Is $V = \{(a + 2b + 1, 2a-3b) | a,b\in\mathbb{R}\}$ a subspace of $\mathbb{R}^2$? Why or why not? I am studying for a linear algebra final and going over the first exam. I just now retried this problem and I am making the same mistake I did the first time. I get that the set in NOT a subspace, but that is incorrect. Here is how I came to this conclusion:
Closure under addition fails:
Let $(a + 2b + 1, 2a-3b), (a' + 2b' + 1, 2a'-3b') \in V$.
Then $$(a + 2b + 1, 2a-3b) + (a' + 2b' + 1, 2a'-3b') = ( (a + a') + 2(b + b') + 2, 2(a-a') - 3(b-b') ).$$ 
This does not look like it would belong in the set V, because the x coordinate has 2 added to 2, not 1.
Also, closure under scalar multiplication fails:
Let $\alpha \in \mathbb{R}$. Then $\alpha(a + 2b + 1, 2a-3b) = (\alpha a + \alpha 2b) + \alpha, \alpha (2a -3b) )$. Once again, since $\alpha$ is being added to the x coordinate instead of 1, I thought it would not belong in the set V.
However, apparently, both of these conclusions are incorrect. If someone could explain why my work is incorrect, I would be very appreciative.
 A: The flaw in your reasoning is as follows:
Suppose we are given $W=\{(a+1,b+2) \, | \, a,b \in \Bbb{R}\}$ and are asked the same question: is $W$ a subspace of $\Bbb{R}^2$? Following your reasoning, we will consider $u=(e+1,f+2) \in W$ and $v=(c+1,d+2) \in W$ and conclude that 
$$u+v=(e+c+2, f+d+4).$$ 
From this, you may incorrectly conclude that this does not look like vectors in $W$ because the first component $e+c+2$ is not of the form $\color{blue}{a+1}$ (because we have a $2$ and not $1$) and the second component $f+d+4$ is not of the form $\color{red}{b+2}$ (because we have a $4$ and not $2$). This is where your reasoning had a flaw because we can write
\begin{align*}
e+c+2 & = (e+c+1)+1\\
f+d+4 & = (f+d+2)+2.
\end{align*}
Since $(e+c+1)$ and $(f+d+2)$ are both real numbers, so the two components still satisfies the membership criterion of $W$. Hence the sum of the vectors is still in $W$.

Coming to your original question:
Any vector in $V$ can be written as 
$$\begin{bmatrix}a+2b+1\\2a-3b\end{bmatrix}=a\begin{bmatrix}1\\2\end{bmatrix}+b\begin{bmatrix}2\\-3\end{bmatrix}+\begin{bmatrix}1\\0\end{bmatrix}.$$
Observe that $a\begin{bmatrix}1\\2\end{bmatrix}+b\begin{bmatrix}2\\-3\end{bmatrix}$ is nothing but linear combination of two linearly independent vectors in $\Bbb{R}^2$, hence this span is $\Bbb{R}^2$. So all we are doing is taking any vector in $\Bbb{R}^2$ and translating it by the vector $\begin{bmatrix}1\\0\end{bmatrix}$, which is still $\Bbb{R}^2$. 
Thus $V=\Bbb{R}^2$, hence a (trivial) subspace of $\Bbb{R}^2$.
A: As a hint, if
$$\begin{pmatrix}a+2b+1\\ 2a-3b\end{pmatrix}=A\begin{pmatrix}a\\ b\end{pmatrix}+\begin{pmatrix}x_0\\ y_0\end{pmatrix}$$
then what is the matrix $A$? Is it invertible?
