 Solution

ID: 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/a/solution

The printed word “eigenvalues” must read “eigenvectors”: eigenvalues are real numbers, whereas a basis of

R^{n}

consists of vectors. The intended assertion is

A is diagonalizable over R ⟺ R^{n} has a basis of eigenvectors of A .

This wording error is present in the original scan.^[ref]

X^{- 1} A X = D = diag (λ_{1}, \dots, λ_{n})

X

, write

x_{i}

for its columns. The matrix identity

A X = X D

says

A x_{i} = λ_{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

λ_{i}

. The basis property makes

X

A x_{i} = λ_{i} x_{i}

give

A X = X D

, so

X^{- 1} A X = D

. This is precisely diagonalizability over the real numbers.

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