
answered Aug 13 at 10:51
Florian
1,2741617
edited Aug 14 at 21:53
answered Aug 13 at 10:51
Florian
1,2741617
answered Aug 13 at 10:51
Florian
1,2741617
answered Aug 13 at 10:51
Florian
1,2741617
1,2741617
I think $B$ in $(d)$ is symmetric.
– awllower
Aug 13 at 12:38
1
Looks like you’ve swapped (c) and (d) in your list.
– amd
Aug 13 at 18:27
Yes I did, thanks for noticing! Corrected.
– Florian
Aug 14 at 21:54
add a comment |Â
I think $B$ in $(d)$ is symmetric.
– awllower
Aug 13 at 12:38
1
Looks like you’ve swapped (c) and (d) in your list.
– amd
Aug 13 at 18:27
Yes I did, thanks for noticing! Corrected.
– Florian
Aug 14 at 21:54
I think $B$ in $(d)$ is symmetric.
– awllower
Aug 13 at 12:38
I think $B$ in $(d)$ is symmetric.
– awllower
Aug 13 at 12:38
1
1
Looks like you’ve swapped (c) and (d) in your list.
– amd
Aug 13 at 18:27
Looks like you’ve swapped (c) and (d) in your list.
– amd
Aug 13 at 18:27
Yes I did, thanks for noticing! Corrected.
– Florian
Aug 14 at 21:54
Yes I did, thanks for noticing! Corrected.
– Florian
Aug 14 at 21:54
add a comment |Â
up vote
5
down vote
The determinant of $A^TA$ is the product $det Atimesdet A^T=(det A)^2$, so must be $>0$.
In the five options there is only one satisfying this criterion.
Hope this helps
answered Aug 13 at 9:01
awllower
9,85132470
add a comment |Â
up vote
5
down vote
The determinant of $A^TA$ is the product $det Atimesdet A^T=(det A)^2$, so must be $>0$.
In the five options there is only one satisfying this criterion.
Hope this helps
answered Aug 13 at 9:01
awllower
9,85132470
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The determinant of $A^TA$ is the product $det Atimesdet A^T=(det A)^2$, so must be $>0$.
In the five options there is only one satisfying this criterion.
Hope this helps
answered Aug 13 at 9:01
awllower
9,85132470
The determinant of $A^TA$ is the product $det Atimesdet A^T=(det A)^2$, so must be $>0$.
In the five options there is only one satisfying this criterion.
Hope this helps
answered Aug 13 at 9:01
awllower
9,85132470
answered Aug 13 at 9:01
awllower
9,85132470
answered Aug 13 at 9:01
awllower
9,85132470
answered Aug 13 at 9:01
awllower
9,85132470
9,85132470
add a comment |Â
add a comment |Â
up vote
0
down vote
This can be deduced as follows. Note that the product of A with it's transpose is symmetric. Thus one can eliminate C. Also if A is nonsingular, then B is also non singular. Thus, the determinant cannot be zero which eliminates D. Finally remember the determinant of transpose(A) is equal to that of A. This implies that the determinant of B is positive since it is equal to (det(A))^2. Therefore B is the correct answer. Hope this helps.
answered Aug 15 at 5:23
Rohpa
383
add a comment |Â
up vote
0
down vote
This can be deduced as follows. Note that the product of A with it's transpose is symmetric. Thus one can eliminate C. Also if A is nonsingular, then B is also non singular. Thus, the determinant cannot be zero which eliminates D. Finally remember the determinant of transpose(A) is equal to that of A. This implies that the determinant of B is positive since it is equal to (det(A))^2. Therefore B is the correct answer. Hope this helps.
answered Aug 15 at 5:23
Rohpa
383
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This can be deduced as follows. Note that the product of A with it's transpose is symmetric. Thus one can eliminate C. Also if A is nonsingular, then B is also non singular. Thus, the determinant cannot be zero which eliminates D. Finally remember the determinant of transpose(A) is equal to that of A. This implies that the determinant of B is positive since it is equal to (det(A))^2. Therefore B is the correct answer. Hope this helps.
answered Aug 15 at 5:23
Rohpa
383
This can be deduced as follows. Note that the product of A with it's transpose is symmetric. Thus one can eliminate C. Also if A is nonsingular, then B is also non singular. Thus, the determinant cannot be zero which eliminates D. Finally remember the determinant of transpose(A) is equal to that of A. This implies that the determinant of B is positive since it is equal to (det(A))^2. Therefore B is the correct answer. Hope this helps.
answered Aug 15 at 5:23
Rohpa
383
answered Aug 15 at 5:23
Rohpa
383
answered Aug 15 at 5:23
Rohpa
383
answered Aug 15 at 5:23
Rohpa
383
383
add a comment |Â
add a comment |Â
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What have you tried? Did you use the given information that the matrix is nonsingular? And what are the properties of $A^T A$? This might get you started :)
– Jan
Aug 13 at 9:03
1
The title is a little misleading. Maybe consider re-wording it?
– awllower
Aug 13 at 9:04