Why Eigenvalue Signs Reveal a Saddle
A stationary point in three variables can be a local maximum, a local minimum, or a saddle. At a saddle, arbitrarily close points have function values both above and below the value at the stationary point. The Hessian describes the function’s second-order change, and its eigenvalues identify directions in which that change is positive or negative.
This is especially useful when the variables are coupled: cross terms such as \(xy\) or \(xz\) mean that the coordinate axes themselves may not be the natural directions to test. Eigenvectors provide the directions in which the quadratic form separates into independent contributions.
For a twice continuously differentiable function \(f:\mathbb{R}^3\to\mathbb{R}\), the Hessian \(\mathbf{H}_f(\mathbf{a})\) is the matrix of second partial derivatives evaluated at \(\mathbf{a}\). It is symmetric when the second partial derivatives are continuous near \(\mathbf{a}\). A stationary point satisfies \(\nabla f(\mathbf{a})=\mathbf{0}\).
The Directional Meaning of an Eigenvalue
Suppose \(\mathbf{v}\) is a unit eigenvector of the Hessian at a stationary point \(\mathbf{a}\), with eigenvalue \(\lambda\). The quadratic term in the Taylor expansion along the line \(\mathbf{a}+t\mathbf{v}\) is \(\frac12\lambda t^2\). Thus, a positive eigenvalue gives a positive second-order change along its eigenvector, while a negative eigenvalue gives a negative second-order change.
Here \(o(\|\mathbf{h}\|^2)\) is a remainder whose ratio to \(\|\mathbf{h}\|^2\) tends to zero as \(\mathbf{h}\) tends to \(\mathbf{0}\). The first-order term is absent because the point is stationary. This expansion makes precise why the Hessian controls sufficiently small changes in directions where its quadratic term is nonzero.
Why the Indefinite Hessian Guarantees a Saddle
Let \(\mathbf{p}\) and \(\mathbf{n}\) be unit eigenvectors with eigenvalues \(\lambda_+>0\) and \(\lambda_-<0\), respectively. Apply the Taylor expansion first to \(\mathbf{h}=t\mathbf{p}\). Since \(\mathbf{H}_f(\mathbf{a})\mathbf{p}=\lambda_+\mathbf{p}\), the quadratic term is positive. For \(\mathbf{h}=t\mathbf{n}\), it is negative.
Along \(\mathbf{a}+t\mathbf{p}\), the change in function value is \(\frac12\lambda_+t^2+o(t^2)\).
For sufficiently small nonzero \(t\), the absolute value of the remainder is less than \(\frac14\lambda_+t^2\). The change is then greater than \(\frac14\lambda_+t^2>0\).
Along \(\mathbf{a}+t\mathbf{n}\), the change is \(\frac12\lambda_-t^2+o(t^2)\). For sufficiently small nonzero \(t\), the absolute value of the remainder is less than \(\frac14|\lambda_-|t^2\), so the change is less than \(-\frac14|\lambda_-|t^2<0\).
Both kinds of points can be found arbitrarily close to \(\mathbf{a}\), because \(t\) can be chosen nonzero and as small as desired. Therefore every neighborhood contains a point with a greater function value and a point with a smaller one. This proves the theorem.
Worked Examples
Worked Example: A Coupled Quadratic Saddle at the Origin
Classify \(f(x,y,z)=x^2+xy+y^2-z^2\) at the origin.
First compute the gradient:
Thus \(\nabla f(0,0,0)=(0,0,0)\), so the origin is stationary. The Hessian is
The vectors \((1,1,0)^{\top}\), \((1,-1,0)^{\top}\), and \((0,0,1)^{\top}\) are eigenvectors with eigenvalues \(3\), \(1\), and \(-2\), respectively. For instance, multiplication by the Hessian sends \((1,1,0)^{\top}\) to \((3,3,0)^{\top}\), and sends \((0,0,1)^{\top}\) to \((0,0,-2)^{\top}\). The Hessian therefore has eigenvalues of both signs.
Check the function values along two eigenvector directions. Along \((x,y,z)=(t,t,0)\),
Along \((x,y,z)=(0,0,t)\),
Both paths approach the origin as \(t\to0\), and \(f(0,0,0)=0\). These substitutions confirm that the origin is a saddle.
Worked Example: A Shifted Saddle with a Higher-Order Term
Classify \(f(x,y,z)=(x-1)^2+(x-1)(y+1)+(y+1)^2-(z-2)^2+(x-1)^3\) at \(\mathbf{a}=(1,-1,2)\).
