Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 1 (Row spaces and commuting matrices)

Write $A=(a_{ij})$ as an $m\times r$ matrix and $B$ as an $r\times n$ matrix, with rows $b_1,\dots ,b_r$. The rule for matrix multiplication givesEvery row of $AB$ is therefore a linear combination of rows of $B$. Taking linear spans gives

When $A$ is an invertible matrix, $B=A^{-1}(AB)$. Apply i first to $AB$ and then to $A^{-1}(AB)$:Thus the two row spaces are equal. In particular, invertible left matrix multiplication preserves row rank.

For any matrix $Y$ and compatible invertible matrix $P$, the linear map $r\mapsto rP$ maps $\mathrm{rowspace}(Y)$ onto $\mathrm{rowspace}(YP)$: a linear combination of rows of $Y$ is sent to the same linear combination of rows of $YP$. Its inverse function is $s\mapsto s{P}^{-1}$, so these row spaces have equal dimensions. Hence right matrix multiplication by $P$ also preserves row rank.

If $X=P^{-1}YP$, ii now givesThis proves the required matrix similarity invariant without needing to assume equality of row rank and column rank.

For the given real matrix $M$, direct matrix multiplication gives the eigenvectorsPut these eigenvectors into the columns ofSince $\det P=-2\ne 0$, $P$ is an invertible matrix. The three column identities give $MP=PD$, soIn particular, the matrix similarity is established by an explicit change-of-basis matrix.

Write $C=(c_{ij})$ and $(d_1,d_2,d_3)=(0,2,4)$. Because $D$ is a diagonal matrix,Thus $CD=DC$ exactly when $(d_j-d_i)c_{ij}=0$ for every $i,j$. The three eigenvalues are distinct, so every off-diagonal entry of $C$ must be zero; diagonal entries are unrestricted. Consequently the centralizer of a diagonal matrix here isEvery displayed matrix indeed commutes with $D$, since diagonal matrices multiply entry by entry on the diagonal.

Matrix similarity transports commuting operators: with $M=PD{P}^{-1}$,By ii, all such matrices are thereforeFor completeness, matrix multiplication givesThe parameterization is injective, since matrix multiplication by $P^{-1}$ and $P$ recovers the parameter diagonal matrix. Thus there are infinitely many choices. Already the distinct scalar matrices $t{I}_3$, $t\in \mathbb{R}$, give an infinite family.