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I'm trying to prove that $B(X,Y)=\det (X+Y) - \det (X) - \det (Y)$ is a blinear form on the vector space $A$ is from, and also trying to determine if it is an inner product space. I think if I know how $\det$ is a bilinear form, then this will become easier.

EDIT$^1$:

I was thinking, if I let $\det : \mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}$ be a bilinear form defined by $$ \det\pmatrix{\pmatrix{a_{11}\\a_{21}}, \pmatrix{a_{12}\\a_{22}}} =a_{11}a_{22}-a_{21}a_{12},$$ then if I can show that $\det$ is a bilinear form, then this will suffice to show that $B$ is a bilinear form as the sum of bilinear forms is yet again a bilinear form, as is remarked in Friedberg's Linear Algebra (page 423).

Cameron Buie
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Trancot
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  • You should explain where you are having trouble. Certainly you know the definition, so what part has you troubled from here? –  May 29 '13 at 00:56
  • One suggestion: think of the determinant of a 2x2 matrix as representing the (signed) area of the parallelogram spanned by the two rows (or alternately the two columns) of the matrix. This should make its bilinearity (as a function of the vectors representing its rows) much more self-evident; see the discussion of the volume form over on my answer to http://math.stackexchange.com/questions/62318/origin-of-the-dot-and-cross-product/62370#62370 for a little bit more detail on this. – Steven Stadnicki May 29 '13 at 00:57
  • I want to know how $\det$ can be seen as a bilinear form. – Trancot May 29 '13 at 01:04
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    Better to say that $\det$ is a quadratic form, and then only in the case $n=2.$ To get the bilinear form, you are quoting one of the standard polarization identities, except you have left off the traditional factor of $1/2,$ which is permitted. See http://math.stackexchange.com/questions/404038/isometries-of-a-hyperbolic-quadratic-form/404140#404140 for example. – Will Jagy May 29 '13 at 02:29
  • For anybody familiar with Mathematica, would this suffice as a proof? – Trancot May 29 '13 at 23:30

1 Answers1

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I think what your textbook means is something like this:

If $B(X,Y)$ is a finite sum of bilinear forms in $X$ and $Y$, then $B$ is also a bilinear form in $X$ and $Y$.

If this is the case, there is no hope to apply this statement to your question directly, because neither $\det(X+Y),\,\det(X)$ nor $\det(Y)$ is a bilinear form in $X$ and $Y$.

Adding to Will Jagy's comment, if you can show that some $q(x)$ is a quadratic form in $x$, then $b(x,y):=\frac{q(x+y)-q(x)-q(y)}2$ is a (symmetric) bilinear form in $x$ and $y$ with $q(x)\equiv b(x,x)$. In your case, if you can show that $q(X):=2\det(X)$ is a quadratic form in $X$, then you are done. $q(X)$ is obviously a polynomial in the entries of $X$. Can you show that $q(kX)=k^2q(X)$ for all scalar $k$?

Alternatively, let $X=\pmatrix{a&b\\ c&d}$. Can you find a symmetric matrix $M$ such that $2\det(X) = (a,b,c,d)M(a,b,c,d)^T$? If you can, then for $Y=\pmatrix{x&y\\ z&w}$, we can express the (symmetric) bilinear form $B(X,Y)$ as $(a,b,c,d)M(x,y,z,w)^T$.

You may also work from the first principle and evaluate $B(X,Y)$ in terms of the entries of $X$ and $Y$: \begin{align*} &\det\pmatrix{a+x&b+y\\ c+z&d+w} - \det\pmatrix{a&b\\ c&d} - \det\pmatrix{x&y\\ z&w}\\ =&(a+x)(d+w) - (b+y)(c+z) - (ad-bc) - (xw-yz).\tag{1} \end{align*} Can you simplify $(1)$ and show that it is a bilinear form?

Finally, if $B(X,X)$ defines an inner product, $B(X,X)$ should be positive for all $X\neq0$. Is that true?

user1551
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  • +1. We can also express $B$ as $\mbox{tr}(X\cdot \mbox{adj}(Y))$ with the help of the adjugate. – Julien May 29 '13 at 12:55
  • @user1551 I know that your $M$ is $\pmatrix{0&1\-1&0}$, right? – Trancot May 29 '13 at 20:26
  • @user1551 Your statement in $(1)$ reduces to $x_{11}y_{22}+y_{11}x_{22}-x_{21}y_{12}-y_{21}x_{12}$, for $X$ and $Y$ $2\times 2$ matrices. – Trancot May 29 '13 at 20:35
  • @user1551 This happens to be $\det\pmatrix{x_{11}&y_{12}\x_{21}&y_{22}}+\det\pmatrix{y_{11}&x_{12}\y_{21}&x_{22}},$ but I'm not sure where this will lead me. – Trancot May 29 '13 at 20:55
  • @Trancot No, $M$ is a $4\times4$ matrix. Your other two comments are right, though. In last comment, the split of $B(X,Y)$ into the two determinants shows that $B$ is bilinear because each determinant is bilinear in $X$ and $Y$. – user1551 May 29 '13 at 22:14
  • @user1551 What is "split"? – Trancot May 29 '13 at 22:20
  • @Trancot: Your suggested edit to the OP was not appropriate. Please refrain from edits that drastically change the OP in any way, and try to decide if your edits would be more appropriate as comments, answers, or questions of your own. – davidlowryduda May 29 '13 at 23:04
  • @Trancot By a split, I meant your way of writing $B(X,Y)$ as a sum of two determinants. – user1551 May 31 '13 at 07:06