I got this vector space: $V=\{(a,b,2a,a+b) : a,b \in \mathbb R\}$. I need to find and orthonormal basis for $V$ and $V^\perp$.

So I need to find a basis, so I took several vectors like $(1,1,2,2)$ & $(2,3,4,5)$ & $(0,1,0,1)$ & $(1,2,2,3)$ and I've found that the basis for them is: $\{(1,1,2,2),(1,2,2,3)\}$ Now I need to change it into orth' basis - so: $w_1=(1,1,2,2)$ and $w_2=(-0.2,0.8,-0.4,0.6)$

and I have to idea what to do for $V^\perp$...

Am I right? Wrong? Would love to get help here. Thanks.

  • $\begingroup$ help please. thanks. $\endgroup$
    – user153515
    Commented May 26, 2014 at 20:01
  • $\begingroup$ It has been some time since I took an Algebra course, so bear with me. $V$ is an $\mathbb{R}$-vector space, right? Because then I would try to find a basis by setting $a=1, b=0$ for a 1st vector and $a=0, b=1$ for a 2nd one. So $v_1=(1,0,2,1), v_2=(0,1,0,1)$. And indeed for any $v=(a,b,2a,a+b)\in V:~ v=a\cdot v_1+ b\cdot v_2$, so this is a basis. You can change this to a orth.norm. basis by using Gram-Schmidt's method for example (heard of that?). Considering your attempt: Your initial basis is equally valid (dimensional argument), but your $w_1, w_2$ are neither normal nor orthogonal. $\endgroup$
    – Piwi
    Commented May 26, 2014 at 20:21
  • $\begingroup$ So far you are right, @user153515 in the choice of your basis for $\;V\;$, but you must first orthonormalize it: $\;w_1\;$ cannot be since its length is not one (take instead $\;\frac{w_1}{||w_1||}\;$ ), and then apply Gram-Schmidt to this with your $\;w_2\;$ and normalize it... $\endgroup$
    – DonAntonio
    Commented May 26, 2014 at 20:24
  • $\begingroup$ I used the G.S. method for finding what are the new orth.norm. vectors of this kind of basis... edit: oh. so I did mistake for w1, probably. then how am i gonna calculate it? can u help me pls and show me? thanks $\endgroup$
    – user153515
    Commented May 26, 2014 at 20:24

1 Answer 1


Hints assuming the usual, euclidean inner product in $\;V\le\Bbb R^4\;$:


$$q_2:=w_2-\langle w_2,u_1\rangle u_1=(1,2,2,3)-\langle\;(1,2,2,3)\,,\,u_1\;\rangle\,u_1=$$



(I left on purpose the unreduced fractions in order to make the calculations easier).

Obsrve that we already have $\;u_1\perp w_2\;$ , yet we still have to normalize $\;w_2\;$, so we take


and now $\;\{u_1\,,\,u_2\}\;$ is an orthonormal basis of $\;V\;$ .

For $\;V^\perp\;$ : you first must find the general form of $\;x\in V^\perp\iff x\perp w_i\;,\;i=1,2\;$ (why the $\;w_i$'s and not the $\;u_i$' s? To make calculations easier!) , so if $\;x=(a,b,c,d)\;$ , it then must be

$$\begin{align*}x\perp w_1\iff&a+b+2c+2d=0\\{}\\x\perp w_2\iff&a+2b+2c+3d=0\end{align*}$$

Now solve the above homogeneous sytem to find $\;V^\perp\;$ and a basis of it, repeat the G-S method as shown above, serve warm and with some salt...and we're done.

  • $\begingroup$ First I want to thank you. Moreover, the only solution for the homogeneous system is only a=b=c=d=0... so what to do? $\endgroup$
    – user153515
    Commented May 27, 2014 at 4:35
  • 1
    $\begingroup$ @user153515: That's not true. Take for example a=2, b=0, c=-1, d=0 or a=1, b=3, c=1, d=-3. To see that, let's call the first equation (I) and the second one (II). Then (II)-(I): $b+d=0$, so $-b=d$. Using that in (I): $a+b+2c-2b=0$, so $a-b+2c=0$. So whenever a, b and c solve this last equation and $-b=d$, they solve the initial homogeneous system {(I),(II)}. $\endgroup$
    – Piwi
    Commented May 27, 2014 at 9:21
  • $\begingroup$ @user153515, besides what Piwi wrote, you must know that a homogeneous linear system with more variables than equations has a non-trivial solution... $\endgroup$
    – DonAntonio
    Commented May 27, 2014 at 10:58

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