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So I can define $z=t+y$ and if I put the vectors into a matrix I get the following system of equations:

$x=0$
$y=0$
$y+t=0$
$t=0 $

Which clearly only has one solution, the trivial solution, therefore the vectors must be linearly independent and therefore form a basis... however I know that's a contradiction because there's no proper subspace of $\mathbb R^4$ that has $\dim=4$

I wrote the linear combination down and through some algebra, I was able to get a different basis for the subspace in the following form:

$x(1,0,0,0) +y(0,1,1,0)+t(0,0,1,1)$

This is also linearly independent and forms a basis for the subspace, and the $\dim=3$ which makes more sense.... but I was only able to reach this basis through intuition and algebra and not a methodical approach... any pointers?

Ron Turetzky
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1 Answers1

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I can tell you how to find a basis of $U$, not the basis; there is no such thing.

You can simply take $\bigl\{(1,0,0,0),(0,1,0,-1),(0,0,1,1)\bigr\}$. It is linearly independent at it spans $U$. Therefore, it is a basis of $U$. I know that it spans $U$ because if $(x,y,z,t)\in U$, then $t=-y+z$ and therefore\begin{align}(x,y,z,t)&=(x,y,z,-y+z)\\&=x(1,0,0,0)+y(0,1,0,-1)+z(0,0,1,1).\end{align}

José Carlos Santos
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  • True, there is no singular basis, and your solution is equivalent to mine. However, was there a methodical, non-intuitive way you used to reach this solution? It seems to me you used the same method I did (writing down the linear combination, factoring out the scalars, and combining the vectors). – Ron Turetzky Feb 01 '20 at 10:23
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    Yes, there is a systematic way: you consider the conditions that you were given (there is a single one in your example) and you see them as a system of linear equations. Then you solve that system, expressing some of the variables as linear combinations of the other ones (in my answer, that would be $t=-y+z$). Finally, you get a basis by picking each of the remaining free variables ($x$, $y$, and $z$, in my answer) and seeing what you get when each variable is $1$ and the remaining ones are $0$. You will get a basis then. – José Carlos Santos Feb 01 '20 at 10:57
  • @RonTuretzky For a systematic method, see this answer. In this case, the coefficient matrix of the system has only one row, so you can proceed more or less directly to reading a basis from it. – amd Feb 01 '20 at 21:01