1. ## Isomorphic Vector Spaces

Here is a question, I have came up with an awnser, although im not really sure if my proof is sufficient.

Let $M_{\infty}(R)$ denote the vector space of infinite matrices with finitely many non zero enteries.

Let $R^{\infty}$ be the vector space of infinite sequences in which only finitely many elements are nonzero.

Prove that $R^{\infty}$ and $M_{\infty}(R)$ are isomorphic.

My attempt at the solution:

We know two vector spaces are isomorphic, if and only if their dimensions are equal.

Let $m =$ the finite number of non zero entires in $M_{\infty}(R)$. This is equal to the dimension of $M_{\infty}(R)$.

Let $r =$ the finite number of non zero elements in $R^{\infty}$. This is equal to the dimension of $R^{\infty}$

Since r and m are just any finite number, we can let $r = m$, and then since the dimensions are equal, they are isomorphic.

Is this proof enough? If not a point in the right direction would be great! Thanks.

2. Originally Posted by joe909
Here is a question, I have came up with an awnser, although im not really sure if my proof is sufficient.

Let $M_{\infty}(R)$ denote the vector space of infinite matrices with finitely many non zero enteries.

Let $R^{\infty}$ be the vector space of infinite sequences in which only finitely many elements are nonzero.

Prove that $R^{\infty}$ and $M_{\infty}(R)$ are isomorphic.

My attempt at the solution:

We know two vector spaces are isomorphic, if and only if their dimensions are equal.

Let $m =$ the finite number of non zero entires in $M_{\infty}(R)$. This is equal to the dimension of $M_{\infty}(R)$.

Let $r =$ the finite number of non zero elements in $R^{\infty}$. This is equal to the dimension of $R^{\infty}$

Since r and m are just any finite number, we can let $r = m$, and then since the dimensions are equal, they are isomorphic.

Is this proof enough? If not a point in the right direction would be great! Thanks.

There is no fixed number of non-zero entries either in $M_\infty(\mathbb{R}) \,\,\,or\,\,\,in\,\,\,\mathbb{R}^\infty$ ! So your numbers m, r are meaningless.

In fact, both space are infinite dimensional, but the gist of the matter is that both have the same cardinal as dimension, which is = $\aleph_0$ over the reals $\mathbb{R}$ and, thus, they both are isomorphic.

Tonio

3. Originally Posted by joe909
Here is a question, I have came up with an awnser, although im not really sure if my proof is sufficient.

Let $M_{\infty}(R)$ denote the vector space of infinite matrices with finitely many non zero enteries.

Let $R^{\infty}$ be the vector space of infinite sequences in which only finitely many elements are nonzero.

Prove that $R^{\infty}$ and $M_{\infty}(R)$ are isomorphic.

My attempt at the solution:

We know two vector spaces are isomorphic, if and only if their dimensions are equal.
No, we do not know that! That is only true for finite dimensional vector spaces.

Let $m =$ the finite number of non zero entires in $M_{\infty}(R)$. This is equal to the dimension of $M_{\infty}(R)$.
That doesn't even make sense. Individual matrices in $M_\infty$ have different numbers of non zero entries. There is no one "m" for every matrix. Since there is no upper bound to how many non-zero entries such a matrix can have, this vector space is infinite dimensional.

Let $r =$ the finite number of non zero elements in $R^{\infty}$. This is equal to the dimension of $R^{\infty}$
Again, that does not make sense. Different sequences have different numbers of non-zero entries. Since there is no upper bound to the number of non-zero entries a sequence can have, this vector space is infinite dimensional.

Since r and m are just any finite number, we can let $r = m$, and then since the dimensions are equal, they are isomorphic.

Is this proof enough? If not a point in the right direction would be great! Thanks.
No, you have apparently misunderstood the what is meant by "finite number of non-zero entries. You will need to find an actual isomorphism from $M_\infty$ to $R^\infty$. Think about making the entries in a matrix into a "list" by zig-zagging through the matrix.