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Show that f(A)=Det(A), where A is a matrix (nxn), is a quadratic form of fourth variables.
Asked
May 13 '19 at 21:16
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May 13 '19 at 21:37
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I tried with induction, but it doesn't works.
determinant
edited May 13 '19 at 21:37
Arthur
199,419
asked May 13 '19 at 21:16
Jordan Hernandez
1
For $2\times 2$ matrices it works.
–
logarithm
May 13 '19 at 21:18
No, matrices n x n.
–
Jordan Hernandez
May 13 '19 at 21:19
Note that $f(rA)=r^nf(A)$. This is, $f$ is homogeneous of degree $n$. Quadratic forms are homogeneous of degree $2$.
–
logarithm
May 13 '19 at 21:20
What are the variables?
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herb steinberg
May 13 '19 at 21:20
Now, there is a image with the question.I do not know if it's well raised
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Jordan Hernandez
May 13 '19 at 21:24
Where it says $n$ it should say $2$. The matrix $S$ will have a $1/2$ at positions $(1,2)$ and $(2,1)$ and a $-1/2$ at positions $(3,4)$ and $(4,3)$.
–
logarithm
May 13 '19 at 21:27
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