Can a symmetric map be split into three simple steps?

Source scope: §3.10. Worked numerical examples, experiments, and practice are teaching supplements unless identified as source results.

Rotate the directions you measure

For $A=\begin{bmatrix}2&1\\1&2\end{bmatrix}$, the unit directions $q_1=(1,1)/\sqrt2$ and $q_2=(-1,1)/\sqrt2$ have eigenvalues three and one. In these directions the map only scales. Their orthogonality lets us change coordinates without distorting length.

A matrix is symmetric if $A=A^T$, meaning $A_{ij}=A_{ji}$ for every entry.

Theorem 2 (Spectral Theorem, source). Every real symmetric $n\times n$ matrix admits an orthonormal basis of $\mathbb R^n$ consisting of its eigenvectors. The source states this theorem without a proof. In particular its eigenvalues are real.

Derive the factorization

Write the orthonormal eigenvectors as columns of $Q$ and their eigenvalues as $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$. Since $Aq_i=\lambda_iq_i$ for each column,

Right-multiply by $Q^T$. Orthonormality gives $QQ^T=I$, hence

This is the eigendecomposition, also called spectral decomposition. Read right to left: $Q^T$ expresses the input in eigenvector coordinates; $\Lambda$ scales those coordinates; $Q$ returns to the original coordinates. These are changes of coordinates and a mathematical scaling, not a rotation of the viewing camera.

For input $(1,0)$ in the example, $Q^Tx=(1,-1)/\sqrt2$, scaling gives $(3,-1)/\sqrt2$, and multiplication by $Q$ gives $(2,1)$, matching $Ax$ directly.

Repeated eigenvalues can allow several choices of orthonormal eigenbasis. The map remains the same. A general nonsymmetric matrix need not admit this orthogonal factorization.

Explore in three dimensions

Transfer

If $A=5I$ in two dimensions, is its orthonormal eigenbasis unique?