Contents

Push a spring. It pushes back. You do work, it stores energy. Push harder, it stores more. Never does a spring hand you free energy on the way in. That one sentence is the whole story of symmetric positive definite (SPD) matrices.

Picture a marble in a bowl πŸ₯£ vs. on a saddle 🐴. In the bowl, every nudge costs energy β€” the marble rolls back. On a saddle, one direction costs energy, the perpendicular direction gives it away. SPD = bowl. Indefinite = saddle. That’s it.

Two requirements, each with a physical meaning:

  • Symmetric ($k_{ij} = k_{ji}$): the push at joint $i$ from moving joint $j$ equals the push at $j$ from moving $i$. For most stiffness matrices you’ll meet β€” linear elastic and hyperelastic systems β€” this follows from Maxwell–Betti reciprocity. (Non-associated plasticity, frictional contact, and follower loads are the exceptions where the tangent stiffness can be non-symmetric.)
  • Positive definite ($\mathbf v^\top \mathbf K \mathbf v > 0$ for every $\mathbf v \neq \mathbf 0$): every possible deformation stores positive energy. No free lunch.

See also: The geometric projector tensor Β Β·Β  Tensor visualization (rank 2, 3, 4)

Where SPD shows up

Pretty much everywhere you store energy, spread uncertainty, or fit a model:

Field Matrix The β€œenergy” being positive
FEM Stiffness $\mathbf K$ Strain energy
FEM Mass $\mathbf M$ Kinetic energy
Micromechanics Elasticity tensor $\mathbb C$ Strain energy density
Statistics Covariance $\boldsymbol\Sigma$ (PSD) Variance in any direction
Surrogates Gram matrix $\mathbf G$ Unique least-squares fit
Dynamics Damping $\mathbf C$ (PSD) Dissipation, never injection

Three things can go wrong:

  • One zero eigenvalue β†’ singular. The structure has a free direction it can move without deforming β€” a rigid-body mode or a mechanism. (Think: a chair with one leg missing that slides sideways for free.)
  • One negative eigenvalue β†’ indefinite. There’s a direction where deforming gives back energy. In real structures this is buckling β€” the instant a column buckles, its effective stiffness goes negative.
  • All positive β†’ SPD. Everything is fine. Solve and go home.

Energy test: what goes wrong

The strain energy is $U = \tfrac12\mathbf v^\top \mathbf K \mathbf v$. We need $U > 0$ for every $\mathbf v$, not just the one we happen to try. The cleanest way to guarantee that: demand every eigenvalue $\lambda_i > 0$. One negative $\lambda$ is enough to break the universe.

Stiffness matrix entries
Displacement vector
Presets
Matrix status
Live strain energy $U = \tfrac{1}{2}\mathbf v^\top\mathbf K\mathbf v$
varies with animation
Can U go negative for some v?

Two-node spring element anchored to a wall. Dashed circles mark the rest positions. Amber arrows = displacement v; coloured arrows = restoring force βˆ’Kv. SPD: nodes oscillate stably back to rest (green). Singular: nodes drift β€” no restoring force in one direction (yellow). Indefinite: displacement grows without bound and the structure goes unstable (red flash).

Play with it

Drag the sliders. Try to break the bowl into a saddle.

Matrix entries
Presets
Status

The 2Γ—2 symmetric matrix $\mathbf A$. Off-diagonal entries are forced equal.

Quadratic form $f(\mathbf v) = \mathbf v^\top \mathbf A\, \mathbf v$. Bowl = SPD. Saddle = indefinite.

Level set $\mathbf v^\top \mathbf A\, \mathbf v = 1$. SPD (green) β†’ ellipsoid, semi-axes $= 1/\sqrt{\lambda_i}$. Singular (yellow) β†’ disc (one axis β†’ ∞). Indefinite (red) β†’ hyperboloid saddle, no closed surface. Drag to orbit.

The ellipsoid tells you everything

Draw the set of points where $\mathbf v^\top \mathbf A\, \mathbf v = 1$. Its shape is the matrix:

  • Axis directions = eigenvectors (the natural grain of the material).
  • Axis lengths = $1/\sqrt{\lambda_i}$. Big $\lambda$ β†’ short axis (stiff direction β€” hard to squash). Small $\lambda$ β†’ long axis (soft direction β€” easy to squash).
  • SPD 🟒 β†’ a proper ellipsoid. Isotropic ($\lambda_1 = \lambda_2$) β†’ a sphere.
  • Singular 🟑 β†’ one $\lambda = 0$, so one axis $= \infty$. The β€œellipsoid” opens into an infinite cylinder β€” a free direction in which deforming costs nothing.
  • Indefinite πŸ”΄ β†’ one $\lambda < 0$. The level set is a hyperboloid (a saddle surface) β€” it runs off to infinity through the negative direction.

The 2Γ—2 quick check

For $\begin{bmatrix} a & b \ b & d \end{bmatrix}$:

\[\text{SPD} \iff a > 0 \;\text{and}\; ad - b^2 > 0.\]

That’s two numbers. Done. (The trace $a + d$ is then automatically positive.)

Spectral decomposition in one line

Every symmetric $\mathbf A = \mathbf Q \boldsymbol\Lambda \mathbf Q^\top$. In the eigenbasis, the quadratic form collapses to $\sum_i \lambda_i w_i^2$ β€” a weighted sum of squares. That’s the whole theory.

Equivalent tests for SPD

Four ways to say the same thing β€” pick whichever is cheapest to check:

  1. All eigenvalues $> 0$.
  2. All leading principal minors $> 0$ (Sylvester).
  3. Cholesky $\mathbf A = \mathbf L \mathbf L^\top$ exists with $L_{ii} > 0$.
  4. $\mathbf v^\top \mathbf A \mathbf v > 0$ for all $\mathbf v \neq \mathbf 0$.

References

[1] Strang, G. (2016). Introduction to Linear Algebra (5th ed.) β€” Ch. 6, positive definiteness. [2] Horn & Johnson (2013). Matrix Analysis (2nd ed.). [3] Golub & Van Loan (2013). Matrix Computations (4th ed.) β€” Ch. 8, symmetric eigenproblem. [4] Bathe, K.-J. (1996). Finite Element Procedures β€” Ch. 4, stiffness/mass definiteness and rigid-body modes.