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A quadratic form is a homogenous polynomial $q(x_1, \dots , x_n) = \sum_{j=1}^n \sum_{i=1}^n x_i x_j a_{ij}$. Let $A := (a)_{ij}$. We define $q$ to be non-singular (or non-degenerate) if its associated bilinear form $B$ is non-singular where $$ B(v,w) := \frac{1}{2} (q(v + w) - q(v) - q(w)) $$

Why is singularity defined in terms of an associated bilinear form? Why not say $q$ is singular if and only if $A$ is?

Edit We may assume that $A$ is symmetric.

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The matrix $A$ is not uniquely determined by $q$. You should require that $A$ is symmetric. –  Andrea Apr 6 '12 at 7:57

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up vote 2 down vote accepted

Two points: (1)The expression $q(x_1,\cdots,x_n)=\sum_{i,j} a_{ij}x_ix_j$ is ambiguous in that it does not define $a_{ij}$ uniquely, as Andrea said. One of the matrices $A=(a_{ij})$ which determines $q$ in this sense (and the only symmetric one which does this) is the matrix of the associated bilinear form in the same basis. We can indeed characterize singular quadratic forms as those whose matrix (in this sense) is singular, in any (equivalently, in some) basis of the vector space. (2) Why do we define a "singular" (or "degenerate", which I'm more used to) quadratic form in terms of its associated bilinear form, rather than simply going to the associated matrix? Well, mainly because in Linear Algebra we'd rather have intrinsic definitions whenever they are at our disposal. That is, if some concept can be characterized in a coordinate-free way (without having to express the object at hand in terms of any basis of the vector space), we'll prefer these definitions.

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Thank you very much for this nice answer! Would it be much of a bother if you could add the connection between $B(v,w)$ and $A$? I thought $B$ should be singular if and only if $A$ is. So I fill $B(x,y) = x^T A y$ into $q(v,w) = \frac{1}{2}(w^T A v + v^T A w)$. And how do I proceed from there? –  Rudy the Reindeer Apr 6 '12 at 15:52
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oops! $q(v,w)$ does not make any sense because $q$ only has a (vector) argument! I suppose you mean $B(v,w)$ instead of $q(v,w)...$ The connection between $B$ and $A$ is very simple (provided $A$ is symmetric, which amounts to the fact that $a_{ij}=B(\bar e_i,\bar e_j)$ where $\{\bar e_i\}$ is the basis of choice): $B(\bar v,\bar w)=v^{T}Aw=w^{T}Av$ where $v$ is the column matrix which contains the coordinates of $\bar v$ in that basis, and the same for $w$ and $\bar w.$ Actually this is how the matrix $A$ is usually defined in the first place. –  Xabier Domínguez Apr 6 '12 at 16:20
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By the way, you keep highlighting the expression which gives $B$ in terms of $q$ and maybe you're not as aware of the inverse relation, which is far more natural: $q(\bar v)=B(\bar v, \bar v)$. This is a coordinate-free definition of a quadratic form as something which grows out of a bilinear form. –  Xabier Domínguez Apr 6 '12 at 16:29
    
Yes, you're right, I meant $q(x) = x^T A x$ and $B(v,w) = \frac{1}{2}(w^TAv + v^T Aw)$. So the matrix of $B$ in any basis coincides with the matrix $A$ of $q$ if and only if we require $A$ to be symmetric? Do I understand this correctly? What do the bars $\bar{v}$ mean? (as opposed to $v$) Thank you! –  Rudy the Reindeer Apr 7 '12 at 12:29
    
There are several ways to look at this. If you don't want to mess with vectors and basis and you like to think of a quadratic form as a homogeneous polynomial with degree 2 in $\mathbb{R}^n$, then really there are infinitely many matrices $A$ for which $q(x)=x^TAx.$ Only one of them is symmetric, and you can work out its elements easily from the coefficients of the polynomial. But soon you'll want to change coordinates (usually because you want to diagonalize this polynomial, i. e. remove the terms in $x_ix_j$ with $i\not=j$) and for that you need a more abstract theory. (see next comment) –  Xabier Domínguez Apr 7 '12 at 14:12

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