Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 2 / 17

A left eigenvector satisfies $(M-\lambda I{)}^T\mathbf{x}=0$. Here $M$ is a symmetric matrix, so its left and right eigenspaces have the same components. The characteristic polynomial factors asThe eigenvalues and all corresponding nonzero row eigenvectors areFor example, in the $xy$ block, equal components give the eigenvalue $3$ and opposite components give $-1$, while the $z$ coordinate is unaffected.

The determinant is multiplicative under matrix multiplication:This also holds when $B$ is singular, in which case both sides vanish.

Multiplying an $n\times n$ matrix by $\alpha$ scales every row, giving a factor ${\alpha }^n$ in its determinant. Also $|A^{-1}|=1/|A|$, since $A{A}^{-1}=I$. HenceOnly $A$ needs to be invertible; the result remains valid when $\alpha =0$.

Let $D_{\beta }=\mathrm{diag}({\beta }_1,\dots ,{\beta }_n)$ and $D_{\gamma }=\mathrm{diag}({\gamma }_1,\dots ,{\gamma }_n)$. The specified entries mean $C=D_{\beta }A{D}_{\gamma }$. Multiplication of determinants therefore givesThis is also the direct row- and column-scaling rule, and it includes zero scaling factors.

The eigenvector equations $Q{\mathbf{e}}_i={\lambda }_i{\mathbf{e}}_i$ combine column by column into $QE=ED$. Since $|E|\ne 0$, multiplication by the inverse matrix gives the diagonalizationThe nonzero determinant of $E$ ensures that its columns form an eigenbasis.

Successive substitution in the matrix recurrence relation gives $V_k=V_0{Q}^k$. In the matrix power of $Q=ED{E}^{-1}$, every adjacent $E^{-1}E$ cancels, so $Q^k=E{D}^k{E}^{-1}$. ConsequentlyHere $D^k=\mathrm{diag}({\lambda }_1^k,\dots ,{\lambda }_n^k)$ for $k\ge 1$, and $D^0=I$. The formula also gives $V_0$ correctly at $k=0$.