1
$\begingroup$

"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?

$\endgroup$
  • 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
0
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.