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Why do we use determinant $det A$ to specify linearly dependence? [on hold]
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Why this works? And does it effect the determinant test(dependence test) writing place the coordinant to the matrix that we are taking determinant of?
linear-algebra
edited Aug 2 at 17:52
copper.hat
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asked Aug 2 at 16:41
Jale'de jale uff ne jale
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put on hold as off-topic by TheGeekGreek, John Ma, Isaac Browne, Adrian Keister, Leucippus Aug 3 at 0:04
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up vote
-1
down vote
favorite
Why this works? And does it effect the determinant test(dependence test) writing place the coordinant to the matrix that we are taking determinant of?
linear-algebra
edited Aug 2 at 17:52
copper.hat
122k557155
asked Aug 2 at 16:41
Jale'de jale uff ne jale
203
put on hold as off-topic by TheGeekGreek, John Ma, Isaac Browne, Adrian Keister, Leucippus Aug 3 at 0:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, John Ma, Isaac Browne, Adrian Keister, Leucippus
1
See: math.stackexchange.com/questions/79356/…
– Moo
Aug 2 at 16:44
How can one explain this idea to a high school student
– Jale'de jale uff ne jale
Aug 2 at 16:44
wow why - votes?
– Jale'de jale uff ne jale
Aug 2 at 16:48
3
Downvotes are not a personal attack.
– saulspatz
Aug 2 at 16:59
1
@saulspatz: They are negative in some sense. Certainly not an attack, but they carry negative sentiment, generally without explaining why. Thankfully MSE is not quite as judgmental as, say SO.
– copper.hat
Aug 2 at 17:52
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show 4 more comments
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Why this works? And does it effect the determinant test(dependence test) writing place the coordinant to the matrix that we are taking determinant of?
linear-algebra
edited Aug 2 at 17:52
copper.hat
122k557155
asked Aug 2 at 16:41
Jale'de jale uff ne jale
203
Why this works? And does it effect the determinant test(dependence test) writing place the coordinant to the matrix that we are taking determinant of?
linear-algebra
edited Aug 2 at 17:52
copper.hat
122k557155
asked Aug 2 at 16:41
Jale'de jale uff ne jale
203
edited Aug 2 at 17:52
copper.hat
122k557155
edited Aug 2 at 17:52
copper.hat
122k557155
edited Aug 2 at 17:52
copper.hat
122k557155
122k557155
asked Aug 2 at 16:41
Jale'de jale uff ne jale
203
asked Aug 2 at 16:41
Jale'de jale uff ne jale
203
asked Aug 2 at 16:41
Jale'de jale uff ne jale
203
203
put on hold as off-topic by TheGeekGreek, John Ma, Isaac Browne, Adrian Keister, Leucippus Aug 3 at 0:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, John Ma, Isaac Browne, Adrian Keister, Leucippus
put on hold as off-topic by TheGeekGreek, John Ma, Isaac Browne, Adrian Keister, Leucippus Aug 3 at 0:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheGeekGreek, John Ma, Isaac Browne, Adrian Keister, Leucippus
1
See: math.stackexchange.com/questions/79356/…
– Moo
Aug 2 at 16:44
How can one explain this idea to a high school student
– Jale'de jale uff ne jale
Aug 2 at 16:44
wow why - votes?
– Jale'de jale uff ne jale
Aug 2 at 16:48
3
Downvotes are not a personal attack.
– saulspatz
Aug 2 at 16:59
1
@saulspatz: They are negative in some sense. Certainly not an attack, but they carry negative sentiment, generally without explaining why. Thankfully MSE is not quite as judgmental as, say SO.
– copper.hat
Aug 2 at 17:52
 |Â
show 4 more comments
1
See: math.stackexchange.com/questions/79356/…
– Moo
Aug 2 at 16:44
How can one explain this idea to a high school student
– Jale'de jale uff ne jale
Aug 2 at 16:44
wow why - votes?
– Jale'de jale uff ne jale
Aug 2 at 16:48
3
Downvotes are not a personal attack.
– saulspatz
Aug 2 at 16:59
1
@saulspatz: They are negative in some sense. Certainly not an attack, but they carry negative sentiment, generally without explaining why. Thankfully MSE is not quite as judgmental as, say SO.
– copper.hat
Aug 2 at 17:52
1
1
See: math.stackexchange.com/questions/79356/…
– Moo
Aug 2 at 16:44
See: math.stackexchange.com/questions/79356/…
– Moo
Aug 2 at 16:44
How can one explain this idea to a high school student
– Jale'de jale uff ne jale
Aug 2 at 16:44
How can one explain this idea to a high school student
– Jale'de jale uff ne jale
Aug 2 at 16:44
wow why - votes?
