What does a matrix determinant depend on? [closed]
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I know the concept. I know the geometric explanation of the transformed volume. But the determinant still troubles me. The geometric explanation does not help when we are solving a non-vector problem, except that it could be assumed that the most different the columns in a matrix are from each other the greater the determinant, as the cube gets a higher volume when the vectors that form it are far away. That brings to the question: what really does the determinant depend on (ie. how large the numbers in the matrix are or how much the differ from each other)?
In my mind, the answer to this question should lead to an intuitive idea of the determinant different from the geometrical explanation.
Many thanks in advance.
linear-algebra linear-transformations numerical-linear-algebra
asked Jan 25 at 23:17
chancarchancar
81
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closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, Shailesh, ancientmathematician, mrtaurho Jan 26 at 12:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I know the concept. I know the geometric explanation of the transformed volume. But the determinant still troubles me. The geometric explanation does not help when we are solving a non-vector problem, except that it could be assumed that the most different the columns in a matrix are from each other the greater the determinant, as the cube gets a higher volume when the vectors that form it are far away. That brings to the question: what really does the determinant depend on (ie. how large the numbers in the matrix are or how much the differ from each other)?
In my mind, the answer to this question should lead to an intuitive idea of the determinant different from the geometrical explanation.
Many thanks in advance.
linear-algebra linear-transformations numerical-linear-algebra
asked Jan 25 at 23:17
chancarchancar
81
$endgroup$
closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, Shailesh, ancientmathematician, mrtaurho Jan 26 at 12:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
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What you are asking is hard to grasp, IMO. I'd like to draw your attention to this ripened post on determinants
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– Hanno
Jan 25 at 23:28
1
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Thanks for the input. Basically I'm asking if the determinant depends on how different the columns of a matrix are form each other.
$endgroup$
– chancar
Jan 25 at 23:46
1
$begingroup$
In order to answer this, you should explain/define, what " ... how different the columns of a matrix are form [typo?] each other" does signify, so that one can work on and with it.
$endgroup$
– Hanno
Jan 25 at 23:58
add a comment |
$begingroup$
I know the concept. I know the geometric explanation of the transformed volume. But the determinant still troubles me. The geometric explanation does not help when we are solving a non-vector problem, except that it could be assumed that the most different the columns in a matrix are from each other the greater the determinant, as the cube gets a higher volume when the vectors that form it are far away. That brings to the question: what really does the determinant depend on (ie. how large the numbers in the matrix are or how much the differ from each other)?
In my mind, the answer to this question should lead to an intuitive idea of the determinant different from the geometrical explanation.
Many thanks in advance.
linear-algebra linear-transformations numerical-linear-algebra
asked Jan 25 at 23:17
chancarchancar
81
$endgroup$
I know the concept. I know the geometric explanation of the transformed volume. But the determinant still troubles me. The geometric explanation does not help when we are solving a non-vector problem, except that it could be assumed that the most different the columns in a matrix are from each other the greater the determinant, as the cube gets a higher volume when the vectors that form it are far away. That brings to the question: what really does the determinant depend on (ie. how large the numbers in the matrix are or how much the differ from each other)?
In my mind, the answer to this question should lead to an intuitive idea of the determinant different from the geometrical explanation.
Many thanks in advance.
linear-algebra linear-transformations numerical-linear-algebra
linear-algebra linear-transformations numerical-linear-algebra
asked Jan 25 at 23:17
chancarchancar
81
asked Jan 25 at 23:17
chancarchancar
81
asked Jan 25 at 23:17
chancarchancar
81
asked Jan 25 at 23:17
chancarchancar
81
asked Jan 25 at 23:17
chancarchancar
81
81
closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, Shailesh, ancientmathematician, mrtaurho Jan 26 at 12:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Lord Shark the Unknown, metamorphy, Shailesh, ancientmathematician, mrtaurho Jan 26 at 12:09
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
$begingroup$
What you are asking is hard to grasp, IMO. I'd like to draw your attention to this ripened post on determinants
$endgroup$
– Hanno
Jan 25 at 23:28
1
$begingroup$
Thanks for the input. Basically I'm asking if the determinant depends on how different the columns of a matrix are form each other.
