Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 2 / 11

Solving $(A-4I)\mathbf{v}=0$ gives $v_2=0$ and $v_1=v_3$. A convenient normalized eigenvector isThe displayed nonzero vector proves that four is an eigenvalue. Every nonzero multiple represents the same eigendirection.

A vector perpendicular to ${\mathbf{e}}_1$ has the form $\mathbf{v}=(s,t,-s)$. Direct matrix multiplication givesThus every nonzero such vector is an eigenvector with $\boxed{\lambda =-2}$. The zero vector must be excluded from the literal claim about “any” perpendicular vector, since an eigenvector is nonzero. The eigenspace is two-dimensional. Equivalently, $A=-2I+6{\mathbf{e}}_1{\mathbf{e}}_1^{\mathsf{T}}$ exhibits the spectrum of a scalar matrix plus a rank-one projector.

Choose two perpendicular unit directions in the $-2$ eigenspace:Their pairwise dot products vanish and their Euclidean norms are one. They form an orthonormal basis of eigenvectors, with eigenvalues $4,-2,-2$, as guaranteed for a real symmetric matrix by the spectral theorem for real symmetric matrices.

The trace is the sum of the eigenvalues, counting multiplicity, while the determinant is their product:A direct expansion along the middle row checks the latter: $(-2)(1-9)=16$.

Using the ordered orthonormal basis chosen in (c), the column matrix isOther signs or another orthonormal choice within the repeated eigenspace are equally valid, provided the same choice is retained in the remaining parts.

The $(i,j)$ entry of $B^{\mathsf{T}}B$ is ${\mathbf{e}}_i\cdot {\mathbf{e}}_j={\delta }_{ij}$. Hence $B$ is an orthogonal matrix, whose matrix inverse equals its transpose:This obtains the inverse without unnecessary cofactor calculations.

The eigenvector equations give $AB=B\mathrm{diag}(4,-2,-2)$. Multiplying on the left by $B^{-1}$ therefore givesThis is an orthogonal diagonalization: the linear map acts independently by multiplication by its eigenvalues in the new orthonormal basis. The repeated eigenvalue explains the freedom to rotate the last two basis directions.