Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / 14

By the spectral theorem for real symmetric matrices, choose the normalized eigenvectors to form an orthonormal basis. With $P_n={\mathbf{e}}_n{\mathbf{e}}_n^{\mathsf{T}}$, orthonormality gives $P_n{P}_m={\delta }_{nm}{P}_n$ and ${\sum }_n{P}_n=I$. For every vector $\mathbf{x}$, expand $\mathbf{x}={\sum }_n({\mathbf{e}}_n^{\mathsf{T}}\mathbf{x}){\mathbf{e}}_n$ and apply $A$:Since this equality holds for every $\mathbf{x}$, the spectral decomposition isEach $P_n$ is the rank-one orthogonal projection onto its eigenvector direction. In a degenerate eigenspace, normalized eigenvectors must be chosen mutually orthogonal; normalization alone does not guarantee this resolution of the identity.

The matrix trace is linear, andTaking the trace of the spectral decomposition therefore proves

If every eigenvalue is nonzero, define $B={\sum }_n{\lambda }_n^{-1}{P}_n$, where $P_n={\mathbf{e}}_n{\mathbf{e}}_n^{\mathsf{T}}$. Using the mutually orthogonal projection identity $P_n{P}_m={\delta }_{nm}{P}_n$,The same calculation gives $BA=I$. Thus the matrix inverse is

The inverse fails to exist exactly when at least one eigenvalue is zero. In that case a nonzero eigenvector lies in the kernel of a linear map, so $A$ is not one-to-one. Equivalently, its determinant ${\prod }_n{\lambda }_n$ is zero. Conversely, if every eigenvalue is nonzero, the preceding construction gives an inverse.

The third coordinate is decoupled, giving the unit eigenvector $(0,0,1{)}^{\mathsf{T}}$ with eigenvalue $1$. In the first two coordinates, the symmetric direction $(1,1)$ has eigenvalue $3-1=2$, and the antisymmetric direction $(1,-1)$ has eigenvalue $3+1=4$. Their normalized forms areThey are an orthonormal basis of ${\mathbb{R}}^3$; independent sign reversals are equally valid choices.

Insert the eigenvalues and eigenvectors into the inverse spectral decomposition. The first two projections contributeAdding the third projection givesMultiplication by $A$ gives the identity; in particular the upper block satisfies $\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)\left(\begin{array}{cc}3& 1\\ 1& 3\end{array}\right)=8I$.

Summing the diagonal entries givesThe eigenvalues of the inverse are the reciprocals $1/2,1/4,1$ of those of $A$, whose sum is also $7/4$. This is precisely what the trace identity predicts.