# Indetifying a vector space

I'm trying to identify which sets are vector spaces. I know that a set $V$ is a vector space if its elements (vectors) have addition and scalar multiplication so that the result is also within the set $V$. I'm trying to identify the following sets.

1. Set of real $m\times n$ matrices $\mathbb{R}^{m\times n}$

If we do operations for two matrices it always yields and matrix within $\mathbb{R}^{m\times n}$. So this is a vector space.

1. Line $\{(x,4x) \in \mathbb{R}^2 \: | \: x\in\mathbb{R}\}$

EDIT: Is a vector subspace of $\mathbb{R}^2$.

1. $\mathbb{R}^n = \{(x_1, \ldots, x_n) \ \vert \ x_i \in \mathbb{R}\}$

Clearly a vector space because for all $n$-vectors in it the operations hold.

1. Curve $\{(x,x^2) \in \mathbb{R}^2 \: | \: x\in\mathbb{R}\}$

Addition does not hold on this set $(1,1) + (2,4) = (3,5) \notin V$ so it is not a vector space.

1. Line $\{(x,3x-2) \in \mathbb{R}^2 \: | \: x\in\mathbb{R}\}$. Same as the other line and not an vector space.

EDIT: Under the vector space axioms addition $\mathbf a + \mathbf b \in V$ when $\mathbf a, \mathbf b \in V$ and $\lambda \mathbf a \in V$ thus if $\mathbf b = -1\cdot \mathbf a$ then $\mathbf a + \mathbf b = \mathbf 0 \notin V$. So additive indentity does not hold and the set is not a vector space.

1. Set of matrices in $\{ M\in\mathbb{R}^{n\times n}\: | \: \det(M)=1\}$.

\begin{align*} \text{det}\left(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) = 4 \neq 1\end{align*}

OR

\begin{align*} \text{det}\left(\lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = \lambda \neq 1 \text{ when } \lambda \neq 1\end{align*}

EDIT: So addition from said set yields a matrix with non 1 determinant. Not a vector space.

• Your reasoning that $\tt 3.$ isn't a vector space isn't correct, $\mathbf{x}+\mathbf{y}=(3,12)$ has the form $(\alpha,4\alpha)$, so it is certainly in that set. – Workaholic Feb 21 '16 at 12:40
• Number 2. seems a vector subspace of $\mathbb{R}^2$ to me. Take two elements in the set, $(\lambda, 4\lambda)$ and $(\mu, 4\mu)$, then $(\lambda, 4\lambda) + (\mu, 4\mu) = (\lambda+\mu , 4(\lambda+\mu))$ which is of the required form. Also if $k\in \mathbb{R}$ then $k(\lambda, 4\lambda)= (k\lambda, 4k\lambda)$ is of the required form. An easy intuition is that this set is a straight line that contains the 0-vector, so it should indeed be a vector space. – Martingalo Feb 21 '16 at 12:40
• Reasoning for 5 is wrong although deduction is correct. 5-is not a vector space as the additive identity is not present in the set. – Qwerty Feb 21 '16 at 12:42
• 6 is clearly not a vector space. Take 2 identity matrices, add them and take their determinant – Qwerty Feb 21 '16 at 12:44
• To qualify as a vector space , don't miss the fact that the additive identity should also be a part of the space – Qwerty Feb 21 '16 at 12:47

Hints:

1. The deduction is correct, although you should actually check all the vector space axioms to conclude that it is one.

2. Your reasoning that $\tt 3.$ isn't a vector space isn't correct, $\mathbf{x}+\mathbf{y}=(3,12)$ has the form $(\alpha,4\alpha)$, so it is certainly in that set. $-$ Edit 1: That's correct, assuming that you already know that $\mathbf{R}^2$ is a vector space.

3. Same comment as in 1.

6. I assume that the field you're working over is $\mathbf{R}$. So take an element $c\in\mathbf{R}$, and suppose that $M$ is in that set, does it follow that $\det(cM)=1$? (i.e. does it follow that it is closed under scalar multiplication?) $-$ Edit 1: You showed that it isn't closed under addition, hence not a vector space, and that's correct.
Number (2) seems a vector subspace of $\mathbb{R}^2$ to me. Take two elements in the set, $(\lambda, 4\lambda)$ and $(\mu, 4\mu)$, then $(\lambda, 4\lambda) + (\mu, 4\mu) = (\lambda+\mu , 4(\lambda+\mu))$ which is of the required form. Also if $k\in \mathbb{R}$ then $k(\lambda, 4\lambda)= (k\lambda, 4k\lambda)$ is of the required form. An easy intuition is that this set is a straight line that contains the 0-vector, so it should indeed be a vector space.
For the last one (6) just use the following fact about determinants of matrices. If $A$ i an $n\times n$-matrix and $\lambda\in \mathbb{R}$ a real scalar, then $$\det (\lambda A) =\lambda^n \det(A) = \lambda^n,$$ where the latter follows if $A$ is in the set you defined.
Now, if $\lambda \neq 1$ then $\det(\lambda A) = \lambda^n \neq 1$ and hence $\lambda A$ is not in the set you defined, and hence it cannot be a vector space.
A last comment, since $a,b\in V$ then $a+b\in V$ and $a\in V$, $\lambda \in \mathbb{R}$ then $\lambda a\in V$, this in particular implies that $0\in V$ (the zero vector should always belong to the space, take $a\in V$ then $(-1)*a =-a\in V$ and hence $a-a=0\in V$). Then, since checking this is fairly quick, you may do this for number (5), where $(0,0)\notin \{(x,3x-2)\in \mathbb{R}^2|x\in \mathbb{R}\}$.