Is this set of vectors a subspace of $R^3$, find basis and dimension

Question

I'm working textbook problems and one of them is: Consider the set of the following form. Determine whether the set is a subspace of $R^3$. If it is, give its basis and dimension. The set is $(a, a, a+3)$.

The constant in there is confusing to me. The book approach for something without that is to factor out a to get $a(1, 1, 1)$, and then this vector(s) $(1, 1, 1)$ can be used to determine whether find the span and linear independence. Is splitting into the vectors $a(1, 1, 1)$ and $(0,0,3)$ how you do this? Very confused. The book doesn't give any examples. If I do that, the dimension $= 2$, and I suppose the basis would be just those $2$ vectors. Is that the right approach?

Answer 1

Here is what I think you are asking: let ${\bf R}^3$ be a vector space. Consider the set $H=\{(a,a,a+3): a\in {\bf R}\}$. Determine if $H$ is a vector subspace of ${\bf R}^3$. In case $H$ is a vector subspace, find a base and the dimension.

You mention that the constant (I suppose that "$a$" and if this is the case, it should be called "parameter") confuses you. So perhaps it is advisable to write the set $H$ in a different appearance, but of course, equivalents. Setting $x:=a, y:=a, z:=a+3$ with $a\in {\bf R}$ you can write $H$ as $$H=\{(x,y,z)\in {\bf R}^3: x=a, y=a, z=a+3, a\in {\bf R}\}$$ One should always try to translate mathematics into a common language. In this case, we are saying that every vector $(x,y,z)\in {\bf R}^3$ belonging to $H$ must satisfy the conditions $x=a, y=a$ and $z=a+3$ for every real number $a$.

If we want know whether $H$ is a vector subspace, we can use the following characterization:

Let $V$ be a vector space over a field $F$ and let $H$ be a subset of $V$. Then, $H$ is a vector subspace of $V$ if and only if
i) $H\not=\emptyset$.
ii) For all $h_1,h_2\in H$, we have $h_1+h_2\in H$.
iii) For all scalar $k\in F$ and $h\in H$, we have $kh\in H$.

We have $H$ is subset of ${\bf R}^3$ by definition. So, we need look at the conditions $i), ii)$ and $iii)$. The first question is : is $H\not=\emptyset$? To answer this question it is convenient (since every vector space has to have the zero vector) to determine if the zero vector $(0,0,0)$ is in $H$, that is just the comment given by geetha290krm. However for the vector zero to be in $H$ it must satisfy the condition imposed by $H$, then you must answer if we can write the vector $(0,0,0)$ as $$0=a, \quad 0=a,\quad 0=a+3$$ You can think about this a bit and eventually conclude that the vector $(0,0,0)$ is not in $H$. Therefore, the above theorem is not satisfied and we can then conclude that $H$ is not a vector subspace of ${\bf R}^3$ and since it is not a vector subspace, the other questions does not make sense.