Bilinear forms

The dot product $\mathbf x\cdot\mathbf y$ is the simplest way to combine two vectors into a single number, and it is linear in each vector separately. A bilinear form is the general version of that idea: any rule $B(\mathbf x,\mathbf y)$ built from two vectors that stays linear in each argument on its own. Every such rule, no matter how it is first written, collapses into one matrix sandwich $\mathbf x^TA\mathbf y$ . Set the two vectors equal and $B$ becomes a quadratic form $Q(\mathbf x)=\mathbf x^TA\mathbf x$ — the matrix language behind expressions like $2x^2+3xy-y^2$ , and the tool that tells you, without plotting anything, whether such an expression is always positive, always negative, or switches sign. That single question — sign definiteness — is what makes bilinear forms worth learning: it decides whether a critical point is a minimum, a maximum, or a saddle, and it is the algebraic seed of everything eigenvalues will later explain.

The core insight

A bilinear form $B(\mathbf x,\mathbf y)=\mathbf x^TA\mathbf y$ reduces, when $\mathbf y=\mathbf x$ , to a quadratic form $Q(\mathbf x)=\mathbf x^TA\mathbf x$ whose sign for every nonzero $\mathbf x$ is decided by a symmetric matrix $A$ , tested compactly through leading principal minors.

The core idea

From two vectors to one matrix. A bilinear form on $\mathbb{R}^n$ is a function $B:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}$ that is linear in each slot separately:

B(\alpha\mathbf u_1+\beta\mathbf u_2,\ \mathbf v)=\alpha B(\mathbf u_1,\mathbf v)+\beta B(\mathbf u_2,\mathbf v),

where: $\mathbf u_1,\mathbf u_2,\mathbf v$ are vectors in $\mathbb{R}^n$ and $\alpha,\beta$ are scalars; the same linearity holds when the second argument is split instead of the first. This one property forces every bilinear form into a fixed shape,

B(\mathbf x,\mathbf y)=\mathbf x^TA\mathbf y=\sum_{i=1}^n\sum_{j=1}^n a_{ij}x_iy_j,

where: $A$ is the unique $n\times n$ matrix with entries $a_{ij}=B(\mathbf e_i,\mathbf e_j)$ , obtained by feeding in standard basis vectors. A bilinear form's matrix need not be symmetric — only once the two arguments are forced equal does symmetry become the natural convention.

Quadratic forms and the symmetry convention. Setting $\mathbf y=\mathbf x$ gives the quadratic form $Q(\mathbf x)=\mathbf x^TA\mathbf x$ . Because $\mathbf x^TA\mathbf x=\mathbf x^TA^T\mathbf x$ for any square $A$ (a scalar equals its own transpose), replacing $A$ by $\tfrac12(A+A^T)$ never changes $Q$ , so by convention the matrix of a quadratic form is always taken symmetric, with cross-term coefficients split evenly between the two off-diagonal positions.

Sign definiteness. A quadratic form is classified by the sign $Q(\mathbf x)$ takes for $\mathbf x\neq\mathbf 0$ : positive definite if $Q(\mathbf x)>0$ always; negative definite if $Q(\mathbf x)<0$ always; positive/negative semidefinite if $Q(\mathbf x)\ge0$ (resp. $\le0$ ) with equality somewhere nonzero; indefinite if $Q$ takes both signs. For a symmetric $A$ this sign pattern matches the sign pattern of $A$ 's eigenvalues — all positive gives positive definite, all negative gives negative definite, mixed signs give indefinite — though the full eigenvalue argument belongs to the diagonalization topic that follows this one.

Leading principal minors. Eigenvalues are expensive to compute by hand, so a faster $2\times2$ / $3\times3$ check uses the determinants of the top-left $k\times k$ blocks of $A$ , called leading principal minors $M_1,M_2,\dots,M_n$ . For $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ , $Q$ is positive definite iff $M_1=a>0$ and $M_2=\det A=ac-b^2>0$ ; negative definite iff $a<0$ and $\det A>0$ ; indefinite iff $\det A<0$ . The general rule (Sylvester's criterion) extends this to $n\times n$ : positive definite iff every $M_k>0$ ; negative definite iff the signs alternate starting negative ( $M_1<0,M_2>0,M_3<0,\dots$ ).

