I have been given $5$ vectors $(u_1, u_2, u_3, u_4 \text{ and } u_5)$:

$u_1= \langle1,-1,2,1\rangle, u_2 = \langle1,2,1,-1\rangle, u_3 = \langle-1,-8,1,5\rangle, u_4 = \langle1,1,1,1\rangle, u_5 = \langle-7,2,4,1\rangle$.

I have a vector space $V$ that is the span of the $5$ vectors. So

$V = Span\{u_1, u_2, u_3, u_4\}$. The vector space $V$ has $3$ dimensions.

I have to check if $V^\perp = Span\{u_5\}$. How do I do that?

First I wrote $u_5$ as $v = \langle-7t, 2t, 4t, t\rangle$. Next I found that this vector is orthogonal to all the other vectors $(u_1, u_2, u_3, u_4 \text{ and } u_5)$. I'm not sure how to continue. I'm not even sure if I'm doing it right.

  • $\begingroup$ I don't understand your question. How could $u_5$ be orthogonal to a space containing $u_5$? $\endgroup$
    – user137731
    Mar 8, 2015 at 15:00
  • 1
    $\begingroup$ Were you intending to say that $V = \mbox{span}\{u_1,u_2,u_3,u_4\}$ and that you found that $u_5$ is perpendicular to each of $u_1, u_2, u_3, u_4$? $\endgroup$ Mar 8, 2015 at 15:03
  • $\begingroup$ @EricTressler Yes $\endgroup$
    – droman07
    Mar 8, 2015 at 15:13
  • $\begingroup$ @droman07 Then you should edit your question so that it's correct. $\endgroup$ Mar 8, 2015 at 15:34
  • $\begingroup$ @EricTressler just for completeness, that's not quite the definition of span maths.kisogo.com/… - is that worth mentioning though (I address this to you because you answered) - in the case of a finite set they are of course the same. $\endgroup$
    – Alec Teal
    Mar 8, 2015 at 17:15

2 Answers 2


You check that u5 is orthogonal to u1 u2 u3 and u4 as a proof. Or you can equivalently prove that u5 is orthogonal to the 3 dimensions you found of V. Make sure to use the given inner product.


I am assuming you meant that $V = \mbox{span}\{u_1,u_2,u_3,u_4\}$.

If $u_5 \perp u_i$ for all $i \in \{1,2,3,4\}$, then $u_5 \perp \mbox{span}\{u_1,u_2,u_3,u_4\} = V$. The span of a set of vectors is just the set of all finite linear combinations of them, and so if the inner product (dot product) of $u_5$ with each of these vectors is 0, then its inner product with any vector in their span is 0 as well.


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