Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 4 / b / iii

Table of contents
- Solution iii

Solution

Let

S

be the vector subspace of symmetric matrices. A basis consists of

E_{ii}

for

1 \leq i \leq n

and

E_{ij} + E_{ji}

for

i < j

, so

dim S = n + (2 n) = \frac{n ( n + 1 )}{2} .

The dimension formula for a sum of subspaces now gives

dim (V \cap S) = dim V + dim S - dim (V + S) \geq dim V + \frac{n ( n + 1 )}{2} - n^{2} = dim V - (2 n) > 0.

Choose

0 \neq = A \in V \cap S

. It belongs to

V

and is a symmetric matrix, so part (a)(i) makes it a diagonalisable matrix. The preceding example shows that the strict threshold is sharp.

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