Past exam of the mathematics course of the University of Oxford / 2014 / Prelims / Mathematics I / 4

A real number $\lambda$ is an eigenvalue of $A$ if there is a nonzero vector $v$ with $Av=\lambda v$; such a $v$ is an eigenvector for $\lambda$. The nonzero condition is essential.

For distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_r$, choose corresponding nonzero eigenvectors $v_1,\dots ,v_r$. Prove their linear independence by induction on $r$. The claim for one vector is immediate. Given a relation ${\sum }_{i=1}^r{a}_i{v}_i=0$, apply $A-{\lambda }_{r}I$ to getBy the induction hypothesis, every coefficient here is zero. Distinctness gives $a_i=0$ for $i<r$, and the original relation then gives $a_r=0$ since $v_r\ne 0$. Thus the eigenvectors are linearly independent.

With $n$ distinct real eigenvalues, the $n$ real eigenvectors form a basis of ${\mathbb{R}}^n$. Put them into the columns of $P$. Then $P$ is invertible andThis proves diagonalizability directly.

Suppose $A^T=A$. For eigenvectors $v_i,v_j$ with distinct real eigenvalues, symmetry givesThus $({\lambda }_i-{\lambda }_j)v_i^T{v}_j=0$, and the eigenvectors are orthogonal. Divide each by its nonzero Euclidean norm to obtain an orthonormal basis. The matrix $P$ having these normalized columns satisfies $P^{T}P=I$, so it is an orthogonal matrix and $P^{-1}=P^T$. As in part (a),This proves the needed orthogonal diagonalization from symmetry and distinct eigenvalues, without invoking an unproved spectral theorem.

For the given symmetric matrix, $(1,1{)}^T$ is an eigenvector with eigenvalue $1$, and $(1,-1{)}^T$ is an eigenvector with eigenvalue $27$. Normalize them to obtainTake the real cube roots of the two diagonal entries and transform back:It is a real symmetric matrix, and $C^3=P\mathrm{diag}(1,27)P^T=A$ because $P^{T}P=I$. Direct multiplication also gives $C^2=\left(\begin{array}{cc}5& -4\\ -4& 5\end{array}\right)$ and then $C^3=\left(\begin{array}{cc}14& -13\\ -13& 14\end{array}\right)$.