$endgroup$
– chancar
Jan 25 at 23:46
1
$begingroup$
In order to answer this, you should explain/define, what " ... how different the columns of a matrix are form [typo?] each other" does signify, so that one can work on and with it.
$endgroup$
– Hanno
Jan 25 at 23:58
add a comment |
2
$begingroup$
What you are asking is hard to grasp, IMO. I'd like to draw your attention to this ripened post on determinants
$endgroup$
– Hanno
Jan 25 at 23:28
1
$begingroup$
Thanks for the input. Basically I'm asking if the determinant depends on how different the columns of a matrix are form each other.
$endgroup$
– chancar
Jan 25 at 23:46
1
$begingroup$
In order to answer this, you should explain/define, what " ... how different the columns of a matrix are form [typo?] each other" does signify, so that one can work on and with it.
$endgroup$
– Hanno
Jan 25 at 23:58
2
2
$begingroup$
What you are asking is hard to grasp, IMO. I'd like to draw your attention to this ripened post on determinants
$endgroup$
– Hanno
Jan 25 at 23:28
$begingroup$
What you are asking is hard to grasp, IMO. I'd like to draw your attention to this ripened post on determinants
$endgroup$
– Hanno
Jan 25 at 23:28
1
1
$begingroup$
Thanks for the input. Basically I'm asking if the determinant depends on how different the columns of a matrix are form each other.
$endgroup$
– chancar
Jan 25 at 23:46
$begingroup$
Thanks for the input. Basically I'm asking if the determinant depends on how different the columns of a matrix are form each other.
$endgroup$
– chancar
Jan 25 at 23:46
1
1
$begingroup$
In order to answer this, you should explain/define, what " ... how different the columns of a matrix are form [typo?] each other" does signify, so that one can work on and with it.
$endgroup$
– Hanno
Jan 25 at 23:58
$begingroup$
In order to answer this, you should explain/define, what " ... how different the columns of a matrix are form [typo?] each other" does signify, so that one can work on and with it.
$endgroup$
– Hanno
Jan 25 at 23:58
add a comment |
1 Answer
1
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oldest
votes
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To attempt to answer the question you asked in the comments, the answer is "no". The determinant is, roughly, a measure of how close a square matrix is to being singular, but doesn't really help us to tell how close the columns are to each other.
Take for example, the matrices
$$begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}.$$
They have the same columns, but their determinants differ by $2$. If order matters here, take instead,
$$begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 end{pmatrix},$$
which both have determinant $1$, have the same columns, but in a totally different order.
Another example:
$$begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix} text{ and } begin{pmatrix} 100 & 100 \ -100 & -100 end{pmatrix}$$
have the same determinant $0$, since both are singular. The columns, as you can see, are very different. The determinant of their difference is also $0$, in case you were thinking $det(A - B) = 0$ might yield some measure of similarity between the columns.
It is true, however, that the determinant is continuous. Small changes in a matrix will result in small (possibly null) changes in the determinant. But knowing the determinant of two matrices tells you next to nothing about how close they are.
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
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Great Explanation, thanks!
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– chancar
Jan 26 at 9:14
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@chancar Then you should upvote as well, IMHO ...
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– Hanno
Jan 26 at 9:27
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I'm new and not allowed to upvote yet. But the north remembers...
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– chancar
Jan 26 at 18:19
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I see, not being aware of it. .. At least some push (+5) towards the rep frontier is possible $ddotsmile$
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– Hanno
Jan 27 at 11:14
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To attempt to answer the question you asked in the comments, the answer is "no". The determinant is, roughly, a measure of how close a square matrix is to being singular, but doesn't really help us to tell how close the columns are to each other.
