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 $M\cap K=M\cap H=\{0\}$.
Quesion about creation of subspace with some properties
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0I am agree to @MarioCarneiro . Also pay attention with condition $dim(K)=dim(H)$ , it will be a question on Iran mathematical competition for University students in the helical year 1368, that you can find its answer in the book was written by "Mr. Bamdad YaHaghi" and also in Linear Algebra written by "Mr. Nikookar". – 2012-12-19
3 Answers
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 $M + K = M+H$.
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.
Not true. Take $K = \{0\}$ and $H = V$.
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1Also $V\neq\{0\}$. – 2012-12-19