Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 4 (Subspaces of diagonalisable matrices)

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

Let $A$ be a skew-symmetric matrix. For every real vector $x$, the real scalar $x^{T}Ax$ equals its own transpose, butHence $x^{T}Ax=0$. If $Av=\lambda v$ is a real eigenvector equation with $v\ne 0$, it follows thatso $\lambda =0$. If $A$ is a diagonalisable matrix over $\mathbb{R}$, it has a real basis of eigenvectors, all with eigenvalue zero. Thus $A$ vanishes on a basis and $\boxed{A=0_n}$.

The field matters: a nonzero real skew-symmetric matrix can be a diagonalisable matrix over the complex numbers, for example $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ with eigenvalues $i,-i$.

The matrix units $E_{ij}$, $1\le i,j\le n$, form a basis of $M_n(\mathbb{R})$: each matrix has the unique expression $A={\sum }_{i,j}{a}_{ij}{E}_{ij}$. There are $n^2$ of them, so

Take the vector subspace of skew-symmetric matrices,It is closed under vector addition and scalar multiplication, and a basis isIndeed, its diagonal entries vanish, and its entries below the diagonal are forced by those above it. Hence $\boxed{\dim K=\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. Part (a)(ii) proves that the only diagonalisable matrix in $K$ is $0_n$.

Let $S$ be the vector subspace of symmetric matrices. A basis consists of $E_{ii}$ for $1\le i\le n$ and $E_{ij}+E_{ji}$ for $i<j$, soThe dimension formula for a sum of subspaces now givesChoose $0\ne A\in V\cap S$. It belongs to $V$ and is a symmetric matrix, so part (a)(i) makes it a diagonalisable matrix. The preceding example shows that the strict threshold is sharp.

Write $r=(r_1,\dots ,r_n{)}^T$. The linear map represented by $A$ has the formThe image of a linear map here is contained in the one-dimensional linear span of $e_i$, so its matrix rank is at most one. The rank-nullity theorem yieldsThus the solution space of $Ax=0$, namely the zero eigenspace when zero is an eigenvalue, contains at least $n-1$ linearly independent vectors. If $r=0$ its dimension is $n$; otherwise it is $n-1$. When $n=1$ the asserted lower bound is zero, even if zero is not an eigenvalue.

Since $A{e}_i=r_i{e}_i$, the vector $e_i$ is an eigenvector with nonzero eigenvalue $r_i$. The assumption $r_i\ne 0$ also makes $r\ne 0$, so the kernel has dimension $n-1$. Choose its basis $u_1,\dots ,u_{n-1}$.

The vector $e_i$ is outside this kernel. Hence $(u_1,\dots ,u_{n-1},e_i)$ is a basis of ${\mathbb{R}}^n$ consisting of eigenvectors, with corresponding eigenvalues $0,\dots ,0,r_i$. In that basis, $A$ is the diagonal matrixThis proves that $A$ is a diagonalisable matrix, including the case $n=1$.

Yes. Consider the following $n^2$ matrices:Each has only row $i$ nonzero and has $i$th diagonal entry $1$, so part (c)(ii) proves that it is a diagonalisable matrix. The collection is a spanning set, because it contains every $E_{ii}$ andThus it spans the matrix unit basis. A spanning set of $n^2$ vectors in an $n^2$-dimensional vector space is a basis. Alternatively, in a vanishing linear combination of $\mathcal{B}$, each off-diagonal entry first forces its own coefficient to be zero, and then each diagonal entry forces the remaining coefficient to be zero.