Why is the volume of a parallelepiped linear in each row in the matrix representation?
In this question, the determinant of a matrix is explained to be a measure of the volume of a parallelepiped formed by using the columns in a matrix as vectors.
It is also noted that the determinant is linear in each row. We can represent this geometrically as the volume of a parallelpiped being the sum of the volumes of two other ones. Is there a geometric proof that the volume corresponding to these matrices is linear in this manner?
matrices determinant volume
asked Nov 21 at 20:52
Casebash
5,63834070
add a comment |
In this question, the determinant of a matrix is explained to be a measure of the volume of a parallelepiped formed by using the columns in a matrix as vectors.
It is also noted that the determinant is linear in each row. We can represent this geometrically as the volume of a parallelpiped being the sum of the volumes of two other ones. Is there a geometric proof that the volume corresponding to these matrices is linear in this manner?
matrices determinant volume
asked Nov 21 at 20:52
Casebash
5,63834070
The determinant, by cinstruction, ismultilinear in its rows (or its columns as well), independently of any geometric consideration.
– Bernard
Nov 21 at 20:57
@Bernard: Sure, but it seems like there should be a geometric representation. Maybe I need to reword the question.
– Casebash
Nov 21 at 21:16
1
I think there is a geometric representation. Note that the determinant does not change upon adding multiples of one row/column to another. This corresponds to a shear mapping of the paralelepiped, so the volume does not change under this operation. Then, you use the Gram-Schmidt process to express the original vectors which form the base,width and height (etc if there are more than 3 dimensions) in terms of a determinant which results from applying the operation of adding multiples of one row/column to another.
– Matematleta
Nov 21 at 21:55
add a comment |
In this question, the determinant of a matrix is explained to be a measure of the volume of a parallelepiped formed by using the columns in a matrix as vectors.
It is also noted that the determinant is linear in each row. We can represent this geometrically as the volume of a parallelpiped being the sum of the volumes of two other ones. Is there a geometric proof that the volume corresponding to these matrices is linear in this manner?
matrices determinant volume
asked Nov 21 at 20:52
Casebash
5,63834070
In this question, the determinant of a matrix is explained to be a measure of the volume of a parallelepiped formed by using the columns in a matrix as vectors.
It is also noted that the determinant is linear in each row. We can represent this geometrically as the volume of a parallelpiped being the sum of the volumes of two other ones. Is there a geometric proof that the volume corresponding to these matrices is linear in this manner?
matrices determinant volume
matrices determinant volume
asked Nov 21 at 20:52
Casebash
5,63834070
asked Nov 21 at 20:52
Casebash
5,63834070
edited Nov 21 at 21:18
asked Nov 21 at 20:52
Casebash
5,63834070
asked Nov 21 at 20:52
Casebash
5,63834070
asked Nov 21 at 20:52
Casebash
5,63834070
5,63834070
The determinant, by cinstruction, ismultilinear in its rows (or its columns as well), independently of any geometric consideration.
– Bernard
Nov 21 at 20:57
@Bernard: Sure, but it seems like there should be a geometric representation. Maybe I need to reword the question.
– Casebash
Nov 21 at 21:16
1
I think there is a geometric representation. Note that the determinant does not change upon adding multiples of one row/column to another. This corresponds to a shear mapping of the paralelepiped, so the volume does not change under this operation. Then, you use the Gram-Schmidt process to express the original vectors which form the base,width and height (etc if there are more than 3 dimensions) in terms of a determinant which results from applying the operation of adding multiples of one row/column to another.
– Matematleta
Nov 21 at 21:55
add a comment |
The determinant, by cinstruction, ismultilinear in its rows (or its columns as well), independently of any geometric consideration.
– Bernard
Nov 21 at 20:57
@Bernard: Sure, but it seems like there should be a geometric representation. Maybe I need to reword the question.
– Casebash
Nov 21 at 21:16
1
I think there is a geometric representation. Note that the determinant does not change upon adding multiples of one row/column to another. This corresponds to a shear mapping of the paralelepiped, so the volume does not change under this operation. Then, you use the Gram-Schmidt process to express the original vectors which form the base,width and height (etc if there are more than 3 dimensions) in terms of a determinant which results from applying the operation of adding multiples of one row/column to another.
– Matematleta
Nov 21 at 21:55
The determinant, by cinstruction, ismultilinear in its rows (or its columns as well), independently of any geometric consideration.
– Bernard
Nov 21 at 20:57
The determinant, by cinstruction, ismultilinear in its rows (or its columns as well), independently of any geometric consideration.
– Bernard
Nov 21 at 20:57
@Bernard: Sure, but it seems like there should be a geometric representation. Maybe I need to reword the question.
– Casebash
Nov 21 at 21:16
@Bernard: Sure, but it seems like there should be a geometric representation. Maybe I need to reword the question.
– Casebash
Nov 21 at 21:16
1
1
I think there is a geometric representation. Note that the determinant does not change upon adding multiples of one row/column to another. This corresponds to a shear mapping of the paralelepiped, so the volume does not change under this operation. Then, you use the Gram-Schmidt process to express the original vectors which form the base,width and height (etc if there are more than 3 dimensions) in terms of a determinant which results from applying the operation of adding multiples of one row/column to another.
– Matematleta
Nov 21 at 21:55
I think there is a geometric representation. Note that the determinant does not change upon adding multiples of one row/column to another. This corresponds to a shear mapping of the paralelepiped, so the volume does not change under this operation. Then, you use the Gram-Schmidt process to express the original vectors which form the base,width and height (etc if there are more than 3 dimensions) in terms of a determinant which results from applying the operation of adding multiples of one row/column to another.
– Matematleta
Nov 21 at 21:55
add a comment |
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The determinant, by cinstruction, ismultilinear in its rows (or its columns as well), independently of any geometric consideration.
– Bernard
Nov 21 at 20:57
@Bernard: Sure, but it seems like there should be a geometric representation. Maybe I need to reword the question.
– Casebash
Nov 21 at 21:16
1
I think there is a geometric representation. Note that the determinant does not change upon adding multiples of one row/column to another. This corresponds to a shear mapping of the paralelepiped, so the volume does not change under this operation. Then, you use the Gram-Schmidt process to express the original vectors which form the base,width and height (etc if there are more than 3 dimensions) in terms of a determinant which results from applying the operation of adding multiples of one row/column to another.
– Matematleta
Nov 21 at 21:55