Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices

In the setting of Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices, let $n\in\mathbb{N}$, $A\in\mathcal{M}_{n}(\mathbb{R})$, $\lambda\in\mathbb{R}$ and $\mu,\nu,x\in\mathbb{R}^{n}$, with square real matrices, their entries and entrywise operations, the transpose, the matrix product, the matrix-vector product, the identity matrix $I_{n}$, the standard basis vectors $e_{k}$, invertible matrices, the determinant $\det$, the components of points of $\mathbb{R}^{n}$ as in Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations §membership, the dot product, the diagonal matrices $\operatorname{diag}(\mu)$, eigenvectors and eigenvalues, the set $\mathrm{O}(n)$ of orthogonal matrices and the set $\mathcal{S}(n)$ of symmetric matrices.

¶ The following are equivalent: $\lambda$ is an eigenvalue of $A$; $A-\lambda I_{n}$ is not invertible; $\det(A-\lambda I_{n})=0$.

¶ $\operatorname{diag}(\mu)^{\top}=\operatorname{diag}(\mu)$; $\operatorname{diag}(\mu)\operatorname{diag}(\nu)=\operatorname{diag}(\mu\nu)$, where $\mu\nu\in\mathbb{R}^{n}$ is the pointwise product, $(\mu\nu)_{k}=\mu_{k}\nu_{k}$; $(\operatorname{diag}(\mu)x)_{k}=\mu_{k}x_{k}$ for every $k\in[n]$; $x\cdot(\operatorname{diag}(\mu)x)=\sum_{k=1}^{n}\mu_{k}x_{k}^{2}$; and $\operatorname{diag}(1)=I_{n}$, where $1=(1,\dots,1)\in\mathbb{R}^{n}$ is the constant map with value $1$.

¶ For every $k\in[n]$, $e_{k}$ is an eigenvector of $\operatorname{diag}(\mu)$ with eigenvalue $\mu_{k}$; and every eigenvalue of $\operatorname{diag}(\mu)$ equals $\mu_{k}$ for some $k\in[n]$.

¶ Let $Q\in\mathrm{O}(n)$, and write $QAQ^{\top}$ for $(QA)Q^{\top}$. Then $x$ is an eigenvector of $A$ with eigenvalue $\lambda$ if and only if $Qx$ is an eigenvector of $QAQ^{\top}$ with eigenvalue $\lambda$; and $\lambda$ is an eigenvalue of $A$ if and only if $\lambda$ is an eigenvalue of $QAQ^{\top}$.

¶ If $A\in\mathcal{S}(n)$, $\lambda'\in\mathbb{R}$, $\lambda'\neq\lambda$, $x$ is an eigenvector of $A$ with eigenvalue $\lambda$, and $y\in\mathbb{R}^{n}$ is an eigenvector of $A$ with eigenvalue $\lambda'$, then $x\cdot y=0$.