– Jale'de jale uff ne jale
Aug 2 at 16:48
wow why - votes?
– Jale'de jale uff ne jale
Aug 2 at 16:48
3
3
Downvotes are not a personal attack.
– saulspatz
Aug 2 at 16:59
Downvotes are not a personal attack.
– saulspatz
Aug 2 at 16:59
1
1
@saulspatz: They are negative in some sense. Certainly not an attack, but they carry negative sentiment, generally without explaining why. Thankfully MSE is not quite as judgmental as, say SO.
– copper.hat
Aug 2 at 17:52
@saulspatz: They are negative in some sense. Certainly not an attack, but they carry negative sentiment, generally without explaining why. Thankfully MSE is not quite as judgmental as, say SO.
– copper.hat
Aug 2 at 17:52
 |Â
show 4 more comments
1 Answer
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If you view $det$ as a function of the columns, it is linear in each column separately. Let me write $det A = d(a_1,...,a_n)$.
If two of the $a_k$ are the same, then
$d(...,a_k,...,a_k,...) = 0$.
Suppose, to be concrete, the last column can be written as a linear combination of the others, $a_n = sum_k<n lambda_k a_k$.
Then by linearity in the last column we have $d(a_1,...,a_n) = sum_k<n lambda_k d(a_1,...,a_k)$ and since each element of the sum repeats one column
we have $d(a_1,...,a_k)=0$ and so $d(a_1,...,a_n) = 0$.
Hence if the columns are linearly dependent, we have $det A = 0$.
Now suppose that you know that the columns $a_k$ are linearly independent. My
proof here is not as satisfactory as it relies on Gaussian elimination and the
fact that $det (AB) = det A det B$.
By post multiplying $A$ by appropriate matrices (elementary column operations, whose determinants are
non zero) we can obtain a column echelon form, that is, we can reduce $A$ to
an upper triangular form $T$ whose diagonal elements are all non zero. That is,
$A C_1cdots C_p = T$. Since $det C_k neq 0$ and $det T neq 0$ we see that $det A = det T det C_1 cdots det C_p neq 0$.
Hence $det A neq 0$ iff the columns are linearly independent.
answered Aug 2 at 18:27
copper.hat
122k557155
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
up vote
1
down vote
If you view $det$ as a function of the columns, it is linear in each column separately. Let me write $det A = d(a_1,...,a_n)$.
If two of the $a_k$ are the same, then
$d(...,a_k,...,a_k,...) = 0$.
Suppose, to be concrete, the last column can be written as a linear combination of the others, $a_n = sum_k<n lambda_k a_k$.
Then by linearity in the last column we have $d(a_1,...,a_n) = sum_k<n lambda_k d(a_1,...,a_k)$ and since each element of the sum repeats one column
we have $d(a_1,...,a_k)=0$ and so $d(a_1,...,a_n) = 0$.
Hence if the columns are linearly dependent, we have $det A = 0$.
Now suppose that you know that the columns $a_k$ are linearly independent. My
proof here is not as satisfactory as it relies on Gaussian elimination and the
fact that $det (AB) = det A det B$.
By post multiplying $A$ by appropriate matrices (elementary column operations, whose determinants are
non zero) we can obtain a column echelon form, that is, we can reduce $A$ to
an upper triangular form $T$ whose diagonal elements are all non zero. That is,
$A C_1cdots C_p = T$. Since $det C_k neq 0$ and $det T neq 0$ we see that $det A = det T det C_1 cdots det C_p neq 0$.
Hence $det A neq 0$ iff the columns are linearly independent.
answered Aug 2 at 18:27
copper.hat
122k557155
add a comment |Â
up vote
1
down vote
If you view $det$ as a function of the columns, it is linear in each column separately. Let me write $det A = d(a_1,...,a_n)$.
If two of the $a_k$ are the same, then
$d(...,a_k,...,a_k,...) = 0$.
Suppose, to be concrete, the last column can be written as a linear combination of the others, $a_n = sum_k<n lambda_k a_k$.
Then by linearity in the last column we have $d(a_1,...,a_n) = sum_k<n lambda_k d(a_1,...,a_k)$ and since each element of the sum repeats one column
we have $d(a_1,...,a_k)=0$ and so $d(a_1,...,a_n) = 0$.
