Isomorphism Clarification and Identification

Question

I understand that to claim isomorphism, algebraic properties must be preserved...so....a) closed under multiplication and closed under addition.

However, I am unsure how to apply these condition-testing properties in the context of polynomial/space based questions

For example: Which of the following is isomorphic to a subspace of R^3x4

1. P9
2. P11
3. Upper triangular matrices in R^2x3
4. R12

How can I show closed under addition/multiplication in each of these contexts?

Answer 1

The simpler way to find out if two finite dimensional vector spaces are isomorphic, as others said, is to find out what if the dimension of the vector space that you're studying. So we can state a little "theorem"

Two finite dimensional vector spaces $V,W$ are isomorphic iff $$\dim(V)=\dim(W)$$ Here are some references.

After stating this we can easily find out which of your vector spaces is isomorphic to $\mathbb{R}^{3\times 4}$. First of all we have that $$\dim(\mathbb{R}^{3\times 4}) = 12$$ and it's easy to see because to write down a $3$ by $4$ matrix you need $12$ indipendent components.

Now let's see the other dimensions $$\begin{align} P_9 &&\rightarrow&&\dim(P_9)=9\tag1\\ P_{11} &&\rightarrow&&\dim(P_{11})=11\tag2\\ \text{upper triangular }\mathbb{R}^{2\times 3}&&\rightarrow&&\dim(\text{upper triangular }\mathbb{R}^{2\times 3})=5\tag3\\ \mathbb{R}^{12}&&\rightarrow&&\dim(\mathbb{R^{12}})=12\tag4 \end{align}$$

For the first and second one the dimension is trivial to find: a polynomial of degree less and equal to $n$ has $n$ coefficients, so the vector space is $n$-dimensional. The third one is more tricky: one would think that the dimension would be $6$ but an upper triangular $2$ by $3$ matrix is of the form $$\left(\begin{matrix}a_{11}&a_{12}&a_{13}\\0&a_{22}&a_{23}\end{matrix}\right)$$ so the element $a_{21}=0$ always. The only independent components ar the $5$ left. The fifth is just $\mathbb{R}^n$ so it's dimension is $n$.

Clearly the only isomorphic space to $\mathbb{R}^{3\times2}$ in this list is $\mathbb{R}^{12}$. If you want to find an isomorphism, which is not requested if the only thing you want is to find if to vector spaces are isomorphic, you could use this isomorphism $$\phi:\mathbb{R}^{3\times4}\rightarrow \mathbb{R}^{12} \\ \left(\begin{matrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\end{matrix}\right)\mapsto (a_{11},a_{12},a_{13},a_{14},a_{21},a_{22},a_{23},a_{24},a_{31},a_{32},a_{33},a_{34})$$ i.e. you take all the entries in the matrix and map them to a vector that has as components the $a_{ij}$ of the matrix.

Answer 2

Leibnewtz is right for finite dimensional vector spaces. You basically just need to count the number of degrees of freedom in your space to determine the dimension. For example, an element of $R^{(3,4)}$ contains 12 degrees of freedom, so it has dimension $12$. So it is obvious that $R^{12}$ is isomorphic to $R^{(3,4)}$. You could also easily construct the isomorphism between these spaces, for example, by stacking the columns of an element of $R^{(3,4)}$ into a single long column vector, and check that this map preserves algebraic properties.