Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 2

For real scalars $\alpha ,\beta$ and matrices $A,C,B$, distributivity of matrix multiplication and linearity of the trace giveThus $\phi$ is linear in its first argument.

Expand the trace by matrix entries:Hence $\phi (A,B)=\phi (B,A)$. Combined with part (a), this makes $\phi$ a symmetric bilinear form.

For $n\ge 2$, take the matrix whose leading $2\times 2$ block is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$ and whose other entries are zero. Its square has leading block $-I_2$, soThis contradicts the positive definiteness required of an inner product. Even the nonzero matrix unit $E_{12}$ has $\phi (E_{12},E_{12})=0$.

The printed assertion has an exception: when $n=1$, $\phi (a,b)=ab$ is an inner product on $\mathbb{R}$. Thus the requested failure requires $n\ge 2$, a restriction absent from the printed question.

Interpret diagonalizability over the specified underlying field $\mathbb{R}$. Write $A=P\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)P^{-1}$ with real eigenvalues. Invariance of the trace under similarity givesIf $A\ne 0$ is diagonalizable, at least one eigenvalue is nonzero, proving $\phi (A,A)>0$. Real diagonalizability is essential: the rotation block from part (c) is diagonalizable over $\mathbb{C}$ but has eigenvalues $i,-i$ and negative trace of its square.

Let $E_{ji}$ be the matrix unit having its only nonzero entry, one, in position $(j,i)$. Direct multiplication gives $\mathrm{tr}(A{E}_{ji})=a_{ij}$. If $A\in R$, every such pairing vanishes, forcing every entry $a_{ij}$ to be zero. Conversely, zero pairs to zero with every matrix. ThusThe matrix trace pairing is therefore a nondegenerate bilinear form, despite its failure to be positive definite when $n\ge 2$.