Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 1 / 5A (Vectors and Matrices)

The matrix trace is the sum of the diagonal entries:Using the definitions of matrix multiplication and matrix transpose,This is the symmetry of the real Frobenius inner product.

The matrix trace calculation in (i) givesA sum of nonnegative real numbers is zero exactly when each is zero. Thus $\boxed{\mathrm{tr}(A^{T}A)=0⟺A=0}$, so the real Frobenius inner product is positive-definite.

Write $\alpha =\mathrm{tr}(A^{T}A)$, $\beta =\mathrm{tr}(B^{T}B)$ and $\gamma =\mathrm{tr}(A^{T}B)$. If $B\ne 0$, then $\beta >0$, and applying the same nonnegativity to $A-tB$ givesTaking $t=\gamma /\beta$ yields $\alpha -{\gamma }^2/\beta \ge 0$. If $B=0$, the claimed inequality is immediate. Hence the Cauchy-Schwarz inequality here isWhen $B\ne 0$, equality holds exactly when $\mathrm{tr}((A-(\gamma /\beta )B{)}^T(A-(\gamma /\beta )B))=0$, hence exactly when $A=(\gamma /\beta )B$. Together with the case $B=0$, the complete equality condition is $\boxed{A,B\text{ are linearly dependent}}$, including the case that either is zero.

Writing $X=\left(\begin{array}{cc}x& y\\ z& w\end{array}\right)$ and using the Frobenius inner product, the three linear equations areThe first two differ by twice the third; equivalently $A-B=2C$. Thus there are only two independent constraints. Substituting $z=-w$ gives $y=2w-x$, so every element of the vector subspace has the unique expressionThe two coefficient matrices satisfy all three constraints. Their linear independence follows by looking first at entry $(1,1)$ and then at entry $(2,2)$. Therefore a basis for this orthogonal complement is