# Quesion about creation of subspace with some properties

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 $M\cap K=M\cap H=\{0\}$.

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it would be helpful if we know what your attempt was. – dineshdileep Dec 19 '12 at 8:25
I think you need $\dim(K)=\dim(H)$ for this to be true. Indeed, if $M\cap H=\{0\}$, then $\dim(M+H)=\dim(M)+\dim(H)$, so that is a simple proof that if such an $M$ exists, then $\dim(K)=\dim(H)$. – Mario Carneiro Dec 19 '12 at 8:32
I 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". – AmirHosein SadeghiManesh Dec 19 '12 at 12:34

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 $M + K = M+H$.

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+1 neat answer. – Babak S. Dec 19 '12 at 10:35

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.

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

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Also $V\neq\{0\}$. – Mario Carneiro Dec 19 '12 at 8:27