Past exam of the mathematics course of the University of Cambridge / 2007 / ia / Paper 3 / 7D (Algebra and Geometry)

Allow an eigenvector $v\ne 0$ to have complex entries, since real matrices can initially have complex eigenvalues. Write $v^{\ast }$ for its conjugate transpose. Reality of $A$ gives $A^{\ast }=A^T$; symmetry then gives $A^{\ast }=A$. Thus $A$ is a Hermitian matrix. The scalar $v^{\ast }Av$ equals its complex conjugate, because $(v^{\ast }Av{)}^{\ast }=v^{\ast }{A}^{\ast }v=v^{\ast }Av$. If $Av=\lambda v$, thensince $v^{\ast }v>0$. This proves that all eigenvalues are real. The equality $A^{\ast }=A^T$ is precisely where reality was used; a complex symmetric matrix need not be Hermitian, as $iI$ illustrates.

If $Au=\lambda u$ and $Av=\mu v$, then $u^{\ast }Av=\mu u^{\ast }v$, but also $u^{\ast }Av=(Au{)}^{\ast }v=\overline{\lambda }{u}^{\ast }v=\lambda u^{\ast }v$. Consequently $(\mu -\lambda )u^{\ast }v=0$. Distinct eigenvalues force $u^{\ast }v=0$. For real eigenvectors this is ordinary Euclidean orthogonality. These arguments establish real symmetric spectral orthogonality without assuming a diagonalization in advance.

A real orthogonal matrix satisfies $P^{T}P=I$, equivalently $P^{-1}=P^T$. If $A$ is symmetric,so orthogonal similarity preserves symmetry. An arbitrary real invertible change of basis need not do so. For example,The final matrix is not symmetric. Orthogonal similarity preserves real symmetry; general similarity does not.

TakeIts characteristic polynomial is ${\lambda }^2+1$, whose roots are $i$ and $-i$. Thus this real matrix has no real eigenvalues.

Take the Jordan blockIts only eigenvalue is one. An eigenvector satisfies $(C-I)v=0$, so its second coordinate vanishes; the eigenspace is one-dimensional. A diagonalizable matrix of size two requires a basis of two independent eigenvectors, which this matrix lacks even over $\mathbb{C}$.

UseIt has independent complex eigenvectors $(1,-i{)}^T$ and $(1,i{)}^T$, with eigenvalues $i$ and $-i$. Taking these as columns gives an invertible complex matrix $P$ with $P^{-1}DP=\mathrm{diag}(i,-i)$, so $D$ is a diagonalizable matrix over $\mathbb{C}$. It cannot be diagonalized over $\mathbb{R}$, since a real diagonal matrix would have real eigenvalues, whereas similarity preserves the characteristic polynomial.

TakeThe real eigenvalues are one and two, with eigenlines spanned by $(1,0{)}^T$ and $(1,1{)}^T$ respectively. These independent vectors form a real eigenbasis, so $E$ is a diagonalizable matrix over $\mathbb{R}$. Their inner product is one, and any two nonzero vectors on those eigenlines have nonzero inner product. Rescaling can normalize their lengths but cannot make them orthogonal. Hence no orthonormal eigenbasis exists.