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added the 2 to the last R^2...
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edit approved Jul 22, 2015 at 18:45
Vincent
edit approved Jul 22, 2015 at 18:45
Vincent
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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 wellcorrectly, 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 $$ \forall \lambda,a,b \in \mathbb R, \quad \lambda \cdot (a,b) \in \mathbb R. $$$$ \forall \lambda,a,b \in \mathbb R, \quad \lambda \cdot (a,b) \in \mathbb R^2. $$ Is it obvious now?

Hope that helps,

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 well, 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 $$ \forall \lambda,a,b \in \mathbb R, \quad \lambda \cdot (a,b) \in \mathbb R. $$ Is it obvious now?

Hope that helps,

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,

added 538 characters in body
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Patrick Da Silva
Patrick Da Silva
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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 well, 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 $$ \forall \lambda,a,b \in \mathbb R, \quad \lambda \cdot (a,b) \in \mathbb R. $$ Is it obvious now?

Hope that helps,

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.

Hope that helps,

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 well, 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 $$ \forall \lambda,a,b \in \mathbb R, \quad \lambda \cdot (a,b) \in \mathbb R. $$ Is it obvious now?

Hope that helps,

Source Link
Patrick Da Silva
Patrick Da Silva
  • 42.8k
  • 6
  • 93
  • 146

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.

Hope that helps,