b (The dimension threshold)

E_{ij}

1 \leq i, j \leq n

, form a basis of

M_{n} (R)

: each matrix has the unique expression

A = \sum_{i, j} a_{ij} E_{ij}

. There are

n^{2}

of them, so

dim M_{n} (R) = n^{2} .

ii

K = {A \in M_{n} (R) : A^{T} = - A} .

{E_{ij} - E_{ji} : 1 \leq i < j \leq n} .

Indeed, its diagonal entries vanish, and its entries below the diagonal are forced by those above it. Hence

dim K = (2 n)

. Part (a)(ii) proves that the only diagonalisable matrix in

K

0_{n}

Let

S

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} .

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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