Hi I need to solve next exercise :"Complement the set $(1,0,1),(1,1,0)$ to the basis of $\mathbb{R}^3$"

I know that I have to get $(0,1,1)$, but I do not know how.

  • 2
    $\begingroup$ Every $3$ linearly independent vectors will form a basis for a $3$ dimensional space. So all you need is to verify that those vectors are linearly independent. What is the problem? $\endgroup$
    – freakish
    Dec 4, 2023 at 13:07
  • $\begingroup$ I dont how to do that, i didnt attend lecture unfortunately :( $\endgroup$ Dec 4, 2023 at 13:11
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    $\begingroup$ "I know that I have to get $(0,1,1)$" : precisely, this is wrong. This is a possible solution, but there are infinitely many other ones. Do you know what a basis is ?... $\endgroup$ Dec 4, 2023 at 13:34
  • $\begingroup$ @TheSilverDoe i solved it by finding complement for (1,0,1),(1,1,0),(1,0,0). Is this valid? $\endgroup$ Dec 4, 2023 at 17:01

1 Answer 1


As some have mentioned before, since the dimension of the space you are considering is three, you only need to find three linearly independent vectors. This is a consequence of the famous Steinitz Theorem. How can we find it? You could propose a system of equations and solve it (using, as other users have already mentioned, the definition of a linear combination). Another way in this case is to see that the determinant containing the three vectors is non-zero.

This is a spoilerFor example, the vectors (1,0,1),(1,1,0),(0,1,0) work. There are many options.

  • $\begingroup$ Wow, thats sounds great , could you explain it little more, i dont understand it really good, i already solved exercise by finding complement for (1,0,1),(1,1,0),(1,0,0) $\endgroup$ Dec 4, 2023 at 16:47
  • $\begingroup$ What is the part that you do not fully understand? Yeah, (1,0,0) also works as it does (0,1,0) or (0,1,1). There are a LOT of choices for the third vector in the basis provided that all of them are linearly independent, following the Steinitz Theorem. $\endgroup$ Dec 4, 2023 at 18:43

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