The Spectral Theorem

Recall that an orthogonal matrix $Q$ is a square matrix with orthonormal columns, equivalently $Q^{T}Q=I$, so that $Q^{-1}=Q^{T}$. A matrix $A$ is orthogonally diagonalizable when there is an orthogonal $Q$ for which $Q^{T}AQ=Q^{-1}AQ=D$ is diagonal (Definition 9.1; the course notes call this matrix $S$). Rearranging gives the spectral decomposition $A=QDQ^{T}=QDQ^{-1}.$ Reading the equation $AQ=QD$ column by column shows exactly what $Q$ and $D$ contain: the $i$-th column $\mathbf{q}_i$ of $Q$ satisfies $A\mathbf{q}_i=d_i\mathbf{q}_i$, so the columns of $Q$ are eigenvectors of $A$ and the diagonal entries of $D$ are the corresponding eigenvalues. Because the columns of $Q$ are orthonormal, this says precisely that $A$ is orthogonally diagonalizable if and only if $A$ has an orthonormal eigenbasis -- an orthonormal basis of $\mathbb{R}^{n}$ made entirely of eigenvectors of $A$.

The Spectral Theorem (Theorem 9.1) is one of the cleanest classification results in linear algebra: a real $n\times n$ matrix $A$ is orthogonally diagonalizable if and only if $A$ is symmetric ($A^{T}=A$). One direction is a one-line computation (Remark 9.1): if $A=QDQ^{T}$ with $Q$ orthogonal and $D$ diagonal, then $A^{T}=(QDQ^{T})^{T}=(Q^{T})^{T}D^{T}Q^{T}=QDQ^{T}=A,$ using $D^{T}=D$. So every orthogonally diagonalizable matrix is automatically symmetric. The striking direction is the converse -- that symmetry alone guarantees an orthonormal eigenbasis -- proved by induction on $n$, peeling off one real unit eigenvector at a time and restricting $A$ to its orthogonal complement, which stays symmetric (Lemma 9.2). The practical payoff is enormous: to decide whether a real matrix can be orthogonally diagonalized you do not compute a single eigenvalue -- you just check whether $A^{T}=A$.

Two facts about a symmetric $A$ make the hard direction work. First, all eigenvalues are real (Proposition 9.2): there are $n$ of them, counted with multiplicity, so the characteristic polynomial never forces complex roots. Second, eigenvectors from distinct eigenvalues are automatically orthogonal (Lemma 9.1). The proof computes $\mathbf{v}_1\cdot A\mathbf{v}_2$ two ways for eigenvectors with $\lambda_1\neq\lambda_2$: on one hand $\mathbf{v}_1\cdot A\mathbf{v}_2=\lambda_2(\mathbf{v}_1\cdot\mathbf{v}_2)$, and on the other, using $A^{T}=A$, $\mathbf{v}_1\cdot A\mathbf{v}_2=(A\mathbf{v}_1)\cdot\mathbf{v}_2=\lambda_1(\mathbf{v}_1\cdot\mathbf{v}_2)$; subtracting gives $(\lambda_1-\lambda_2)(\mathbf{v}_1\cdot\mathbf{v}_2)=0$, so $\mathbf{v}_1\cdot\mathbf{v}_2=0$. These two facts drive Proposition 9.1: if a symmetric $A$ is diagonalizable then it is orthogonally diagonalizable -- its distinct eigenspaces are already mutually perpendicular, so you only normalise the eigenvectors in each one-dimensional eigenspace and run Gram--Schmidt inside any higher-dimensional eigenspace to finish an orthonormal eigenbasis.

Remark 9.3 turns the theory into a recipe for a symmetric $A$: (1) find the eigenvalues from $\det(A-\lambda I)=0$; (2) for each eigenvalue find its eigenvectors; (3) orthonormalise -- normalise within each one-dimensional eigenspace, and apply Gram--Schmidt inside any repeated eigenspace (distinct eigenspaces are already orthogonal by Lemma 9.1); (4) place the orthonormal eigenvectors as the columns of $Q$ and the matching eigenvalues, in the same order, on the diagonal of $D$. Then $A=QDQ^{T}$. A crucial caveat for checking answers: $Q$ is far from unique. You may replace any column $\mathbf{q}_i$ by $-\mathbf{q}_i$ (still a unit eigenvector), reorder the columns (reordering $D$ to match), or, inside a repeated eigenspace, rotate to any other orthonormal basis. The diagonal matrix $D$ is unique up to the order of its entries -- it is literally the list of eigenvalues -- but there is no single 'correct' $Q$. This is why the reconstructed product $A=QDQ^{T}$ and the (ordered) eigenvalues are well-defined answers, while '$Q$' is not.