Set \(X=x-1\), \(Y=y+1\), and \(Z=z-2\). Then \(f=X^2+XY+Y^2-Z^2+X^3\). Its first partial derivatives with respect to \(X,Y,Z\) are \(2X+Y+3X^2\), \(X+2Y\), and \(-2Z\). All vanish at \(X=Y=Z=0\), so \(\mathbf{a}\) is stationary. The Hessian there is
As in the previous calculation, the eigenvalues are \(3\), \(1\), and \(-2\). We can also verify the signs directly, including the cubic term. Along \(X=Y=t/\sqrt{2}\), \(Z=0\), the change from the value at \(\mathbf{a}\) is
For \(0<|t|<1\), the parenthesis is positive: it is greater than \(\frac32-\frac{1}{2\sqrt{2}}>0\). Along \(X=Y=0\), \(Z=t\), the change is exactly \(-t^2<0\) for \(t\ne0\). These two paths verify that points arbitrarily close to \(\mathbf{a}\) have function values on both sides of \(f(\mathbf{a})=0\). Hence \(\mathbf{a}\) is a saddle.
Worked Example: Classifying a Fully Coupled Quadratic Form
Let \(\mathbf{x}=(x,y,z)^{\top}\) and \(f(\mathbf{x})=\frac12\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\), where
This symmetric matrix has nonzero off-diagonal entries, so the variables are coupled. Since \(\nabla f(\mathbf{x})=\mathbf{A}\mathbf{x}\), the origin is stationary. To find the eigenvalues, use the vectors \(\mathbf{u}=(1,1,1)^{\top}\), \(\mathbf{v}=(1,-1,0)^{\top}\), and \(\mathbf{w}=(1,1,-2)^{\top}\). Direct multiplication gives
Thus the eigenvalues are \(2\), \(1\), and \(-3\). There are positive and negative eigenvalues. For a direct check, along \(\mathbf{x}=t\mathbf{u}\),
because \(\mathbf{u}^{\top}\mathbf{u}=3\). Along \(\mathbf{x}=t\mathbf{w}\),
because \(\mathbf{w}^{\top}\mathbf{w}=6\). Since \(f(\mathbf{0})=0\), the origin is a saddle. The direct substitutions agree with the signs of the corresponding eigenvalues.
A Common Mistake: Using Only the Determinant
The determinant is the product of the eigenvalues, but its sign alone does not tell us whether both positive and negative eigenvalues occur. For example, consider \(g(x,y,z)=-x^2-y^2-z^2\). Its gradient vanishes at the origin and its Hessian is \(\operatorname{diag}(-2,-2,-2)\), so its determinant is \((-2)(-2)(-2)=-8\).
A mistaken conclusion would be: “The determinant is negative, so the Hessian is indefinite and the origin is a saddle.” The correction is to inspect the eigenvalues: all three are negative. Indeed, \(g(x,y,z)<g(0,0,0)=0\) at every nonzero point, so the origin is a strict local maximum. In three variables, a negative determinant does not distinguish one negative eigenvalue from three negative eigenvalues. To establish indefiniteness using eigenvalues, verify that at least one eigenvalue is positive and at least one is negative.
What the Test Establishes
An indefinite Hessian gives a decisive local classification at a stationary point: the point is a saddle. The reason is not simply that the quadratic form takes both signs somewhere; the Taylor remainder must also be controlled. Along eigenvector directions with nonzero eigenvalues, that remainder is smaller than the leading quadratic change sufficiently near the point, so it cannot reverse either sign.
This test applies to twice continuously differentiable functions near the stationary point. It is particularly effective for coupled functions, where eigenvectors may reveal increasing and decreasing directions that are not obvious from the coordinate axes. The conclusion depends on finding eigenvalues of both signs; a determinant or a few coordinate-axis substitutions alone may fail to reveal the relevant directions.
Check Your Understanding
Use eigenvalue signs and directional changes to answer these questions.
- What does it mean for a symmetric Hessian to be indefinite?
- Why does a positive Hessian eigenvalue give a direction of local increase at a stationary point?
- For \(f(x,y,z)=x^2+xy+y^2-z^2\), what are the Hessian eigenvalues, and which two paths verify the saddle classification?
- Why can a negative determinant in three variables not, by itself, establish that a Hessian is indefinite?
- What condition on the Taylor remainder ensures that the positive and negative quadratic changes determine the signs nearby?