I do not understand why elementary operations are made to find Eigen values. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Invariance of traceFind Eigen values of given matrix with nonfactorable polynomialFind eigen value without expand and factorCompute $f_A(lambda)$ without factoring cubic polynomial?Linear Algebra - seemingly incorrect result when looking for a basisJordan Canonical form questionJordan form of the matrix $left(beginsmallmatrix 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 1 & 1 endsmallmatrixright)$Relation of eigenvector, eigenvalue and diagonalizationFind the eigen values and the associated eigen vectors of the matrix,$~~~~$ $A=beginpmatrix 3~~1~~1\ 2~~4~~2\ 1~~1~~3 endpmatrix$Linear Algebra: $2times 2$ matrix yields only 1 eigenvalue
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I do not understand why elementary operations are made to find Eigen values.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Invariance of traceFind Eigen values of given matrix with nonfactorable polynomialFind eigen value without expand and factorCompute $f_A(lambda)$ without factoring cubic polynomial?Linear Algebra - seemingly incorrect result when looking for a basisJordan Canonical form questionJordan form of the matrix $left(beginsmallmatrix 1 & 1 & 0 \ 0 & 1 & 0 \ 0 & 1 & 1 endsmallmatrixright)$Relation of eigenvector, eigenvalue and diagonalizationFind the eigen values and the associated eigen vectors of the matrix,$~~~~$ $A=beginpmatrix 3~~1~~1\ 2~~4~~2\ 1~~1~~3 endpmatrix$Linear Algebra: $2times 2$ matrix yields only 1 eigenvalue
$begingroup$
We where just introduced to eigen values and the following bugs me.
Polynomial(lambda) = det(A - lambda I) =
beginpmatrix
1 - lambda & 1 & 1\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they do Line 1 - line 2 = beginpmatrix
-lambda & lambda & 0\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they put lambda out of the matrix and compute the det.
Now what I don't understand isn't how to find the eigen value but more about what is the objective of doing the elementary operation. Is there a common pattern like the gauss-jordan REF.
linear-algebra
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
$endgroup$
add a comment |
$begingroup$
We where just introduced to eigen values and the following bugs me.
Polynomial(lambda) = det(A - lambda I) =
beginpmatrix
1 - lambda & 1 & 1\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they do Line 1 - line 2 = beginpmatrix
-lambda & lambda & 0\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they put lambda out of the matrix and compute the det.
Now what I don't understand isn't how to find the eigen value but more about what is the objective of doing the elementary operation. Is there a common pattern like the gauss-jordan REF.
linear-algebra
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
$endgroup$
1
$begingroup$
The objective of doing the elementary operations is to make the matrix upper triangle. Which makes the determinant equal to the product of the diagonal entries which is convenient.
$endgroup$
– Justin Stevenson
Apr 2 at 0:41
$begingroup$
it is not really needed. There is that "rule of Sarrus" that allows us to write out a 3 by 3 determinant. However, when the entries have the extra $lambda,$ I guess it gets long, so they gave an alternative
$endgroup$
– Will Jagy
Apr 2 at 0:42
add a comment |
$begingroup$
We where just introduced to eigen values and the following bugs me.
Polynomial(lambda) = det(A - lambda I) =
beginpmatrix
1 - lambda & 1 & 1\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they do Line 1 - line 2 = beginpmatrix
-lambda & lambda & 0\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they put lambda out of the matrix and compute the det.
Now what I don't understand isn't how to find the eigen value but more about what is the objective of doing the elementary operation. Is there a common pattern like the gauss-jordan REF.
linear-algebra
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
$endgroup$
We where just introduced to eigen values and the following bugs me.
Polynomial(lambda) = det(A - lambda I) =
beginpmatrix
1 - lambda & 1 & 1\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they do Line 1 - line 2 = beginpmatrix
-lambda & lambda & 0\
1 & 1-lambda & 1\
1& 1& 1-lambda
endpmatrix
Then they put lambda out of the matrix and compute the det.
Now what I don't understand isn't how to find the eigen value but more about what is the objective of doing the elementary operation. Is there a common pattern like the gauss-jordan REF.
linear-algebra
linear-algebra
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
asked Apr 2 at 0:27
Dr.StoneDr.Stone
516
516
1
$begingroup$
The objective of doing the elementary operations is to make the matrix upper triangle. Which makes the determinant equal to the product of the diagonal entries which is convenient.
