Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 6

Complete the square in the quadratic form:The real invertible linear transformationsets $x_1+2x_2=y_1$, $x_2=y_2$, $\sqrt{6}{x}_3=y_3$, giving $Q_1=y_1^2+y_2^2+y_3^2$. The same transformation givesIts signature is $(2,1)$. By Sylvester's law of inertia no real invertible change of variables can turn this second form into a sum of three positive squares.

The trace of a square matrix is $\mathrm{tr}A={\sum }_i{A}_{ii}$. Expanding a product and interchanging its finite sums givesHence every commutator has trace zero. Taking the trace of $AB-BA=cI$ gives $0=cn$. Over the complex numbers, with $n\ge 1$, this forcesFor that value, choose $A=B=0$, or any commuting pair. For every nonzero $c$ the trace contradiction proves nonexistence. The finite matrix size is essential to this trace argument.

Differentiate the eigenvalue equation along the assumed smooth normalized eigenbranch:Since $A$ is real symmetric, $x^{T}A=\lambda x^T$. Multiplying the differentiated equation by $x^T$ and using $x^{T}x=1$ cancels the terms containing $x^{\prime }$:Thus the finite-dimensional Hellmann–Feynman theorem givesEquivalently, differentiating the normalization gives $2x^T{x}^{\prime }=0$ and eliminates those terms from the derivative of $\lambda =x^{T}Ax$. The proof needs the smooth eigenbranch assumed in the question; the smallest eigenvalue need not be differentiable at an arbitrary crossing without that assumption.