Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 18

A row's squared Euclidean norm is $[(1+\sqrt{3}{)}^2+1+(1-\sqrt{3}{)}^2]/6=3/2$, and the dot product of the two rows is $[2(1-3)+1]/6=-1/2$. Thus the row Gram matrix isFor the column Gram matrix, the first and third squared column norms are $4/3$, the middle one is $1/3$, and the respective cross-products are $1/3,-2/3,1/3$. HenceThe two products are square matrices of different sizes; their unequal dimensions are consistent with $A$ being rectangular.

For a matrix with equal diagonal entries, the symmetric and antisymmetric directions $(1,1)$ and $(1,-1)$ are eigenvectors. Direct multiplication of $A{A}^{\mathsf{T}}$ gives eigenvalues $3/2-1/2=1$ and $3/2+1/2=2$. Dividing each direction by its Euclidean norm givesThe two normalized eigenvectors are orthogonal; either may independently be replaced by its negative.

In $A^{\mathsf{T}}A$, every row sums to one, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The antisymmetric direction $(1,0,-1)$ gives eigenvalue $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the kernel of a linear map. Normalizing all three gives the complete systemThey form an orthonormal basis of ${\mathbb{R}}^3$. The extra zero eigenvalue reflects the one-dimensional kernel of a linear map for this rank-two rectangular matrix. The question's symbols ${\sigma }_m$ denote the eigenvalues of $A^{\mathsf{T}}A$; the nonzero singular values of $A$ are their square roots, namely $1$ and $\sqrt{2}$.

Using the signs chosen above, explicit multiplication givesThus the nonzero cases obey $A{\mathbf{v}}_m=\sqrt{{\sigma }_m}{\mathbf{u}}_m$. This relates the right singular vectors and left singular vectors in a singular value decomposition. For example, if $U=({\mathbf{u}}_1{\mathbf{u}}_2)$ and $V=({\mathbf{v}}_1{\mathbf{v}}_2{\mathbf{v}}_3)$, then $A=U\Sigma V^{\mathsf{T}}$ with $\Sigma =\left(\begin{array}{ccc}1& 0& 0\\ 0& \sqrt{2}& 0\end{array}\right)$.