An expression in which every term has degree two, such as three x squared plus two x y plus three y squared, is a quadratic form. The cross term makes it hard to read, but turning the axes onto the eigenvectors of its symmetric matrix removes the cross term, and what is left shows at once its sign, its largest and smallest values and the shape of its level curves.

A hill seen from the wrong side looks lopsided, but walk round to face it squarely and it is a simple mound with a steep direction and a gentle one. The eigenvectors are the direction to face it from.

A quadratic form on $\mathbb{R}^n$ is $Q(\mathbf{x})=\mathbf{x}^{T}A\mathbf{x}$ with $A$ symmetric; the coefficient of a cross term $x_ix_j$ is shared equally between $a_{ij}$ and $a_{ji}$. The form is positive definite when $Q(\mathbf{x})>0$ for every $\mathbf{x}\neq\mathbf{0}$, negative definite when $Q(\mathbf{x})<0$, and indefinite when it takes both signs. The theorem below, the principal axes theorem, decides all of this from the eigenvalues.

• $3x^2+2xy+3y^2$ has the matrix $\begin{pmatrix}3&1\\1&3\end{pmatrix}$: the $2xy$ is split into $1$ and $1$.
• Positive definite exactly when every eigenvalue is positive; indefinite exactly when there are eigenvalues of both signs.
• On unit vectors, $Q$ is largest at an eigenvector of the largest eigenvalue, where it equals that eigenvalue, and smallest at one of the smallest.
• In two variables, $Q(\mathbf{x})=1$ is an ellipse when both eigenvalues are positive, with half-axes $1/\sqrt{\lambda}$ along the eigenvectors, and a hyperbola when their signs differ.

Points of the curve $5x^2+8xy+5y^2=9$. Its matrix $\begin{pmatrix}5&4\\4&5\end{pmatrix}$ has eigenvalues $9$ and $1$ with eigenvectors along $(1,1)$ and $(1,-1)$. In those axes the curve is $9y_1^2+y_2^2=9$: an ellipse with half-axis $1$ along $(1,1)$ and $3$ along $(1,-1)$.

Diagonalise $A=QDQ^{T}$ by the spectral theorem and substitute. With $\mathbf{y}=Q^{T}\mathbf{x}$, the coordinates of $\mathbf{x}$ in the orthonormal basis of eigenvectors, the form becomes $\mathbf{y}^{T}D\mathbf{y}$, and a diagonal matrix leaves only squares. The new axes are rotated or reflected, never stretched, because $Q$ is orthogonal.

The spectral theorem.

Substitute for $A$.

Name the new coordinates: the coordinates of $\mathbf{x}$ in the orthonormal basis of eigenvectors.

A diagonal matrix leaves only the squares, each weighted by its eigenvalue.

Put the coefficients of the squares on the diagonal and split each cross term's coefficient in half between the two mirror positions.

In the principal axes the form is $\lambda_1y_1^2+\cdots+\lambda_ny_n^2$. It is positive for every non-zero input exactly when every $\lambda_i$ is positive.

• $Q(\mathbf{x})=\mathbf{x}^{T}A\mathbf{x}$ with $A$ symmetric, cross terms split in half.
• In the eigenvector axes, $Q=\sum\lambda_iy_i^2$.
• Positive definite exactly when every eigenvalue is positive; indefinite when the signs are mixed.
• On unit vectors the form ranges between the smallest and the largest eigenvalue.

Great work! You now know how to:

• write the symmetric matrix of a quadratic form;
• remove the cross terms by turning to the eigenvector axes;
• decide definiteness from the eigenvalues;
• find the extreme values on unit vectors and the shape of a level curve.