Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 3 / 10H (Linear Algebra)

The matrix trace is the sum of diagonal entries, $\mathrm{tr}A={\sum }_{i=1}^n{A}_{ii}$. For any two square matrices,Apply this cyclic property of the trace with $B=T$ and $C=A{T}^{-1}$:Thus matrix similarity preserves the matrix trace; no assumption that either matrix is diagonalizable is needed.

Suppose $v=(v_1,\dots ,v_r{)}^t$ belongs to the kernel of the displayed matrix $\Lambda$. Define the polynomial $q(z)={\sum }_{k=1}^r{v}_k{z}^k$. Then $q({\lambda }_j)=0$ for all $j$, and $q(0)=0$. These are $r+1$ distinct roots of a polynomial of degree at most $r$. Repeated application of the factor theorem therefore makes $q$ identically zero, so every $v_k=0$. Hence $\Lambda$ is injective and, by finite-dimensional rank-nullity theorem, invertible. Its transpose is invertible with inverse $({\Lambda }^{-1}{)}^t$, proving both kernels are trivial.

For the second matrix, membership of the all-ones vector in the kernel meansGroup the distinct nonzero values as ${\mu }_1,\dots ,{\mu }_r$, with multiplicities $m_1,\dots ,m_r>0$. Zero values contribute nothing to these positive powers. The first $r$ equations areor $\Lambda (\mu {)}^{t}m=0$. The proved transpose-kernel statement implies $m=0$, contradicting its positive entries if $r>0$. Therefore $r=0$ andConversely, all zero values plainly give a zero matrix. This proves the zero power sums force a finite complex multiset to vanish criterion, including repetitions and zero values rather than assuming all eigenvalues distinct.

Use the permitted triangularization to write $A=TU{T}^{-1}$ with $U$ upper triangular and diagonal ${\lambda }_1,\dots ,{\lambda }_n$. Products of upper-triangular matrices are upper triangular, and their diagonal entries multiply, soThe result of part b makes every ${\lambda }_j$ zero. Thus $U$ is strictly upper triangular. To see directly that $U^n=0$, a potentially nonzero entry in a product of $n$ factors requires a chainof $n+1$ indices in $\{1,\dots ,n\}$, which is impossible. HenceThis is nilpotence from vanishing traces of powers. It proves nilpotence without substituting an unproved diagonalization for the allowed triangularization.