Visual walkthrough — Quadratic forms — positive definite, negative definite, indefinite

Everything below is built from zero. If a symbol appears, it was drawn first.

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Step 1 — What "${\mathbf{x}}^{\top }A\mathbf{x}$" actually means

WHAT. We feed one arrow $\mathbf{x}$ into the machine and get back a single number $Q(\mathbf{x})$.

WHY a single number. We are asking "how high is the surface above this direction?" Height is one number. The whole story of definiteness is: is that number positive, negative, or does its sign flip as the arrow turns?

PICTURE. Below, the arrow $\mathbf{x}$ (amber) sits in the plane. Multiplying by $A$ first bends it into a new arrow $A\mathbf{x}$ (cyan). Then ${\mathbf{x}}^{\top }(A\mathbf{x})$ measures how much $A\mathbf{x}$ points along the original $\mathbf{x}$ — that overlap is the output number.

$$Q(\mathbf{x})=\underset{\text{arrow laid flat}}{\underbrace{{\mathbf{x}}^{\top }}}\underset{\text{bends the arrow}}{\underbrace{A}}\underset{\text{the arrow}}{\underbrace{\mathbf{x}}}=\text{one number}$$

If the two arrows roughly agree in direction, the number is positive. If $A\mathbf{x}$ points backwards relative to $\mathbf{x}$, the number is negative. That sign is the entire game.

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Step 2 — The problem: cross-terms make the sign hard to read

WHAT. There are three terms. The $x^2$ and $y^2$ terms have honest signs ($x^2\ge 0$ always). But the middle term $2bxy$ can be positive or negative depending on the quadrant the arrow lands in.

WHY it's a problem. Suppose $a>0$ and $c>0$ — you might guess "always positive." But if $b$ is large and negative, the cross-term $2bxy$ can drag the total below zero for some arrows. You cannot read the sign off the coefficients directly. The cross-term $xy$ couples the two directions.

PICTURE. The surface below is tilted and twisted — the $xy$ term is the shear that skews the bowl so its lowest valleys don't line up with the $x$- or $y$-axis. We need to un-twist it.

Recall Why "

$2b$" and not "$b$"? The off-diagonal $b$ appears in both the top-right and bottom-left of $A$. When you multiply out ${\mathbf{x}}^{\top }A\mathbf{x}$, the $xy$ contribution shows up twice — once as $A_{12}xy$, once as $A_{21}yx$ — giving $2bxy$. ::: That is exactly why (parent note) the off-diagonal is half the cross-coefficient.

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Step 3 — The rescue: symmetric matrices have perpendicular eigen-directions

WHAT. For a symmetric $A$ (our case, from Symmetric Matrices), the Spectral Theorem guarantees two things at once:

1. the eigenvalues ${\lambda }_1,{\lambda }_2$ are real numbers, and
2. the eigenvectors can be chosen perpendicular to each other.

WHY this saves us. Perpendicular special directions mean we can use them as a brand-new pair of axes. Along these axes there is no twist — the cross-term vanishes. The twisted bowl of Step 2 becomes an untwisted bowl.

PICTURE. The two eigen-directions (cyan) are drawn on the twisted surface. Notice they line up exactly with the valley and ridge of the shape — the natural axes the surface "wants."

$$A\underset{\text{special arrow}}{\underbrace{{\mathbf{v}}_i}}=\underset{\text{stretch factor}}{\underbrace{{\lambda }_i}}{\mathbf{v}}_i{\mathbf{v}}_1\perp {\mathbf{v}}_2$$

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Step 4 — Rotate to the eigen-axes (the change of coordinates)

WHAT. We define new coordinates $\mathbf{y}=Q_{\text{rot}}^{\top }\mathbf{x}$. This is just "the same arrow, described using the eigen-axes instead of the standard axes."

WHY it's harmless. A rotation never changes an arrow's length, and it never turns a non-zero arrow into the zero arrow. So $\mathbf{x}\ne 0$ if and only if $\mathbf{y}\ne 0$. Whatever we conclude about signs for $\mathbf{y}$ is exactly true for $\mathbf{x}$.

