# Linear Algebra: Subspaces

In my text it's not very clear as to what the procedure is for determining when a vector is a subspace in say the subset $R^3$ in this example: Consider the vector of the following form $\begin{bmatrix} a \\ b \\ 1 \end{bmatrix}$ is this a subspace of the subset $R^3$?

My best understanding is that I do something like this: $u = \begin{bmatrix} a_1 \\ b_1 \\ 1 \end{bmatrix}$ and $v=\begin{bmatrix} a_2 \\ b_2 \\ 1 \end{bmatrix}$

(Quick Question: $\oplus$ and $\odot$ what does it actually mean? Since I'm currently just using it based on the other examples without an understanding for what it really means) Now $u\oplus v =\begin{bmatrix} a_1 + a_2 \\ b_1 + b_2 \\ 2 \end{bmatrix}$ Now since there is no way to use this to express $c \odot \begin{bmatrix} a_1 \\ b_1 \\ 1 \end{bmatrix}$ it's not a subspace then.

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Regarding the notation $\bigoplus$, it is often used as the notation for direct sum. – yoyostein May 4 '12 at 3:43

First, regarding the terminology: One wouldn't usually speak of a vector being a subspace. A more usual formulation of your question would be: Do the vectors of the form $$\begin{bmatrix} a \\ b \\ 1 \end{bmatrix}$$ with $a,b\in\mathbb R$ form a subspace of $\mathbb R^3$?

Your negative answer to this question is correct. Your argument is also correct but a bit of a detour, since you're using the fact that in a subspace any vector can be expressed as the sum of two vectors in the subspace. That's true, but it's easier to argue directly from the definition. You could use either $u+v$ or $cv$ with $c\in\mathbb R\setminus\{1\}$ to show that this is not a subspace, but you shouldn't compare these two cases with each other, but rather argue that in neither case the result is of the given form (since it doesn't have a $1$ in the third component).

Regarding your notation for the operations, I've never seen that before; I suspect it's specific to the text you're using. I presume that they're using these symbols to distinguish vector addition and scalar multiplication of vectors from addition and multiplication of real numbers.

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Okay, cool thanks for the detailed explanation! – eWizardII Nov 5 '11 at 19:32

Another way to see that it's not a subspace is that a subspace of ${\bf R}^3$ must contain the zero-vector, $(0,0,0)$, and your subset does not.

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Is that a definite way to determine that it's a subspace? Or are there cases were that can be true and it still isn't a subspace in $R^3$ – eWizardII Nov 5 '11 at 19:32
@eWizardII: The set $\{(0,0,0),(1,0,0)\}$ contains the zero vector but is not a subspace. A subspace has to be closed under scalar multiplication and addition. – Jonas Meyer Nov 5 '11 at 19:45
@eWizardII: This is what is called a "necessary, but not sufficient, condition." Containing the zero vector is necessary in order for the set to be a subspace (if it doesn't contain $\mathbf{0}$, then it's definitely not a subspace), but it's not sufficient (it's not enough). Just like being enrolled in the class is necessary in order to actually get an A for the class put in your transcript, but it is not sufficient (if you aren't enrolled, you cannot get the A in your transcript, but being enrolled isn't enough). – Arturo Magidin Nov 5 '11 at 21:36