Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 4 / 8G (Linear Algebra)

Two matrices are equivalent when $A^{\prime }=PAQ$ for invertible matrices $P$ and $Q$ of the appropriate sizes. A change of basis in the codomain multiplies a matrix on the left, while a change of basis in the domain multiplies it on the right. Hence two matrices of the same linear map in two pairs of bases are equivalent.

Conversely, start with the map $\alpha :F^n\to F^m$ having matrix $A$ in the standard bases. Given $A^{\prime }=PAQ$, choose domain and codomain bases whose change-of-coordinate matrices produce $Q$ and $P$; then $A^{\prime }$ represents the same map in those bases. This proves the equivalence.

The column rank of $A$ is the dimension of the span of its columns, and its row rank is the dimension of the span of its rows. If $A$ represents $\alpha :F^n\to F^m$, its column rank is $\mathrm{rank}\alpha$. The transpose represents the dual mapso the row rank is $\mathrm{rank}{\alpha }^{\ast }$.

If $\mathrm{rank}\alpha =r$, thenwhose dimension is $m-r$: extend a basis of $\mathrm{im}\alpha$ to one of $F^m$ and use the dual basis. Rank-nullity now gives $\mathrm{rank}{\alpha }^{\ast }=r$. This proves the equality of row rank and column rank.

For $I\subseteq [m]$, the independent vectors $\{v_i:i\in I\}$ lie in the coordinate subspace supported onwhich has dimension $|S_I|$. Hence $|I|\le |S_I|$, and the stated form of Hall marriage theorem gives an injection $f$ with $f(i)\in \mathrm{supp}(v_i)$.

Let $V$ be the $n\times m$ matrix whose $i$th column is $v_i$. Its column rank is $m$. By equality of row and column rank, it has $m$ linearly independent rows. Let $B$ be their indices. The $m\times m$ submatrix $V_B$ is invertible, so its columns are independent. Those columns are exactly the nonzero coordinates of $B{v}_1,\dots ,B{v}_m$, and therefore these truncated vectors are linearly independent.

Expanding $\det V_B$ gives a permutation $f:[m]\to B$ for whichThus $f(i)\in \mathrm{supp}(v_i)$. Finally, order the coordinates with $B$ first. The matrix whose columns are the $v_i$ together with the $e_j$ for $j\notin B$ is block triangular, with diagonal blocks $V_B$ and an identity matrix. Its determinant is nonzero. Consequentlyis a basis of ${\mathbb{C}}^n$. This is the simultaneous basis exchange from a nonzero minor.