PICTURE. Same arrow, two rulers. On the left, the standard grid (white). On the right, the tilted eigen-grid (cyan) reads off the components $y_1,y_2$.

$$\underset{\text{arrow in eigen-coords}}{\underbrace{\mathbf{y}}}=\underset{\text{rotate the ruler}}{\underbrace{Q_{\text{rot}}^{\top }}}\underset{\text{same arrow}}{\underbrace{\mathbf{x}}}$$

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Step 5 — The twist disappears: pure squares

WHAT. Because $\Lambda$ is diagonal, ${\mathbf{y}}^{\top }\Lambda \mathbf{y}$ has no cross-term. It collapses to:

$$Q=\underset{\text{stretch 1}}{\underbrace{{\lambda }_1}}\underset{\ge 0}{\underbrace{y_1^2}}+\underset{\text{stretch 2}}{\underbrace{{\lambda }_2}}\underset{\ge 0}{\underbrace{y_2^2}}.$$

WHY this is the whole point. Look at what each piece can do:

• $y_1^2$ and $y_2^2$ are squares — they can never be negative.
• So the sign of each term is decided entirely by the sign of its eigenvalue ${\lambda }_i$.

The twist (Step 2) is gone. We have separated the machine into two independent one-dimensional springs.

PICTURE. The un-twisted surface, now aligned with the axes. Along the $y_1$-axis it curves with steepness ${\lambda }_1$; along $y_2$ with steepness ${\lambda }_2$. Two independent parabolas.

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Step 6 — Read off the four cases from the signs

WHAT. Each row is just "add up two signed squares."

WHY each conclusion.

• Both positive: the only way to get $0$ is $y_1=y_2=0$, i.e. the zero arrow. Every other arrow gives $Q>0$ → bowl.
• Mixed signs: point the arrow along ${\mathbf{v}}_1$ (so $y_2=0$) and get ${\lambda }_1{y}_1^2$; along ${\mathbf{v}}_2$ and get ${\lambda }_2{y}_2^2$. One is $+$, the other $-$. Both signs appear from the same machine → saddle.
• One zero: along that flat eigen-direction $Q$ stays exactly $0$ even though the arrow is non-zero → the boundary case, semidefinite not definite.

PICTURE. Three surfaces side by side — bowl (both up), dome (both down), saddle (up one way, down the other) — each with its eigen-axes and the sign of $Q$ marked in amber.

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Step 7 — The degenerate cases, done carefully

WHAT / WHY — every corner covered:

• Both eigenvalues zero ($A=0$): $Q\equiv 0$ for all arrows. The surface is a flat plane. It is technically PSD and NSD.
• One zero, one positive (PSD): flat along one eigen-line (a straight-bottomed trough), rising everywhere else.
• One zero, one negative (NSD): flat along one eigen-line, falling everywhere else.
• Both non-zero, same sign: strictly definite (bowl or dome) — never touches $0$ except at the origin.
• Both non-zero, opposite sign: indefinite — a genuine saddle.

PICTURE. The PSD trough: a valley with a flat straight bottom running along the zero-eigenvalue direction (amber line), not a single lowest point.

Recall Cross-check with the other test

This eigenvalue verdict must agree with Sylvester's leading-minor test (parent §4) and with the Second Derivative Test via the Hessian Matrix. It does: a bowl/PD Hessian ⇒ local minimum; dome/ND ⇒ maximum; saddle/indefinite ⇒ saddle point. Same algebra, two languages. Why can't Sylvester alone catch PSD? ::: Its clean form only certifies strict definiteness; a boundary $D_k=0$ needs the full principal-minor analysis (or just the eigenvalues).

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The one-picture summary

Everything above is one idea: rotate to the eigen-axes, and the twisted machine becomes two independent signed springs; the signs of the springs are the eigenvalues.

$${\mathbf{x}}^{\top }A\mathbf{x}\underset{\text{Spectral Thm}}{\overset{\text{rotate }\mathbf{y}=Q_{\text{rot}}^{\top }\mathbf{x}}{\to }}{\lambda }_1{y}_1^2+{\lambda }_2{y}_2^2\stackrel{\text{read signs}}{\to }\{\text{bowl / dome / saddle / trough}\}$$

Recall Feynman retelling — say it to a 12-year-old

You have a machine: put an arrow in, get a number out. At first the machine looks messy — there's a "twist" term that mixes the two directions, so you can't tell if the number will be positive or negative just by looking.

But a symmetric machine has two magic directions (the eigenvectors) that sit at right angles. If you turn your head so those directions become your new left-right and up-down, the twist vanishes. Now the number is just: (first stretch) times (first distance squared) plus (second stretch) times (second distance squared).

Distances squared are never negative — they're the innocent part. So the only thing that decides the sign of the answer is whether the stretches (the eigenvalues) are positive or negative.

• Both stretches up → every arrow gives a positive number → you're standing at the bottom of a bowl.
• Both down → dome, you're on a hilltop.
• One up, one down → saddle, uphill one way and downhill the other.
• A stretch of exactly zero → a flat groove: walk along it and the number never changes.

That's the whole theorem. Turn your head to the magic directions, then just count plus and minus signs.