$U_1, U_2$ are linear subspaces of $\mathbb{R}^3$. $$ U_1 = \{(x_1, x_2, x_3) \in \mathbb{R}^3 | \ x_1 = x_2 = x_3\}$$ $$ U_2 = \{(x_1, x_2, x_3) \in \mathbb{R}^3 | \ x_1 + \ x_2 = 0\}$$

For both $U_1$ and $U_2$ I want to find a basis and the dimension.

For $U_1$ I assumed one possible basis is $ \{ \left( \begin{array}{c} 1\\ 1\\ 1\\ \end{array} \right) \}$, because there every element of the vector is the same, just as the conditions demands. As we have only one vector in our basis, the dimension of $U_1$ is $1$, hence $dim(U_1)=1$.

For $U_2$ I assumed a possible basis could be $ \{ \left( \begin{array}{c} 1\\ -1\\ 1\\ \end{array} \right) \}$, because $x_1 + x_2 = 1 + (-1) = 0$. This should fulfil the condition. Again, as we have only one vector in our basis, the dimension of $U_2$ is 1, hence $dim(U_2)=1$.

I am fairly new to the topic of linear algebra and therefore I wanted to ask: is the above reasoning correct?

  • $\begingroup$ You have found elements in the two vector spaces, and thus you have found a vector that is a valid element in a basis for each of them, which means that you're off to a good start. However, you haven't shown that you have found all elements. Yes, $\left(\begin{smallmatrix}1\\1\\1\end{smallmatrix}\right)\in U_1$, as is any scalar multiple of that vector, but is that the entirety of $U_1$? Similarily with $\left(\begin{smallmatrix}1\\-1\\1\end{smallmatrix}\right)\in U_2$. $\endgroup$ – Arthur Nov 14 '17 at 13:58
  • 2
    $\begingroup$ $(0,0,1)\in U_2$, but this element is not a scalar multiple of $(1,-1,1)$. You're close! $\endgroup$ – Koto Nov 14 '17 at 14:00
  • $\begingroup$ @Arthur Sir, isn't the basis for $U_1$ given by the OP correct? I thought OP went wrong with only the 2nd part i.e., basis for $U_2$. $\endgroup$ – shwetha Nov 14 '17 at 14:02
  • $\begingroup$ @shwetha Why are you giving away the solution? I was intentionally being vague so that the OP had to actually check that he had indeed found all of $U_1$. If you say any correct answer is correct, and for any incorrect answer ask whether they think it's correct, that's not teaching them to check for themselves. It's teaching them that if you ask, then that means it's wrong. $\endgroup$ – Arthur Nov 14 '17 at 14:04

Your answer for $U_1$ is alright. But for $U_2$ you can do this:

Let $a=(x_1, x_2, x_3)$ be an element of $U_2$, then $$a=(x_1, -x_1, x_3)$$ $$ =(x_1, -x_1, 0) + (0,0,x_3) $$ $$ =x_1(1,-1,0) + x_3(0,0,1) $$

Then $ (1,-1,0);(0,0,1) $ constitute a basis for $U_2$. Therefore the dimension is equal to 2.


Your reasoning for $U_1$ is correct, but for $U_2$ not !

For example we have $(1,-1,1)^T,(0,0,1)^T \in U_2$, hence $\dim U_2 \ge 2$.

But $(1,1,0)^T \notin U_2$, hence $U_2 \ne \mathbb R^3$ and therefore $\dim U_2 = 2$ and $(1,-1,1)^T,(0,0,1)^T $ is a basis for $U_2$.

  • $\begingroup$ How can the reasoning for $U_1$ be correct and the reasoning for $U_2$ be wrong when they are the exact same reasoning? Yes, the answer to $U_1$ is correct, but that doesn't mean that the reasoning is. They still haven't actually checked that their proposed basis spans the entire space, which is the same mistake that they have done for $U_2$. $\endgroup$ – Arthur Nov 14 '17 at 14:06
  • $\begingroup$ @Arthur Could you give me a hint on how to show that my proposed basis actually spans the entire space? I would assume the answer that evaristegd gave is a legitimate way perhaps? $\endgroup$ – Jennifer Nov 14 '17 at 16:08

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.