# Prove that two vectorial subspaces are complementary.

I have two vectorial subspaces of $\mathbb{R}^4$ $U=\{u_1=(2,3,1,1), u_2 = (1,1,5,2),u_3=(0,1,1,1)\}$ and $V=\{v_1(2,1,3,2),v_2(1,1,3,4),v_3(5,2,6,2)\}$. I need to prove that V and U are complementar(i.e. The direct sum between U and V is $\mathbb{R^4}$). I have tried to prove that every vector $x \in \mathbb{R}^4$ can be represented as a linear combination of $v_1,v_2,v_3,u_1,u_2,u_3$, and after that to show that that form of x is unique.

• So, what are you stuck on? – anomaly Sep 8 '16 at 15:40
• It will probably be quicker and less painful if you reduce both these sets to a basis (both should be 2-spaces), and then show that all the elements of one do not belong to the other. – Alfred Yerger Sep 8 '16 at 15:46
• Or even quicker that these 2 2-basis from Alfred are linear independent and hence form a basis of $\mathbb{R}^4$. – ctst Sep 8 '16 at 15:52

Basis for $\;U\;$ :

$$\begin{pmatrix}1&1&5&2\\2&3&1&1\\0&1&1&1\\\end{pmatrix}\stackrel{R_2-2R_1}\longrightarrow\begin{pmatrix}1&1&5&2\\0&1&\!\!-9&\!\!-3\\0&1&1&1\\\end{pmatrix}\stackrel{R_3-R_2}\longrightarrow\begin{pmatrix}1&1&5&2\\0&1&\!\!-9&\!\!-3\\0&0&10&4\\\end{pmatrix}$$

Thus $\;\dim U=3\;$ and the given three vectors are a basis for it:

Basis for $\;V\;$ :

$$\begin{pmatrix}1&1&3&4\\2&1&3&2\\5&2&6&2\end{pmatrix}\stackrel{R_2-2R_1,\,R_3-5R_1}\longrightarrow\begin{pmatrix}1&1&3&4\\0&\!\!-1&\!\!-3&\!\!-6\\0&\!\!-3&\!\!-9&\!\!-18\end{pmatrix}\implies R_3=3R_2$$

and thus $\;\dim V=2\;$ and the first two vectors above are linearly independent and then basis of $\;V\;$

Since $\;\dim U+\dim V=5>4\;$ these both subspaces cannot be direct sum