Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 5

Let $v_1,\dots ,v_n$ be the linearly independent eigenvectors, and form $V=(v_1\cdots v_n)$. This matrix is invertible and $AV=VD$, where $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$. Hencewhich proves diagonalizability.

If $B=P^{-1}AP$, then $P^{-1}{v}_j$ are linearly independent eigenvectors of $B$ with the same eigenvalues. Explicitly, with $W=P^{-1}V$,Thus matrix similarity preserves diagonalizability.

The spectral theorem for a Hermitian matrix supplies a unitary matrix $U$ and real diagonal matrix $D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ with $A=UD{U}^{†}$. This single representation gives all three conclusions below.

For a positive integer $k$, cancellation of $U^{†}U=I$ gives $A^k=U{D}^k{U}^{†}$. The trace is invariant under matrix similarity, since $\mathrm{tr}(XY)=\mathrm{tr}(YX)$, and the determinant is multiplicative. ThereforeThis includes zero eigenvalues without requiring $A$ to be invertible.

The convergent power series defining the matrix exponential gives $e^A=U{e}^D{U}^{†}$, with $e^D=\mathrm{diag}(e^{{\lambda }_1},\dots ,e^{{\lambda }_n})$. Hence the matrix exponential determinant identity isThe diagonalization proof uses the given Hermitian hypothesis; the identity itself holds for every square real or complex matrix.

Applying $A^2=A$ to an eigenvector gives ${\lambda }_j^2={\lambda }_j$, so each eigenvalue is either zero or one. If all are one, the spectral theorem gives $A=I$, hence $\det A=1$ and $\mathrm{tr}A=n$. Otherwise, if $m$ eigenvalues are one, with $0\le m<n$, thenThus the trace of this orthogonal projection equals its rank, including $m=0$ for the zero matrix.