Symmetric Matrices and Quadratic Forms

2 MIN READ · PRACTISE AT YOUR OWN PACE · FREE

A stronger diagonalization theorem

Every real symmetric matrix $A=A^T$ has real eigenvalues and an orthonormal eigenvector basis. Thus $A=Q\Lambda Q^T$ with $Q^TQ=I$ . This is the spectral theorem. Orthogonal coordinate changes preserve lengths and angles, unlike general basis changes.

A quadratic form is $q(x)=x^TAx$ . Its value depends only on the symmetric part $(A+A^T)/2$ , so symmetric matrices capture all real quadratic forms. In eigen-coordinates $y=Q^Tx$ , it becomes $q=\sum_i\lambda_i y_i^2$ .

Visual guide

VISUAL GUIDEOrthogonal principal directions of a quadratic form

Quadratic-form contour

The contour x² + 2xy + 3y² = 1 is an ellipse. A symmetric matrix has perpendicular eigenvectors, which align with the ellipse’s principal axes. Rotating into that basis removes the mixed term.

Worked example: classify an energy

For $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ , eigenvalues are $3$ and $1$ . Its quadratic form $2x_1^2+2x_1x_2+2x_2^2$ becomes $3y_1^2+y_2^2$ in diagonal coordinates. It is strictly positive for every nonzero vector, so $A$ is positive definite.

All positive eigenvalues mean positive definite; all nonnegative mean positive semidefinite. Mixed positive and negative eigenvalues mean indefinite. A zero eigenvalue creates a flat direction, so nonnegative alone is insufficient for strict positivity.

PAUSE & THINKA quick check, not a grade

Try it yourself.

Choose an answer and explain your reasoning to yourself. Use a hint if you get stuck.

A real symmetric matrix has all eigenvalues strictly positive. What can you say about xᵀAx for x≠0?

It is strictly positive.

It can have either sign.

It is always zero.

Hint 1 · Find a starting point

Use an orthonormal eigenbasis.

Hint 2 · Take the next step

The quadratic form becomes a sum of positive eigenvalues times squared coordinates.

Show the reasoning

Answer: It is strictly positive.

At least one coordinate is nonzero, so the weighted sum is positive: A is positive definite.

Before moving on: what would make one of the other answers wrong? Saying why is part of understanding.

Worked example: an optimization model

Consider $E(x)=\tfrac12x^TAx-b^Tx$ with symmetric positive definite $A$ . Its gradient is $Ax-b$ . The unique critical point solves $Ax=b$ , and it is the global minimizer because

E(x_*+h)-E(x_*)=\tfrac12h^TAh>0\quad(h\ne0).

For $A=\operatorname{diag}(2,4)$ and $b=(2,8)^T$ , the minimizer is $(1,2)^T$ . If $A$ were indefinite, a stationary point would not be a minimum.

This connects linear systems to spring energies, least squares and second-derivative tests. In least squares, $A^TA$ is always positive semidefinite and becomes positive definite exactly when $A$ has independent columns.

Practice

Classify $\operatorname{diag}(2,0)$ .
Classify $\operatorname{diag}(1,-3)$ .
Why is $x^TA^TAx\ge0$ for every real $x$ ?

Show worked solutions

Positive semidefinite, not definite, because the vertical direction has zero quadratic value.
Indefinite: the coordinate directions give values of opposite signs.
It equals $(Ax)^T(Ax)=\|Ax\|^2$ .

MAKE IT YOURS

Pause before the next idea.

Can you explain how the visualization connects to this lesson’s goal? If a step still feels uncertain, put this lesson on your review list and try the check again another day.

Optional marks, not a grade. Saved in this browser only. Open notebook →