"Suppose that V is a vector space and $L : V → V$ is a linear map.

(i) Let K ⊂ V be the set of all vectors $v ∈ V$ such that $L(v) = −v$. Show that K is a subspace of V .

(ii) Construct a linear map $M : V → V$ with the property that $K = \left\{v ∈ V | Mv = 0\right\}.$

For part (i), I just showed that K contained the zero vector, was closed under addition and closed under scalar multiplication.

However I'm really not sure what to do for part (ii). Would I have to use matrices?

  • 1
    $\begingroup$ Note that $Lv=-v$ iff $(L+I)v=0$, where $I$ is the identity transformation. $\endgroup$ – symplectomorphic Jun 17 '16 at 4:00

For part (i) you're done with addition and scalar multiplication alone ($0$ then follows), and you can do it in one go with showing that $v_1,v_2 \in K$ and $\lambda$ a scalar implies $v_1 + \lambda \cdot v_2 \in K$.

For (ii) we always have the linear map $I: V \rightarrow V$ which has $I(v) = v$ for all $v \in V$, the identity. Then $L(v) = -v$ iff $L(v) + v = 0$ iff $L(v) + I(v) = 0$ iff $(L+I)(v) = 0$.

So the pointwise sum $M = L+I$ works. It's a standard exercise that the (pointwise) sum of linear maps is again a linear map. No need for matrices, this is true in the purely abstract.

Note that then (i) becomes a consequence of (ii), as $K$ is the kernel of $M$, and I suppose you have shown that every kernel of a linear map is a subspace (again true in general).

  • $\begingroup$ Great, thank you! That makes perfect sense :) $\endgroup$ – maths123 Jun 17 '16 at 21:36

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