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Under an orthogonal change of coordinates represented by $Q$, the component matrix changes byThus $T^{\prime }$ is similar to $T$, so its characteristic polynomial, eigenvalues, and their algebraic multiplicities are unchanged.

If $T$ is symmetric, the spectral theorem for real symmetric matrices makes it diagonalizable, so algebraic and geometric multiplicities agree. Consequently the dimensions of its eigenspaces are also independent of the coordinate frame.

For the stated tensor, choose a unit eigenvector $n$ belonging to the simple eigenvalue $\mu$. Its perpendicular plane is the eigenspace of $\lambda$, soHence

Let $A$ be the adjacency matrix of a graph. Since $G$ is a regular graph of degree $d$, every row of $A$ sums to $d$. Therefore, for $\mathbf{e}=(1,\dots ,1{)}^T$,so $d$ is a graph eigenvalue.

If $Av=dv$, then the quadratic form of the Graph Laplacian $L=dI-A$ givesEvery summand is nonnegative, so $v_i=v_j$ along every edge. The graph is connected, hence all coordinates of $v$ are equal and $v$ is a scalar multiple of $\mathbf{e}$. Thus the $d$-eigenspace is one-dimensional. Since $A$ is a symmetric matrix, the spectral theorem for real symmetric matrices makes it diagonalizable, so the algebraic multiplicity is also one. This proves the constant eigenvector of a regular graph result.

Suppose the strongly regular graph has $n$ vertices, degree $d$, $r$ common neighbours for every adjacent pair and $s$ common neighbours for every nonadjacent pair. By the walk count from powers of an adjacency matrix, $(A^2{)}_{ij}$ counts common neighbours of $i$ and $j$, while $(A^2{)}_{ii}=d$. Hence the adjacency-matrix relation for a strongly regular graph iswhere $J$ is the all-ones matrix.

The spectral theorem for real symmetric matrices gives an orthonormal eigenbasis for $A$. Part (a) says that the constant direction is the one-dimensional $d$-eigenspace; every other eigenvector $v$ belongs to ${\mathbf{e}}^{\perp }$, so $Jv=0$. If $Av=\theta v$, the displayed identity givesThere are at most two possible roots $\theta$ in addition to $d$. Thus $G$ has at most three distinct eigenvalues.

Because every column of $X$ has sample mean zero, every column $u_i$ of $U=XV$ also has sample mean zero. The spectral theorem for real symmetric matrices givesHence for $i\ne j$ the sample covariance is $n^{-1}{u}_i^T{u}_j=0$, while the sample variance of $u_i$ isThus the $u_i$ are pairwise uncorrelated sample principal components, ordered by decreasing sample variance.

By the spectral theorem for real symmetric matrices, write $A=O^{t}DO$ with $O$ orthogonal and $D=\mathrm{diag}({\xi }_1,\dots ,{\xi }_n)$. Part a and the tensor-product property of the Fourier transform giveThusApplying the pullback rule from part c to the orthogonal change of variables, for which $|\det O|=1$, replaces $D$ by $A$ and $D^{-1}$ by $A^{-1}$. Since determinant and signature are invariant under orthogonal conjugation,

Let the columns of a centered design matrix $X\in {\mathbb{R}}^{n\times p}$ be variables and write $X^{T}X=V\Lambda V^T$ by the spectral theorem for real symmetric matrices. The columns $u_i$ of $U=XV$ are the sample principal-component score vectors. They satisfy $u_i^T{u}_j={\Lambda }_{ij}$, so distinct score vectors have zero sample covariance and $u_i$ has sample variance ${\Lambda }_{ii}/n$ under the divisor-$n$ convention.