<p>Yes, because (equivalently) determinants can be computed via row reduction.  Those steps exploit only the bilinearity and the antisymmetry of the determinant.</p>

CC BY-SA 2.5
2010-11-05T02:54:53.667Z
2010-11-05T02:54:53.667Z
whuber
<p>If one calculates the product of diagonal elements of the $U$ matrix in a $LUP$ factorization of a given matrix $A$, one can calculate the determinant of $A$. Also it is known that $LUP$ factorization can be reduced to a series of matrix multiplications which are bilinear forms. Does that mean determinant calculation (multilinear form) can be reduced to a bilinear calculation?</p>

CC BY-SA 2.5
2010-11-05T02:16:23.477Z
Sorry for my ignorance, but what is bilinear calculation?  (I know what a bilinear form is.)
Tsuyoshi Ito
CC BY-SA 2.5
2010-11-05T02:35:41.443Z
Hi Tusyoshi I was implying the same.
Turbo
CC BY-SA 2.5
2010-11-05T03:06:21.827Z
Obviously the deterinant is not a bilinear form in the elements of the matrix (no matter how you divide the elements into two sets), if that is what you are asking.
Tsuyoshi Ito
CC BY-SA 2.5
2010-11-05T03:24:55.987Z
Hi Tsuyoshi. Yes I know that determinant is a multilinear form. However using only a sequence of matrix multiplications one can reduce to $LUP$. From which taking the product of diagonals of $U$ suffice to find the determinant. Multiplication is a bilinear map. I was curious then may be the determinant is actually a product of bilinear forms.
Turbo
CC BY-SA 2.5
2010-11-05T03:25:41.413Z
I feel this is just a tautological question.
Turbo
<ds.algorithms><linear-algebra>
Determinants and bilinear forms
Turbo
# Determinants and bilinear forms

(Asked by: username_0 on 2010-11-04)

If one calculates the product of diagonal elements of the $U$ matrix in a $LUP$ factorization of a given matrix $A$, one can calculate the determinant of $A$. Also it is known that $LUP$ factorization can be reduced to a series of matrix multiplications which are bilinear forms. Does that mean determinant calculation (multilinear form) can be reduced to a bilinear calculation?

**Tags:** ds.algorithms, linear-algebra
> **Comments:**
> **username_1**: Sorry for my ignorance, but what is bilinear calculation?  (I know what a bilinear form is.)  
> **username_0**: Hi Tusyoshi I was implying the same.  
> **username_1**: Obviously the deterinant is not a bilinear form in the elements of the matrix (no matter how you divide the elements into two sets), if that is what you are asking.  
> **username_0**: Hi Tsuyoshi. Yes I know that determinant is a multilinear form. However using only a sequence of matrix multiplications one can reduce to $LUP$. From which taking the product of diagonals of $U$ suffice to find the determinant. Multiplication is a bilinear map. I was curious then may be the determinant is actually a product of bilinear forms.  
> **username_0**: I feel this is just a tautological question.  

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## 1 answers

### Answer by username_2 on 2010-11-05. Score: 1

Yes, because (equivalently) determinants can be computed via row reduction. Those steps exploit only the bilinearity and the antisymmetry of the determinant.


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