Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Digital systems, linear algebra, introduction to formal proof / 4

Write $\hat{A}=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ and $v_i=P^{-1}{e}_i$, where $e_i$ are the standard basis vectors. Since $A{P}^{-1}=P^{-1}\hat{A}$,Thus the columns of $P^{-1}$ form an eigenbasis, and the requested ordered basis is $D=(P^{-1}{e}_1,\dots ,P^{-1}{e}_n)$. Its basis matrix is $P^{-1}$, giving $[t{]}_D=PA{P}^{-1}=\hat{A}$. More completely, for each distinct eigenvalue $\lambda$, its eigenspace isAll nonzero vectors in these eigenspaces are eigenvectors; when an eigenvalue repeats, they need not be individual columns of $P^{-1}$.

Let $PA{P}^{-1}=\mathrm{diag}({\alpha }_1,\dots ,{\alpha }_n)$ and $PB{P}^{-1}=\mathrm{diag}({\beta }_1,\dots ,{\beta }_n)$. ThenBy matrix similarity, every eigenvalue of $A+B$ is ${\alpha }_i+{\beta }_i$ for some $i$. These are corresponding eigenvalues on the common eigenbasis, rather than every possible pairwise sum. This is eigenvalue addition in a common eigenbasis.

With the two diagonal matrices from the preceding part, write $D_A=PA{P}^{-1}$ and $D_B=PB{P}^{-1}$. Their diagonal products agree entry by entry, so $D_A{D}_B=D_B{D}_A$. ThereforeHence simultaneous diagonalization implies that $A,B$ are commuting operators.

Using $BA=AB$ and linearity,This establishes that commuting maps preserve eigenspaces. The vector $Av$ can be zero, so this step alone does not assert that $Av$ is an eigenvector.

Work over the field containing the stated $n$ distinct eigenvalues of $B$; in a real formulation these must be real to obtain real eigenvectors. Choose one eigenvector for each distinct eigenvalue. By independence of eigenvectors for distinct eigenvalues, these $n$ vectors are a basis. Consequently each eigenspace $E_{\lambda }(B)$ has dimension one: in an expansion in this eigenbasis, $Bw=\lambda w$ forces every coefficient with a different eigenvalue to vanish.

Now let $v\ne 0$ be any eigenvector of $B$. By part (d), $Av$ belongs to its one-dimensional eigenspace, so $Av=\mu v$ for some scalar $\mu$. This proves that $v$ is an eigenvector of $A$, including the case $Av=0$, where $\mu =0$. This is the crucial one-dimensional step in simple spectrum gives simultaneous diagonalization.

The $n$ independent eigenvectors $v_1,\dots ,v_n$ of $B$ from part (e) are also eigenvectors of $A$. Put $S=(v_1\cdots v_n)$ and $P=S^{-1}$. ThenThis supplies the required simultaneous diagonalization over the same field as in part (e).

For $n=2$, takeThey are commuting operators because every matrix commutes with the identity matrix. However $A$ has only the eigenvalue zero, and $\ker A=\mathrm{span}\{(1,0{)}^T\}$ has dimension one. It has no eigenbasis, so it cannot appear in any simultaneous diagonalization. Equivalently, a diagonal matrix similar to this nonzero nilpotent matrix would have to be zero, an impossibility. For any prescribed $n\ge 2$, use $A=E_{12}$ and $B=I_n$; its zero eigenspace has dimension $n-1$. For $n=1$ no such example exists.