Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5

A square matrix is a diagonalizable matrix over the chosen field when there are an invertible matrix $P$ and a diagonal matrix $D$ such that $A=PD{P}^{-1}$. If $v_j$ is column $j$ of $P$ and $D_{jj}={\lambda }_j$, the identity $AP=PD$ gives $A{v}_j={\lambda }_j{v}_j$. Invertibility makes these $n$ columns linearly independent eigenvectors.

Conversely, put $n$ linearly independent eigenvectors in the columns of $P$ and their eigenvalues in $D$. Then $P$ is invertible and $AP=PD$, givingRepeated eigenvalues are allowed; the requirement is a complete eigenbasis, not distinctness of the eigenvalues.

A single eigenbasis works for every choice of the parameters. The columns $v_1=(1,0,0{)}^T$, $v_2=(1,1,0{)}^T$, $v_3=(1,1,1{)}^T$ satisfyThey are linearly independent because their column matrix has determinant one. Thus the diagonalization isThis formula includes all coincidences among $a,b,c$, so no special repeated-eigenvalue cases need separate bases.

The diagonalization in part (b) gives $M^k=P\mathrm{diag}(a^k,b^k,c^k)P^{-1}$. HenceSufficiency follows from the scalar powers tending to zero; necessity also follows immediately from the diagonal entries of $M^k$. Under these conditions the matrix geometric series converges, and its sum isWriting $A_0=1/(1-a)$, $B_0=1/(1-b)$ and $C_0=1/(1-c)$, all entries are explicitlyFor example, $B_0-A_0=(b-a)/[(1-a)(1-b)]$ and $C_0-B_0=(c-b)/[(1-b)(1-c)]$. These expressions remain valid when parameter values coincide.

Expand the quadratic form of this Gram matrix:For $\mathbf{c}\ne 0$, linear independence of the real vectors implies ${\sum }_i{c}_i{\mathbf{x}}_i\ne 0$, so its squared Euclidean norm is strictly positive. Thusproving the positive definiteness of a Gram matrix in the independent-vector case.