Figure

Three quadratic-form surfaces labelled positive definite bowl opening upward negative definite bowl opening downward and indefinite saddle

Derivation ASylvester's 2x2 criterion by completing the square

The goal is to derive, from the algebraic expression $Q(x,y)=ax^2+2bxy+cy^2$ alone and nothing beyond completing the square, exactly why positive definiteness requires both $a>0$ and $\det A=ac-b^2>0$ , so that the leading-principal-minors shortcut used throughout this topic is not just memorized but seen to follow from the definition of sign definiteness itself.

Step 1. Write the general $2\times2$ quadratic form with symmetric matrix $A=\begin{pmatrix}a&b\\b&c\end{pmatrix}$ , so $Q(x,y)=\mathbf x^TA\mathbf x=ax^2+2bxy+cy^2$ . Assume first $a\neq0$ , which is needed to complete the square in $x$ .

Step 2. Factor $a$ out of the $x$ -terms and complete the square: $Q=a\left(x^2+\dfrac{2b}{a}xy\right)+cy^2=a\left(x+\dfrac{b}{a}y\right)^2-\dfrac{b^2}{a}y^2+cy^2$ . Combine the remaining $y^2$ terms over the common denominator $a$ : $Q=a\left(x+\dfrac{b}{a}y\right)^2+\dfrac{ac-b^2}{a}y^2$ .

Step 3. Read off the sign directly. $Q$ is now a sum of two pieces: $a$ times a perfect square, plus $\dfrac{\det A}{a}$ times $y^2$ , where $\det A=ac-b^2$ is exactly $M_2$ . For $Q>0$ whenever $(x,y)\neq(0,0)$ , both coefficients must be positive — if $a\le0$ the first term cannot always be positive, and if $\det A\le0$ the second term can go negative or vanish for suitable $y$ . So positive definiteness forces $a>0$ and $\det A>0$ , matching $M_1>0,M_2>0$ .

Step 4. Check the converse and the other cases. If $a>0$ and $\det A>0$ , both terms are nonnegative and the perfect square vanishes only when $x=-\tfrac{b}{a}y$ ; substituting that back leaves $\dfrac{\det A}{a}y^2>0$ unless $y=0$ too, so $Q>0$ for every nonzero $(x,y)$ — positive definite confirmed. Flipping $a<0$ with $\det A>0$ makes both terms nonpositive by the same argument, giving negative definite; and $\det A<0$ makes the two terms fight for opposite signs as $y$ and $x$ vary, giving indefinite regardless of the sign of $a$ . This reproduces every case of Sylvester's $2\times2$ criterion from first principles.

Worked example

Problem. Classify $Q(x,y)=2x^2+6xy-y^2$ using leading principal minors, then confirm the classification by exhibiting vectors where $Q$ takes both signs (if indefinite) or by writing $Q$ as a sum of squares (if definite).

Step 1. Build the symmetric matrix. The coefficient of $x^2$ gives $a_{11}=2$ , the coefficient of $y^2$ gives $a_{22}=-1$ , and the cross term $6xy$ splits evenly into $a_{12}=a_{21}=3$ :

A=\begin{pmatrix}2&3\\3&-1\end{pmatrix}.

Step 2. Compute the leading principal minors. $M_1=a_{11}=2>0$ . $M_2=\det A=(2)(-1)-(3)(3)=-2-9=-11<0$ .

Step 3. Apply the criterion. Since $M_2<0$ , neither the positive-definite pattern ( $M_1>0,M_2>0$ ) nor the negative-definite pattern ( $M_1<0,M_2>0$ ) holds, so $Q$ is indefinite.

