Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 2

The printed word “eigenvalues” must read “eigenvectors”: eigenvalues are real numbers, whereas a basis of ${\mathbb{R}}^n$ consists of vectors. The intended assertion isThis wording error is present in the original scan.^[ref]

If $X^{-1}AX=D=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)$ for an invertible matrix $X$, write $x_i$ for its columns. The matrix identity $AX=XD$ says $A{x}_i={\lambda }_i{x}_i$. The columns of an invertible matrix are a basis, hence they give the required basis of eigenvectors.

Conversely, put a basis of real eigenvectors $x_1,\dots ,x_n$ into the columns of $X$, with corresponding eigenvalues ${\lambda }_i$. The basis property makes $X$ an invertible matrix. The equations $A{x}_i={\lambda }_i{x}_i$ give $AX=XD$, so $X^{-1}AX=D$. This is precisely diagonalizability over the real numbers.

For the first matrix, the characteristic polynomial isThus the real eigenvalues are $4,-1,3$. Corresponding eigenvectors are $(3,1,0{)}^{\mathsf{T}}$, $(-2,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$. They are linearly independent; the determinant of the matrix with these columns is $5$. ThereforeFor example, multiplying the first column by $A$ gives $(12,4,0{)}^{\mathsf{T}}$, four times that column; multiplying the second gives $(2,-1,0{)}^{\mathsf{T}}$, its negative. These also check the chosen eigenvectors directly.

For the second matrix,The quadratic factor has negative discriminant $-3$, so $1$ is the only real eigenvalue. Solving $Av=v$ gives $v=(a,a,a{)}^{\mathsf{T}}$, a one-dimensional eigenspace. It cannot provide a basis of ${\mathbb{R}}^3$, so

Let $\lambda$ be the single eigenvalue. By diagonalizability, there is a basis $v_1,\dots ,v_n$ of eigenvectors, and every corresponding eigenvalue is $\lambda$. For any $v={\sum }_i{a}_i{v}_i$, linearity givesThus $\boxed{T=\lambda I}$, a scalar multiple of the identity map.

A counterexample to the conclusion without diagonalizability isIts characteristic polynomial is $(t-2{)}^2$, but $(J-2I)(a,b{)}^{\mathsf{T}}=(b,0{)}^{\mathsf{T}}$. The eigenspace for $2$ is therefore $\mathrm{span}\{(1,0{)}^{\mathsf{T}}\}$, of dimension one. There is no basis of eigenvectors in dimension two, so $J$ is not diagonalizable.