Question about the determinant of an upper triangular matrix Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Getting the determinant value of the original matrix from its upper triangular matrixHow is diagonal matrices a subspace of upper triangular matrices?Upper Triangular Matrix DefinitionTransforming a matrix to upper triangular formWhat is the transformation matrix “by definition”?How to derive the Vandermonde Determinant?Various matrix manipulations effect on determinantDeterminant of Partitioned Upper Triangular MatrixHow would I answer the following question about the determinant of a matrix?Determinant of Triangular Matrix
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Question about the determinant of an upper triangular matrix
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Getting the determinant value of the original matrix from its upper triangular matrixHow is diagonal matrices a subspace of upper triangular matrices?Upper Triangular Matrix DefinitionTransforming a matrix to upper triangular formWhat is the transformation matrix “by definition”?How to derive the Vandermonde Determinant?Various matrix manipulations effect on determinantDeterminant of Partitioned Upper Triangular MatrixHow would I answer the following question about the determinant of a matrix?Determinant of Triangular Matrix
$begingroup$
I was watching an online lecture on the properties of determinants and at 27:00 in this lecture: https://www.youtube.com/watch?v=srxexLishgY&t=4s, the professor did a step which I don't understand.
If $A = beginpmatrixd1 & 0 & 0\ 0 & d2 & 0\ 0 & 0 & d3endpmatrix $, then according to the lecturer, $A=d1 *d2 *d3 beginpmatrix1 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 1endpmatrix$, this was then used to show that the determinant of an upper triangular matrix can be written as the product of the leading diagonal
. However I thought if a matrix is multiplied by a number, all elements of the matrix are multiplied by it. Is this just wrong?
linear-algebra matrices determinant
$endgroup$
add a comment |
$begingroup$
I was watching an online lecture on the properties of determinants and at 27:00 in this lecture: https://www.youtube.com/watch?v=srxexLishgY&t=4s, the professor did a step which I don't understand.
If $A = beginpmatrixd1 & 0 & 0\ 0 & d2 & 0\ 0 & 0 & d3endpmatrix $, then according to the lecturer, $A=d1 *d2 *d3 beginpmatrix1 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 1endpmatrix$, this was then used to show that the determinant of an upper triangular matrix can be written as the product of the leading diagonal
. However I thought if a matrix is multiplied by a number, all elements of the matrix are multiplied by it. Is this just wrong?
linear-algebra matrices determinant
$endgroup$
add a comment |
$begingroup$
I was watching an online lecture on the properties of determinants and at 27:00 in this lecture: https://www.youtube.com/watch?v=srxexLishgY&t=4s, the professor did a step which I don't understand.
If $A = beginpmatrixd1 & 0 & 0\ 0 & d2 & 0\ 0 & 0 & d3endpmatrix $, then according to the lecturer, $A=d1 *d2 *d3 beginpmatrix1 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 1endpmatrix$, this was then used to show that the determinant of an upper triangular matrix can be written as the product of the leading diagonal
. However I thought if a matrix is multiplied by a number, all elements of the matrix are multiplied by it. Is this just wrong?
linear-algebra matrices determinant
$endgroup$
I was watching an online lecture on the properties of determinants and at 27:00 in this lecture: https://www.youtube.com/watch?v=srxexLishgY&t=4s, the professor did a step which I don't understand.
If $A = beginpmatrixd1 & 0 & 0\ 0 & d2 & 0\ 0 & 0 & d3endpmatrix $, then according to the lecturer, $A=d1 *d2 *d3 beginpmatrix1 & 0 & 0\ 0 & 1 & 0\ 0 & 0 & 1endpmatrix$, this was then used to show that the determinant of an upper triangular matrix can be written as the product of the leading diagonal
. However I thought if a matrix is multiplied by a number, all elements of the matrix are multiplied by it. Is this just wrong?
linear-algebra matrices determinant
linear-algebra matrices determinant
asked Apr 2 at 16:37
Vishal JainVishal Jain
62
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$begingroup$
you are right about multiplying a matrix by a constant.
The professor in the video is also right, because he is calculating the determinant of the matrix.
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
you are right about multiplying a matrix by a constant.
The professor in the video is also right, because he is calculating the determinant of the matrix.
$endgroup$
add a comment |
$begingroup$
you are right about multiplying a matrix by a constant.
The professor in the video is also right, because he is calculating the determinant of the matrix.
$endgroup$
add a comment |
$begingroup$
you are right about multiplying a matrix by a constant.
The professor in the video is also right, because he is calculating the determinant of the matrix.
$endgroup$
you are right about multiplying a matrix by a constant.
The professor in the video is also right, because he is calculating the determinant of the matrix.
answered Apr 2 at 16:40
MoonKnightMoonKnight
1,673611
1,673611
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