Why Is Isomorphism Confusing in Linear Algebra?

[MidgetDwarf]

I am having problems in my linear algebra class. The class is taught rather poorly. There is only but 3 students left. The instructor is of no help. I tried reading my txtbook and following a few videos (even went to office hours). However I am not understanding Isomorphism.

I know that a transformation from V to W is a linear transformation from a vector space V to a vector space W, when V maps district vectors into W. (Thinking of a function that is 1 to 1)

@and onto if every vector in W is the image of at least one vector in V.I am using Anton Elementary Linear algebra txt.

From my understanding. The first property is to prove 1 to 1. The second is to prove onto?

There is an example in my book. P 466.

Let V= R^inffinty. Be the sequence space in which u1, u2,..., un,... is an infinite sequence of Real numbers..

Consider the linear "sshifting operators" on V defined by

T1 (u1, u2,..., un,...)=(0, u1, u2,...un,... )

T2 (u1, u2,..., un,...)=(u2, u3,...un,...)

(a) Show that T1 is one to one but not not onto.

what am I supposed to do? I am lost.

Maybe?

Let a=a1,a2,..., an,... and b=b1, b2,..., bn,... be a sequence of infinite real numbers in V.

T1 (a+b)=(0, a1+b1, a2+b2,...an+bn,...)

T1 (a)+T(b)=(0, a1, a2,...an,...)+(0, b1, b2,...bn,...)

Which equals the previous line? How would I show that it fails for multp. By scalar? Thanks

 

[micromass]

You seem to be confusing the terms "one-to-one", "onto" and "linear". They are very distinct.

Let V= R^inffinty. Be the sequence space in which u1, u2,..., un,... is an infinite sequence of Real numbers..

Consider the linear "sshifting operators" on V defined by

T1 (u1, u2,..., un,...)=(0, u1, u2,...un,... )

T2 (u1, u2,..., un,...)=(u2, u3,...un,...)

(a) Show that T1 is one to one but not not onto.

what am I supposed to do? I am lost.

Maybe?

Let a=a1,a2,..., an,... and b=b1, b2,..., bn,... be a sequence of infinite real numbers in V.

T1 (a+b)=(0, a1+b1, a2+b2,...an+bn,...)

T1 (a)+T(b)=(0, a1, a2,...an,...)+(0, b1, b2,...bn,...)

Which equals the previous line? How would I show that it fails for multp. By scalar? Thanks

Now you just checked additivity, which is not what they asked you to do at all. Multiplication by scalars will be satisfied to, it won't fail. The map is linear, but that's not what they asked you.

You need to check that $T_1$ is one-to-one which means that you need to check that if $T(\mathbf{v}) = T(\mathbf{w})$, then $\mathbf{v} = \mathbf{w}$. So assume that $T(\mathbf{v}) = T(\mathbf{w})$ and write out what that means.

For not being onto, you need to find some vector $\mathbf{v}\in \mathbb{R}^\infty$ that is not in the image of $T_1$. So describe the image of $T_1$ and try to find some vector not in there?

 

[MidgetDwarf]

Still a bit confused.(u1, u2, .., un,..).=(0, u1, u2,..., un,...) ? For T (v)=T (w)

Pick 1,1,...,1,.. When I plug it in. I get a distinct image?

Not onto: because there is an element zero in W not in V?

Sorry about the idiocy, I'm preaty much on my own.

 

[micromass]

Let $\mathbf{v} = (v_1,v_2,v_3,...)$ and $\mathbf{w} = (w_1,w_2,w_3,...)$. You want to prove that $\mathbf{v} = \mathbf{w}$. This means that you need to prove that $v_1 = u_1$, $v_2 = u_2$, ...

So assume that $T(v_1,v_2,v_3,...) = T(w_1,w_2,w_3,...)$. Can you do something with this?

 

[WWGD]

For onto, consider

T(u_1,...,u_2)=(0,u_1,...). Can a point (1,v_1,v_2,...) be part of the image?