$endgroup$
– Justin Stevenson
Apr 2 at 0:41
$begingroup$
it is not really needed. There is that "rule of Sarrus" that allows us to write out a 3 by 3 determinant. However, when the entries have the extra $lambda,$ I guess it gets long, so they gave an alternative
$endgroup$
– Will Jagy
Apr 2 at 0:42
add a comment |
1
$begingroup$
The objective of doing the elementary operations is to make the matrix upper triangle. Which makes the determinant equal to the product of the diagonal entries which is convenient.
$endgroup$
– Justin Stevenson
Apr 2 at 0:41
$begingroup$
it is not really needed. There is that "rule of Sarrus" that allows us to write out a 3 by 3 determinant. However, when the entries have the extra $lambda,$ I guess it gets long, so they gave an alternative
$endgroup$
– Will Jagy
Apr 2 at 0:42
1
1
$begingroup$
The objective of doing the elementary operations is to make the matrix upper triangle. Which makes the determinant equal to the product of the diagonal entries which is convenient.
$endgroup$
– Justin Stevenson
Apr 2 at 0:41
$begingroup$
The objective of doing the elementary operations is to make the matrix upper triangle. Which makes the determinant equal to the product of the diagonal entries which is convenient.
$endgroup$
– Justin Stevenson
Apr 2 at 0:41
$begingroup$
it is not really needed. There is that "rule of Sarrus" that allows us to write out a 3 by 3 determinant. However, when the entries have the extra $lambda,$ I guess it gets long, so they gave an alternative
$endgroup$
– Will Jagy
Apr 2 at 0:42
$begingroup$
it is not really needed. There is that "rule of Sarrus" that allows us to write out a 3 by 3 determinant. However, when the entries have the extra $lambda,$ I guess it gets long, so they gave an alternative
$endgroup$
– Will Jagy
Apr 2 at 0:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Recall that performing elementary row oeperations does not change the (magnitude of) determinant. In other words, $|det A|=|det A'|$ if $A'$ is the result of elementary row operations on $A$. There is of course no need to perform these operations, since computing directly will lead to the correct result, but doing so would make the determinant easier to compute. All of this has nothing to do with eigenvalues per se, just in computing determinants (easily).
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
$endgroup$
add a comment |
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$begingroup$
Recall that performing elementary row oeperations does not change the (magnitude of) determinant. In other words, $|det A|=|det A'|$ if $A'$ is the result of elementary row operations on $A$. There is of course no need to perform these operations, since computing directly will lead to the correct result, but doing so would make the determinant easier to compute. All of this has nothing to do with eigenvalues per se, just in computing determinants (easily).
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
$endgroup$
add a comment |
$begingroup$
Recall that performing elementary row oeperations does not change the (magnitude of) determinant. In other words, $|det A|=|det A'|$ if $A'$ is the result of elementary row operations on $A$. There is of course no need to perform these operations, since computing directly will lead to the correct result, but doing so would make the determinant easier to compute. All of this has nothing to do with eigenvalues per se, just in computing determinants (easily).
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
$endgroup$
add a comment |
$begingroup$
Recall that performing elementary row oeperations does not change the (magnitude of) determinant. In other words, $|det A|=|det A'|$ if $A'$ is the result of elementary row operations on $A$. There is of course no need to perform these operations, since computing directly will lead to the correct result, but doing so would make the determinant easier to compute. All of this has nothing to do with eigenvalues per se, just in computing determinants (easily).
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
$endgroup$
Recall that performing elementary row oeperations does not change the (magnitude of) determinant. In other words, $|det A|=|det A'|$ if $A'$ is the result of elementary row operations on $A$. There is of course no need to perform these operations, since computing directly will lead to the correct result, but doing so would make the determinant easier to compute. All of this has nothing to do with eigenvalues per se, just in computing determinants (easily).
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
answered Apr 2 at 1:20
YiFanYiFan
5,5652829
5,5652829
add a comment |
add a comment |
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The objective of doing the elementary operations is to make the matrix upper triangle. Which makes the determinant equal to the product of the diagonal entries which is convenient.
$endgroup$
– Justin Stevenson
Apr 2 at 0:41
$begingroup$
it is not really needed. There is that "rule of Sarrus" that allows us to write out a 3 by 3 determinant. However, when the entries have the extra $lambda,$ I guess it gets long, so they gave an alternative
$endgroup$
– Will Jagy
Apr 2 at 0:42