Question 4

(i) We first establish that \mathrm{im}(S + T) \subseteq \mathrm{im}(S) + \mathrm{im}(T). Let w \in \mathrm{im}(S + T), then there exists v \in V such that w = (S + T)(v) = S(v) + T(v). But S(v) \in \mathrm{im}(S) and T(v) \in \mathrm{im}(T), so S(v) + T(v) \in \mathrm{im}(S) + \mathrm{im}(T), hence \mathrm{im}(S + T) \subseteq \mathrm{im}(S) + \mathrm{im}(T).

Now note that from linearity of S,T we have S(0) = T(0) = 0, so 0 \in \mathrm{im}(S + T). Now let v \in \mathrm{im}(S + T) and w \in \mathrm{im}(S + T). We then have v + \lambda w = (S + T)(u_1) + \lambda (S + T)(u_2) for u_1, u_2 \in V. Then (S + T)(u_1) + \lambda (S + T) (u_2) = (S + T)(u_1 + \lambda u_2) from linearity, so v + \lambda w \in \mathrm{im}(S + T). (note that in general if U \le V then \mathrm{dim} (U) \le \mathrm{dim}(V), since otherwise we could form a set of more than \mathrm{dim}(V) linearly independent vectors using the basis of U)

We conclude that \mathrm{im}(S + T) \le \mathrm{im}(S) + \mathrm{im}(T). Note that this implies that \mathrm{dim} \left(\mathrm{im}(S + T)\right) = \mathrm{rank}(S + T) \le \mathrm{dim}(\mathrm{im}(S) + \mathrm{im}(T)).

Note that for vector spaces we have \mathrm{dim}(U + V) \le \mathrm{dim}(U) + \mathrm{dim}(V). (combining the bases of U,V we certainly get a set spanning U + V. Some of these may however be linearly dependent [hence can be omitted from a basis for U+V] hence the \le)

So:

\mathrm{rank}(S + T) \le \mathrm{dim}(\mathrm{im}(S) + \mathrm{im}(T)) \le \mathrm{dim}\left(\mathrm{im}(S)\right) + \mathrm{dim}\left(\mathrm{im}(T)\right) = \mathrm{rank}(S) + \mathrm{rank}(T)

as required.

(ii)

Note that we can write \mathrm{null}(ST) = \mathrm{dim} \left(\mathrm{ker}(ST)\right). With a view to apply the rank-nullity theorem, we want to look at a map with domain \mathrm{ker}(ST). The most useful map for this is actually T restricted to \mathrm{ker}(ST). Denote this restriction T_K with K = \mathrm{ker}(ST). With this map we have:

\mathrm{null}(ST) \le \mathrm{null}(T_K) + \mathrm{rank}(T_K)

We want to bound these two terms above.

Since T_K (x) = T(x) for all x \in K, we have T_K (x) = 0 iff x \in \mathrm{ker}(ST) and T(x) = 0, giving x \in \mathrm{ker}(T). So \mathrm{ker}(T_K) = \mathrm{ker}(ST) \cap \mathrm{ker}(T). But note that if x \in \mathrm{ker}(T), then (ST)(x) = S(T(x)) = S(0) = 0, so \mathrm{ker}(T) \subseteq \mathrm{ker}(ST). So \mathrm{ker}(T_K) = \mathrm{ker}(T). Taking dimensions of both sides we have \mathrm{null}(T_K) = \mathrm{null}(T).

It should scream out that we know want to show \mathrm{null}(S) \ge \mathrm{rank}(T_K). This is clear when we note the definition of nullity/rank. Note that for x \in K we have ST_K (x) = 0 merely from the definition of kernel, so T_K(x) \in \mathrm{ker}(S). That is, \mathrm{im}(T_K) \le \mathrm{ker}(S). We then have \mathrm{dim} \left(\mathrm{im}(T_K)\right) \le \mathrm{dim}\left(\mathrm{ker}(S)\right) so that \mathrm{rank}(T_K) \le \mathrm{null}(S) as we wanted.

Putting these together gives \mathrm{null}(ST) \le \mathrm{null}(S) + \mathrm{null}(T).

1 Like

:star_struck: :star_struck: :star_struck: :star_struck: