Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 17G (Linear Mathematics)

Use the Leibniz definition of the determinant:For the permuted-column matrix, $a_{ij}^{\sigma }=a_{i,\sigma (j)}$. Reindex the sum by $\tau =\sigma \circ \pi$. The sign of a permutation is multiplicative, so $\mathrm{sgn}(\pi )=\mathrm{sgn}(\sigma )\mathrm{sgn}(\tau )$. It follows directly thatFor the matrix transpose, the products in its determinant are ${\prod }_i{a}_{\pi (i),i}$. Set $j=\pi (i)$ to obtain ${\prod }_j{a}_{j,{\pi }^{-1}(j)}$, and use $\mathrm{sgn}({\pi }^{-1})=\mathrm{sgn}(\pi )$. This proves $\det (A^t)=\det A$. These facts also show that equal columns, or equal rows by transposition, force the determinant to be zero: exchanging the equal pair leaves the matrix unchanged but changes the determinant's sign.

Let $M_{ij}$ be the matrix obtained by deleting row $i$ and column $j$, with the empty determinant taken as one for $n=1$. Define the cofactor $C_{ij}=(-1{)}^{i+j}\det M_{ij}$ and the adjugate matrix by $\mathrm{adj}(A{)}_{ij}=C_{ji}$. Grouping the Leibniz sum according to the column chosen in row $i$ gives the cofactor expansion $\det A={\sum }_k{a}_{ik}{C}_{ik}$: deleting that row and its chosen column leaves a permutation of $n-1$ entries, with sign factor $(-1{)}^{i+k}$. ThereforeWhen $i=j$ this is $\det A$. When $i\ne j$ it is the cofactor expansion along row $j$ of the matrix formed by replacing that row with row $i$; its remaining minors are unchanged, and its two equal rows give determinant zero. The column version proves the other product in the same way. Thus the adjugate identity is

For the real matrices $C,D$, let $p(t)=\det (C+tD)$. The determinant definition makes this a real polynomial of degree of a polynomial at most $n$. Since $C+iD$ is invertible, the allowed converse implies $p(i)\ne 0$, so $p$ is not the zero polynomial. It has at most $n$ real roots of a polynomial. Choose a real $\lambda$ outside those roots; the adjugate argument proves that $C+\lambda D$ is invertible.

Finally write the given complex similarity transformation as $P=C+iD$. The relation $AP=PB$, with $A,B$ real, yields $AC=CB$ and $AD=DB$ by comparing real and imaginary parts. Choose the preceding $\lambda$ and put $Q=C+\lambda D$. Then $AQ=QB$, soThis proves real similarity from complex similarity by constructing a real intertwiner, rather than requiring $C$ or $D$ separately to be invertible.