Show That g(y)=proj_x y is a Linear Operator.

Homework Statement

Let x be a fixed nonzero vector in R^3. Show that the mapping g:R^3→R^3 given by g(y)=projxy is a linear operator.

Homework Equations

projxy = $\left(\frac{x\cdot y}{\|x\|}\right)x$

My book defines linear operator as: Let V be a vector space. A linear operator on V is a linear transformation whose domain and codomain are both V.

The Attempt at a Solution

I know how to show g is a linear transformation but I need help proving that g is a linear operator. Do I need to show y and projxy share the same vector space? If so, how would I go about doing that?

Thanks

Homework Helper
x and y are in R^3. The right side is a scalar times x. Doesn't that make it an element of R^3? There's not really much to show here.

x and y are in R^3. The right side is a scalar times x. Doesn't that make it an element of R^3? There's not really much to show here.

In the problem statement where it says "g:R^3→R^3" is that enough to tell us that the vector space mapped to by g is R^3? Or just that any element mapped to by g will be in R^3 but that doesn't necessarily mean the set of all mappings made by g equal R^3?

How do you show 2 vector spaces are equivalent usually? I see why the vector space of $\vec{y}$ is R^3, but is there a way to really prove that the vector space of $\left(\frac{x\cdot y}{\|x\|}\right)x$ is R^3 or is it just really obvious?

thanks for your help by the way

Homework Helper
In the problem statement where it says "g:R^3→R^3" is that enough to tell us that the vector space mapped to by g is R^3? Or just that any element mapped to by g will be in R^3 but that doesn't necessarily mean the set of all mappings made by g equal R^3?

How do you show 2 vector spaces are equivalent usually? I see why the vector space of $\vec{y}$ is R^3, but is there a way to really prove that the vector space of $\left(\frac{x\cdot y}{\|x\|}\right)x$ is R^3 or is it just really obvious?

thanks for your help by the way

You may be misreading the definition of codomain. It doesn't mean that the image of the map is all of R^3. Since it's a projection, it's certainly not. It's a line in R^3 composed of all of the vectors parallel to x. Look up the exact definition of 'codomain'.

You may be misreading the definition of codomain. It doesn't mean that the image of the map is all of R^3. Since it's a projection, it's certainly not. It's a line in R^3 composed of all of the vectors parallel to x. Look up the exact definition of 'codomain'.

Am I correct in saying the domain of domain of g($\vec{y}$) is the set of all $\vec{y}$? And the co-domain of g($\vec{y}$)= $\vec{proj_{x}y}$ is the set of all $\vec{proj_{x}y}$?

And the vector space of $\vec{y}$ is the set of all vectors that can be attained through linear combinations of $\vec{y}$?
And the vector space of $\vec{proj_{x}y}$ is the set of all vectors that can be attained through linear combinations of $\vec{proj_{x}y}$?

So how can we say the vector space of $\vec{proj_{x}y}$ is equal to the vector space of $\vec{y}$ if the vector space of $\vec{proj_{x}y}$ is a line extending from $\vec{x}$ and the vector space of $\vec{y}$ is a straight line extending from $\vec{y}$?

Thanks again

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Am I correct in saying the domain of domain of g($\vec{y}$) is the set of all $\vec{y}$? And the co-domain of g($\vec{y}$)= $\vec{proj_{x}y}$ is the set of all $\vec{proj_{x}y}$?
And the vector space of $\vec{y}$ is the set of all vectors that can be attained through linear combinations of $\vec{y}$?
And the vector space of $\vec{proj_{x}y}$ is the set of all vectors that can be attained through linear combinations of $\vec{proj_{x}y}$?
So how can we say the vector space of $\vec{proj_{x}y}$ is equal to the vector space of $\vec{y}$ if the vector space of $\vec{proj_{x}y}$ is a line extending from $\vec{x}$ and the vector space of $\vec{y}$ is a straight line extending from $\vec{y}$?