Jtdylktuy

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.

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
$$begin{bmatrix}a+2b+1\2a-3bend{bmatrix}=abegin{bmatrix}1\2end{bmatrix}+bbegin{bmatrix}2\-3end{bmatrix}+begin{bmatrix}1\0end{bmatrix}.$$
Observe that $abegin{bmatrix}1\2end{bmatrix}+bbegin{bmatrix}2\-3end{bmatrix}$ 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 $begin{bmatrix}1\0end{bmatrix}$, which is still $Bbb{R}^2$.

Thus $V=Bbb{R}^2$, hence a (trivial) subspace of $Bbb{R}^2$.

As a hint, if
$$begin{pmatrix}a+2b+1\ 2a-3bend{pmatrix}=Abegin{pmatrix}a\ bend{pmatrix}+begin{pmatrix}x_0\ y_0end{pmatrix}$$
then what is the matrix $A$? Is it invertible?