Step 4. Confirm directly with test vectors. At $(x,y)=(1,0)$ : $Q=2(1)^2+6(1)(0)-(0)^2=2>0$ . At $(x,y)=(0,1)$ : $Q=2(0)^2+6(0)(1)-(1)^2=-1<0$ . $Q$ takes both a positive value ( $2$ , at $(1,0)$ ) and a negative value ( $-1$ , at $(0,1)$ ), confirming $Q$ is indefinite, exactly as the negative determinant $M_2=-11$ predicted.

Competitive stretch

Using the derivation above, complete the square directly: $Q=2\left(x+\tfrac32y\right)^2-\tfrac{9}{2}y^2-y^2=2\left(x+\tfrac32y\right)^2-\tfrac{11}{2}y^2$ . The two coefficients $2$ and $-\tfrac{11}{2}$ have opposite signs, which is exactly the indefinite signature — and their product, up to the squared denominator, reproduces $\det A=-11$ , tying the sum-of-squares picture directly back to the minor test.

Common traps

Splitting a cross-term coefficient unevenly (writing $a_{12}=6,a_{21}=0$ for $6xy$ instead of $a_{12}=a_{21}=3$ ) before applying the minor test — the symmetrization must happen first.
Applying leading-principal-minors to a non-symmetric matrix; Sylvester's criterion as used here assumes $A=A^T$ , and a general bilinear form's matrix need not be symmetric at all.
Reading a zero minor as "indefinite" rather than the borderline semidefinite case, which needs specific test vectors rather than a clean sign pattern.
Confusing $B(\mathbf x,\mathbf y)$ , a function of two independent vectors, with $Q(\mathbf x)=B(\mathbf x,\mathbf x)$ , a function of one vector used twice — a quadratic form is a special case, not another name for a bilinear form.

Ask the tutor: want to practice converting a messier three-variable quadratic expression into matrix form and classifying it with the minor test, or work through the eigenvalue connection with a concrete symmetric matrix?

Why must the matrix of a quadratic form be symmetric, when a general bilinear form's matrix need not be?

A quadratic form only ever sees $\mathbf x^TA\mathbf x$ , and for any square matrix that scalar equals $\mathbf x^TA^T\mathbf x$ too, since a $1\times1$ matrix equals its own transpose. That means the symmetric part $\tfrac12(A+A^T)$ produces exactly the same $Q$ as $A$ does, while the antisymmetric part contributes nothing at all. So there is no information lost by always choosing the symmetric representative, and doing so is what makes the leading-principal-minors test and the eigenvalue connection work cleanly. A general bilinear form $B(\mathbf x,\mathbf y)$ has two independent arguments, so this collapsing trick does not apply and its matrix can be anything.

What happens when a leading principal minor is exactly zero instead of positive or negative?

A zero minor signals a borderline, semidefinite case rather than a clean definite or indefinite classification, and the sign-pattern shortcut does not resolve it by itself. For example $Q(x,y)=x^2$ has $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ with $M_1=1>0$ and $M_2=0$ ; $Q\ge0$ always but is zero along the whole $y$ -axis, so it is positive semidefinite, not positive definite. When you hit a zero minor, fall back to testing specific vectors or writing $Q$ as an explicit sum of squares to see whether it ever changes sign or merely touches zero.

How does sign definiteness here connect to what comes next with eigenvalues?

For a symmetric matrix, the sign pattern of $Q(\mathbf x)=\mathbf x^TA\mathbf x$ matches the sign pattern of $A$ 's eigenvalues exactly: all eigenvalues positive gives positive definite, all negative gives negative definite, and mixed signs give indefinite. Leading principal minors are the fast hand-calculation route to that same answer without first finding eigenvalues, which typically need the characteristic polynomial. Once diagonalization of symmetric matrices is covered, the same $A$ can be written in a rotated coordinate system where $Q$ becomes a plain sum $\lambda_1u_1^2+\lambda_2u_2^2+\cdots$ , making the eigenvalue-sign connection visible directly rather than through minors.

Bilinear forms

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads