0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ So, what are you stuck on? $\endgroup$ – anomaly Sep 8 '16 at 15:40
  • $\begingroup$ 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. $\endgroup$ – Alfred Yerger Sep 8 '16 at 15:46
  • $\begingroup$ Or even quicker that these 2 2-basis from Alfred are linear independent and hence form a basis of $\mathbb{R}^4$. $\endgroup$ – ctst Sep 8 '16 at 15:52
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.