Take for example, the matrices
$$begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}.$$
They have the same columns, but their determinants differ by $2$. If order matters here, take instead,
$$begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 end{pmatrix},$$
which both have determinant $1$, have the same columns, but in a totally different order.
Another example:
$$begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix} text{ and } begin{pmatrix} 100 & 100 \ -100 & -100 end{pmatrix}$$
have the same determinant $0$, since both are singular. The columns, as you can see, are very different. The determinant of their difference is also $0$, in case you were thinking $det(A - B) = 0$ might yield some measure of similarity between the columns.
It is true, however, that the determinant is continuous. Small changes in a matrix will result in small (possibly null) changes in the determinant. But knowing the determinant of two matrices tells you next to nothing about how close they are.
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
$endgroup$
$begingroup$
Great Explanation, thanks!
$endgroup$
– chancar
Jan 26 at 9:14
$begingroup$
@chancar Then you should upvote as well, IMHO ...
$endgroup$
– Hanno
Jan 26 at 9:27
$begingroup$
I'm new and not allowed to upvote yet. But the north remembers...
$endgroup$
– chancar
Jan 26 at 18:19
$begingroup$
I see, not being aware of it. .. At least some push (+5) towards the rep frontier is possible $ddotsmile$
$endgroup$
– Hanno
Jan 27 at 11:14
add a comment |
$begingroup$
To attempt to answer the question you asked in the comments, the answer is "no". The determinant is, roughly, a measure of how close a square matrix is to being singular, but doesn't really help us to tell how close the columns are to each other.
Take for example, the matrices
$$begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}.$$
They have the same columns, but their determinants differ by $2$. If order matters here, take instead,
$$begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 end{pmatrix},$$
which both have determinant $1$, have the same columns, but in a totally different order.
Another example:
$$begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix} text{ and } begin{pmatrix} 100 & 100 \ -100 & -100 end{pmatrix}$$
have the same determinant $0$, since both are singular. The columns, as you can see, are very different. The determinant of their difference is also $0$, in case you were thinking $det(A - B) = 0$ might yield some measure of similarity between the columns.
It is true, however, that the determinant is continuous. Small changes in a matrix will result in small (possibly null) changes in the determinant. But knowing the determinant of two matrices tells you next to nothing about how close they are.
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
$endgroup$
$begingroup$
Great Explanation, thanks!
$endgroup$
– chancar
Jan 26 at 9:14
$begingroup$
@chancar Then you should upvote as well, IMHO ...
$endgroup$
– Hanno
Jan 26 at 9:27
$begingroup$
I'm new and not allowed to upvote yet. But the north remembers...
$endgroup$
– chancar
Jan 26 at 18:19
$begingroup$
I see, not being aware of it. .. At least some push (+5) towards the rep frontier is possible $ddotsmile$
$endgroup$
– Hanno
Jan 27 at 11:14
add a comment |
$begingroup$
To attempt to answer the question you asked in the comments, the answer is "no". The determinant is, roughly, a measure of how close a square matrix is to being singular, but doesn't really help us to tell how close the columns are to each other.
Take for example, the matrices
$$begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}.$$
They have the same columns, but their determinants differ by $2$. If order matters here, take instead,
$$begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 end{pmatrix},$$
which both have determinant $1$, have the same columns, but in a totally different order.
Another example:
$$begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix} text{ and } begin{pmatrix} 100 & 100 \ -100 & -100 end{pmatrix}$$
have the same determinant $0$, since both are singular. The columns, as you can see, are very different. The determinant of their difference is also $0$, in case you were thinking $det(A - B) = 0$ might yield some measure of similarity between the columns.
It is true, however, that the determinant is continuous. Small changes in a matrix will result in small (possibly null) changes in the determinant. But knowing the determinant of two matrices tells you next to nothing about how close they are.
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
$endgroup$
To attempt to answer the question you asked in the comments, the answer is "no". The determinant is, roughly, a measure of how close a square matrix is to being singular, but doesn't really help us to tell how close the columns are to each other.
Take for example, the matrices
$$begin{pmatrix} 1 & 0 \ 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 \ 1 & 0 end{pmatrix}.$$
They have the same columns, but their determinants differ by $2$. If order matters here, take instead,
$$begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 end{pmatrix} text{ and } begin{pmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ 1 & 0 & 0 end{pmatrix},$$
which both have determinant $1$, have the same columns, but in a totally different order.
Another example:
$$begin{pmatrix} 0 & 0 \ 0 & 0 end{pmatrix} text{ and } begin{pmatrix} 100 & 100 \ -100 & -100 end{pmatrix}$$
have the same determinant $0$, since both are singular. The columns, as you can see, are very different. The determinant of their difference is also $0$, in case you were thinking $det(A - B) = 0$ might yield some measure of similarity between the columns.
It is true, however, that the determinant is continuous. Small changes in a matrix will result in small (possibly null) changes in the determinant. But knowing the determinant of two matrices tells you next to nothing about how close they are.
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
edited Jan 27 at 12:51
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
answered Jan 26 at 1:06
Theo BenditTheo Bendit
19.5k12353
19.5k12353
$begingroup$
Great Explanation, thanks!
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– chancar
Jan 26 at 9:14
$begingroup$
@chancar Then you should upvote as well, IMHO ...
$endgroup$
– Hanno
Jan 26 at 9:27
$begingroup$
I'm new and not allowed to upvote yet. But the north remembers...
$endgroup$
– chancar
Jan 26 at 18:19
$begingroup$
I see, not being aware of it. .. At least some push (+5) towards the rep frontier is possible $ddotsmile$
$endgroup$
– Hanno
Jan 27 at 11:14
add a comment |
$begingroup$
Great Explanation, thanks!
$endgroup$
– chancar
Jan 26 at 9:14
$begingroup$
@chancar Then you should upvote as well, IMHO ...
$endgroup$
– Hanno
Jan 26 at 9:27
$begingroup$
I'm new and not allowed to upvote yet. But the north remembers...
$endgroup$
– chancar
Jan 26 at 18:19
$begingroup$
I see, not being aware of it. .. At least some push (+5) towards the rep frontier is possible $ddotsmile$
$endgroup$
– Hanno
Jan 27 at 11:14
$begingroup$
Great Explanation, thanks!
$endgroup$
– chancar
Jan 26 at 9:14
$begingroup$
Great Explanation, thanks!
$endgroup$
– chancar
Jan 26 at 9:14
$begingroup$
@chancar Then you should upvote as well, IMHO ...
$endgroup$
– Hanno
Jan 26 at 9:27
$begingroup$
@chancar Then you should upvote as well, IMHO ...
$endgroup$
– Hanno
Jan 26 at 9:27
$begingroup$
I'm new and not allowed to upvote yet. But the north remembers...
$endgroup$
– chancar
Jan 26 at 18:19
$begingroup$
I'm new and not allowed to upvote yet. But the north remembers...
$endgroup$
– chancar
Jan 26 at 18:19
$begingroup$
I see, not being aware of it. .. At least some push (+5) towards the rep frontier is possible $ddotsmile$
$endgroup$
– Hanno
Jan 27 at 11:14
$begingroup$
I see, not being aware of it. .. At least some push (+5) towards the rep frontier is possible $ddotsmile$
$endgroup$
– Hanno
Jan 27 at 11:14
add a comment |
2
$begingroup$
What you are asking is hard to grasp, IMO. I'd like to draw your attention to this ripened post on determinants
$endgroup$
– Hanno
Jan 25 at 23:28
1
$begingroup$
Thanks for the input. Basically I'm asking if the determinant depends on how different the columns of a matrix are form each other.
$endgroup$
– chancar
Jan 25 at 23:46
1
$begingroup$
In order to answer this, you should explain/define, what " ... how different the columns of a matrix are form [typo?] each other" does signify, so that one can work on and with it.
$endgroup$
– Hanno
Jan 25 at 23:58