Hence if the columns are linearly dependent, we have $det A = 0$.
Now suppose that you know that the columns $a_k$ are linearly independent. My
proof here is not as satisfactory as it relies on Gaussian elimination and the
fact that $det (AB) = det A det B$.
By post multiplying $A$ by appropriate matrices (elementary column operations, whose determinants are
non zero) we can obtain a column echelon form, that is, we can reduce $A$ to
an upper triangular form $T$ whose diagonal elements are all non zero. That is,
$A C_1cdots C_p = T$. Since $det C_k neq 0$ and $det T neq 0$ we see that $det A = det T det C_1 cdots det C_p neq 0$.
Hence $det A neq 0$ iff the columns are linearly independent.
answered Aug 2 at 18:27
copper.hat
122k557155
add a comment |Â
up vote
1
down vote
up vote
1
down vote
If you view $det$ as a function of the columns, it is linear in each column separately. Let me write $det A = d(a_1,...,a_n)$.
If two of the $a_k$ are the same, then
$d(...,a_k,...,a_k,...) = 0$.
Suppose, to be concrete, the last column can be written as a linear combination of the others, $a_n = sum_k<n lambda_k a_k$.
Then by linearity in the last column we have $d(a_1,...,a_n) = sum_k<n lambda_k d(a_1,...,a_k)$ and since each element of the sum repeats one column
we have $d(a_1,...,a_k)=0$ and so $d(a_1,...,a_n) = 0$.
Hence if the columns are linearly dependent, we have $det A = 0$.
Now suppose that you know that the columns $a_k$ are linearly independent. My
proof here is not as satisfactory as it relies on Gaussian elimination and the
fact that $det (AB) = det A det B$.
By post multiplying $A$ by appropriate matrices (elementary column operations, whose determinants are
non zero) we can obtain a column echelon form, that is, we can reduce $A$ to
an upper triangular form $T$ whose diagonal elements are all non zero. That is,
$A C_1cdots C_p = T$. Since $det C_k neq 0$ and $det T neq 0$ we see that $det A = det T det C_1 cdots det C_p neq 0$.
Hence $det A neq 0$ iff the columns are linearly independent.
answered Aug 2 at 18:27
copper.hat
122k557155
If you view $det$ as a function of the columns, it is linear in each column separately. Let me write $det A = d(a_1,...,a_n)$.
If two of the $a_k$ are the same, then
$d(...,a_k,...,a_k,...) = 0$.
Suppose, to be concrete, the last column can be written as a linear combination of the others, $a_n = sum_k<n lambda_k a_k$.
Then by linearity in the last column we have $d(a_1,...,a_n) = sum_k<n lambda_k d(a_1,...,a_k)$ and since each element of the sum repeats one column
we have $d(a_1,...,a_k)=0$ and so $d(a_1,...,a_n) = 0$.
Hence if the columns are linearly dependent, we have $det A = 0$.
Now suppose that you know that the columns $a_k$ are linearly independent. My
proof here is not as satisfactory as it relies on Gaussian elimination and the
fact that $det (AB) = det A det B$.
By post multiplying $A$ by appropriate matrices (elementary column operations, whose determinants are
non zero) we can obtain a column echelon form, that is, we can reduce $A$ to
an upper triangular form $T$ whose diagonal elements are all non zero. That is,
$A C_1cdots C_p = T$. Since $det C_k neq 0$ and $det T neq 0$ we see that $det A = det T det C_1 cdots det C_p neq 0$.
Hence $det A neq 0$ iff the columns are linearly independent.
answered Aug 2 at 18:27
copper.hat
122k557155
answered Aug 2 at 18:27
copper.hat
122k557155
answered Aug 2 at 18:27
copper.hat
122k557155
answered Aug 2 at 18:27
copper.hat
122k557155
122k557155
add a comment |Â
add a comment |Â
1
See: math.stackexchange.com/questions/79356/…
– Moo
Aug 2 at 16:44
How can one explain this idea to a high school student
– Jale'de jale uff ne jale
Aug 2 at 16:44
wow why - votes?
– Jale'de jale uff ne jale
Aug 2 at 16:48
3
Downvotes are not a personal attack.
– saulspatz
Aug 2 at 16:59
1
@saulspatz: They are negative in some sense. Certainly not an attack, but they carry negative sentiment, generally without explaining why. Thankfully MSE is not quite as judgmental as, say SO.
– copper.hat
Aug 2 at 17:52