Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Digital systems, linear algebra, introduction to formal proof / 5

For square matrices of the same size,For an upper or lower triangular matrix $T$, its determinant is the product of its diagonal entries:These formulas include singular matrices and zero diagonal entries.

Let $e_i$ be standard coordinate columns. The required elementary matrix isIt is the identity matrix with the extra entry $\alpha$ in position $(j,i)$. Left matrix multiplication implements the row operation: all other rows are unchanged. Since $j>i$, $E$ is a lower triangular matrix with every diagonal entry one. Part (a) therefore gives $\det (E)=1$ andHence adding a multiple of one row to another leaves the determinant unchanged, even if $A$ is singular.

Use $-A=(-I_n)A$. The diagonal matrix $-I_n$ has $n$ diagonal entries equal to $-1$, so the product and triangular matrix formulas from part (a) giveEquivalently, multiplying each of the $n$ rows by $-1$ supplies one factor $-1$ per row.

Write the determinant in its Leibniz formula for determinants, where $S_n$ is the set of permutations of $\{1,\dots ,n\}$:The permutation sign is justified by the given row-interchange property: a permutation matrix is obtained from the identity matrix by row interchanges, and every interchange changes its determinant by a factor $-1$. Reversing a sequence of interchanges constructs the inverse permutation with the same number of interchanges. Consequently $\mathrm{sgn}({\sigma }^{-1})=\mathrm{sgn}(\sigma )$. The Leibniz formula for determinants can itself be obtained by expanding the rows multilinearly in the standard basis: repeated selected coordinate rows contribute zero, and the remaining terms are precisely these signed permutation matrices.

Since $(A^T{)}_{ij}=a_{ji}$, the same formula givesReindex the product with $j=\sigma (i)$ and reindex the sum with $\tau ={\sigma }^{-1}$. Both reindexings are bijections, soThus $\det (A)=\det (A^T)$ for every real square matrix, including singular matrices. This proof does not assume the analogous column-operation rules in order to establish the transpose identity.

For a skew-symmetric matrix, $A^T=-A$. By parts (c) and (d),If $n$ is odd, $\det (A)=-\det (A)$, hence $2\det (A)=0$ over $\mathbb{R}$. Therefore $\det (A)=0$, so $A$ is a singular matrix. The conclusion depends on odd dimension: in dimension two, $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ is skew-symmetric and has determinant one.