Quesion about creation of subspace with some properties

Question

Let $V$ be a vector space with finite dimension and $K, H$ are subspaces of $V$. Prove that there is subspace $M$ of $V$ s.t $M+K=M+H$ and $M\cap K=M\cap H=\{0\}$.

Answer 1

Assume (as was shown to be necessary) $\dim(H) = \dim(K)$. Let $L = H \cap K$.
We can write $H = L \oplus N$ and $K = L \oplus P$ for some subspaces $N$ and $P$, and $\dim N = \dim H - \dim L = \dim K - \dim L = \dim P$. So there is a linear map $T$ from $P$ onto $N$. Let $M = (I+T) P = \{p + Tp: p \in P\}$.

To show $M \cap H = \{0\}$: if $h \in M \cap H$, we can write $h = p + Tp$ for some $p \in P$, but also $h = u + n$ for some $u \in L$ and $n \in N$. Thus $n - Tp = p - u$. But $n - Tp \in N \subseteq H$ while $p - u \in P + L = K$, and $H \cap K = L$ but $N \cap L = \{0\}$. Thus $p - u = 0$. But $p = u \in P \cap L = \{0\}$, so $p = 0$ and $h = 0 + T 0 = 0$.

The proof of $M \cap K = \{0\}$ is similar.

To show $M + K \subseteq M + H$: take any $y \in M + K = M + L + P$. Then $y = p + T p + r + q$ where $p \in P$, $r \in L$ and $q \in P$. Now write this as $y = (p + q) + T(p+q) - T q + r$. We have $p+q \in P$ so $(p+q) + T(p+q) \in M$, $-Tq \in N$ and so $-Tq + r \in N + L = H$, and thus $y \in M + H$.

But since $M \cap H = \{0\}$ and $M \cap K = \{0\}$, $\dim(M+H) = \dim M + \dim H = \dim M + \dim K = \dim(M+K)$, so $M + K = M+H$.

Answer 2

Assume $\dim(H)=\dim(K)$. Let $\{e_1,\dots,e_a\}$ be a basis for $H\cap K$, and let $\{e_1,\dots,e_a,h_1,\dots,h_b\}$ and $\{e_1,\dots,e_a,k_1,\dots,k_b\}$ be bases for $H$ and $K$, respectively (where $a+b=\dim(H)=\dim(K)$). Then $h_i\notin K$, because if it was, then $h_i\in H\cap K$ implies $h_i$ is a linear combination of the $e_i$, so $\{e_1,\dots,e_a,h_1,\dots,h_b\}$ is not a linearly independent set. Similarly, $k_i\notin H$. Thus, let $M=\operatorname{span}(\{h_1+k_1,\dots,h_b+k_b\})$. If $x\in M\cap H-\{0\}$, then

$$x=A_1e_1+\dots+A_ae_a+B_1h_1+\dots+B_bh_b=C_1(h_1+k_1)+\dots+C_b(h_b+k_b)$$ $$A_1e_1+\dots+A_ae_a+(B_1-C_1)h_1+\dots+(B_b-C_b)h_b=C_1k_1+\dots+C_bk_b:=y$$

which expresses $y\in H\cap K=\{0\}$. Thus $C_i=0$, and so $x=0$, a contradiction. Thus $M\cap H=\{0\}$. Similarly, $M\cap K=\{0\}$. But if $x\in H+M$, then

$$\begin{align} x&=A_1e_1+\dots+A_ae_a+B_1h_1+\dots+B_bh_b+C_1(h_1+k_1)+\dots+C_b(h_b+k_b) \\ &=A_1e_1+\dots+A_ae_a+(B_1-C_1)h_1+\dots+(B_b-C_b)h_b+C_1k_1+\dots+C_bk_b\in H+K, \end{align}$$

so $H+M\subseteq H+K$. Conversely, if $x\in H+K$, then

$$\begin{align} x&=A_1e_1+\dots+A_ae_a+B_1h_1+\dots+B_bh_b+C_1k_1+\dots+C_bk_b \\ &=A_1e_1+\dots+A_ae_a+(B_1-C_1)h_1+\dots+(B_b-C_b)h_b+C_1(h_1+k_1)+\dots+C_b(h_b+k_b) \end{align}$$

so $H+M=H+K$. Similarly, $K+M=H+K$.

Note that I had to assume $\dim(H)=\dim(K)$ at the start. Conversely, if $M\cap H=\{0\}$ and $M+H=M+K$, then $\dim(M)+\dim(H)=\dim(M+H)=\dim(M+K)=\dim(M)+\dim(K)$, so $\dim(H)=\dim(K)$ is a necessary and sufficient condition for this construction to exist.

Answer 3

Not true. Take $K = \{0\}$ and $H = V$.