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I am working in the vector space $\mathbb F^{2\times 2}$, which denotes the space of $2\times 2$ matrices over the field $\mathbb F$ (the characteristic of $\mathbb F$ is not equal to $2$) and I would like to proof that: $\det:A\mapsto\det A$ is a quadratic form. Furthermore, I would like to show that $$\det(A+B) - \det(A) - \det(B) = tr(AB^\#)$$ whereas $tr$ denotes the trace of the matrix and $B^\#$ denotes the cofactor matrix of $B$.

My definition of a quadratic form $q$ is that $q(cx) = c^2q(x)$ and that the map $$(x,y) \mapsto q(x+y)-q(x)-q(y)$$ is a bilinear form. I could easily prove the first criterion, but I am really stuck with the second. I also tried to first show the equality above (the trace equality), but for some reason, for matrices $$A= \begin{pmatrix} a & b\\c & d \end{pmatrix} \qquad B = \begin{pmatrix} w & x\\y & z \end{pmatrix}$$

the left side gives me $az+dw-by-cx$ and the right side leaves me with $az+dw-bx-cy$

I am pretty sure that I computed the determinants and the cofactor matrix correctly, so I am genuinely confused where my mistake lies and I also do not know how to prove the bilinear form criterion. Any help is greatly appreciated!

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  • $\begingroup$ "...in a field $\;\Bbb F^{2\times2}\;$ "....and its characteristic isn't $\;2\;$ ...then what in the world is that field, anyway!? $\endgroup$ – DonAntonio Apr 1 '17 at 9:50
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    $\begingroup$ @DonAntonio I suppose M3xr means the algebra of $2\times 2$ matrices over the field $\Bbb F$, where $\operatorname{char}F\ne 2$ $\endgroup$ – Hagen von Eitzen Apr 1 '17 at 9:52
  • $\begingroup$ updated it now, sorry. $\endgroup$ – AxiomaticApproach Apr 1 '17 at 9:53
  • $\begingroup$ @HagenvonEitzen Yup, that is now edited that way. Thanks. $\endgroup$ – DonAntonio Apr 1 '17 at 9:53
  • $\begingroup$ The determinant of a $\;2\times2\;$ matrix is a homogeneous polynomial of degree two in the matrix's entries and thus a quadratic form. $\endgroup$ – DonAntonio Apr 1 '17 at 9:55
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Your computation of the left hand side, $az+dw-by-cx$ is correct. Note that this sum of products can be interpreted as trace of $$\tag1\begin{pmatrix}az-by&?\\?&dw-cx\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix} \begin{pmatrix}z&-x\\-y&w\end{pmatrix}$$ So maybe you check again the definition of $B^\#$. At any rate, the second factor in $(1)$ is clearly linear in $B$, hence $(1)$ is linear in $A$ for fixed $B$ and linear in $B$ for fixed $A$ - in other words, it is bilinear, as desired.

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  • $\begingroup$ First of all, thank you very much! Maybe it's just me, but I am sure that for $B = \begin{pmatrix} w\ x\\y\ z \end{pmatrix}$ $B^\# = \begin{pmatrix} z\ -y\\-x\ w \end{pmatrix}$. I double checked definitions from various sources but I dont really get it. $\endgroup$ – AxiomaticApproach Apr 1 '17 at 10:08
  • $\begingroup$ @M3xr Yes, I meant that maybe your textbook might have a different definition than most of the rest. Alternatively, they may have accidentally left out a ${}^T$ somewhere in the problem statement. $\endgroup$ – Hagen von Eitzen Apr 1 '17 at 10:15
  • $\begingroup$ okay, I think then they just might have forgotten a transpose sign. Thanks! $\endgroup$ – AxiomaticApproach Apr 1 '17 at 10:19

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