I'm have problem proving: Law for Scalar Multiplication :
Vector spaces possess a collection of specific characteristics and properties. Use the definitions in the attached “Definitions” to complete this task.
Define the elements belonging to $\mathbb{R}^2$ as $\{(a, b) | a, b \in \mathbb{ R}\}$. Combining elements within this set under the operations of vector addition and scalar multiplication should use the following notation:
Vector Addition Example: $(–2, 10) + (–5, 0) = (–2 – 5, 10 + 0) = (–7, 10)$
Scalar Multiplication Example: $–10 × (1, –7) = (–10 × 1, –10 × –7) = (–10, 70)$, where –10 is a scalar.
Under these definitions for the operations, it can be rigorously proven that R2 is a vector space.
Prove Closure under Scalar Multiplication - **i need help with this law **
Can someone put it in a proof form?
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$\begingroup$If you're asking if vector spaces are closed under multiplication by a scalar, then yes, it is true. If you're asking why, it's because it's written in the definition of a vector space that it must be true ; there is nothing to prove here. It's true because we assume it is when we speak of a vector space.
EDIT : So if I understand this correctly, you need to show that $\mathbb R^2$ is a vector space and you need help showing that $\mathbb R^2$ is closed under scalar multiplication. Scalar multiplication is defined for $\lambda \in \mathbb R$ and $(a,b) \in \mathbb R^2$ via $$ \lambda \cdot (a,b) \overset{def}= (\lambda a, \lambda b) $$ where $\lambda a$ is the usual multiplication of real numbers. What you want to show is that $$ \forall \lambda,a,b \in \mathbb R, \quad \lambda \cdot (a,b) \in \mathbb R^2. $$ Is it obvious now?
Hope that helps,
$\endgroup$ $\begingroup$Take the examples of all directed arrows (every possible length and every possible angle) originating from a single point.
Now in this set vector addition is like addition of forces in physics: parallelogram law. In this set internally there is addition. Also there is an external operation. Any vector can be "scaled up/down" by any real number. This real number is not part of the set of arrows. But it makes sense to talk of 3.75 times a force. SImply a force directed in the same way but with with strength 3.75 times the original. This is depicted as an arrow of that length in the same direction.
In general any set where we can add them among themselves, and multiply by an external scalar usually the set of real numbers subject to some expected conditions is called a vector space.
$\endgroup$ 2 $\begingroup$If $V$ is a vector space over the field $\mathbf F$, then it must satisfy two properties, namely closure under addition and closure under multiplication.
For closure under multiplication, we demand that if $u \in V$, $a \in \mathbf F$, then $a \mathbf F \in V$. Note that the 'multiplication' needs to be defined beforehand.
Specifically to your example, perhaps you are having trouble with what's meant by closure. According to the definition, you want to prove that $S:\{(a,b)|a,b \in \mathscr R\}$ is a vector space.
You have actually done it. You have shown it is closed under addition as well as multiplication.
Let's take your example,$–10 \times (1, –7) = (–10 \times 1, –10 \times –7) = (–10, 70)$, the original vector$(1,-7) \in S$,right? Also$(-10,70) \in S$ because let $a=-10,b=70$, then you see $(-10,70)$ satisfy the requirement $a,b \in R$. Thus you have proved closure under multiplication.
Edit: I will try to prove the statement
$\mathbb R^2$ is a vector space over the field $\mathbb R$
Proof: if $ a,b \in \mathbb R$, then $(a,b) \in \mathbb R^2$. Let $k \in \mathbb R$ be a scalar, then $k(a,b)=(ka,kb)$. As $k,a,b \in \mathbb R$, $ka,kb \in \mathbb R$, thus $(ka,kb) \in \mathbb R^2$. Therefore scalar multiplication on $\mathbb R^2$ is closed.
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