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My book says "Distinct vector spaces such as $\mathbb{R}^3$ and $\mathcal{M}_{(3,1)}$ can be thought of as being essentially the same--at least with respect to the operations of vector addition and scalar multiplication. Such spaces are said to be isomorphic to each other."

What exactly do they mean when they say isomorphic vector spaces are the same with respect to operations of vector addition and scalar multiplication? I guess I'm just not sure what they're referring to. The only thing I can think of them meaning is how vector spaces all satisfy the same 10 vector space axioms, but that applies to ALL vector spaces, not just isomorphic vector spaces.

Alekos Robotis
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dagny
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2 Answers2

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The best way I can answer this is as follows. If $V\cong W$ as vector spaces over a field $\mathbb{F}$ (say $\mathbb{F}=\mathbb{R},\mathbb{C}$), then there exists a bijective linear transformation $T:V\to W$ called an isomorphism. Linear maps have the following properties aside from being set theoretic maps. $$ T(v+w)=T(v)+T(w)$$ $$ T(kv)=kT(v)$$ for $v,w\in V$ and $k\in \mathbb{F}$. What does this tell us? In some sense, the operations of the two spaces can be understood as being the same. That is, to compute the sum of two vectors in $W$ we can compute the product in $V$ and then "send" it to $W$ using our isomorphism. We can perform the same procedure with scalar multiplication. In this sense, the operations are the same$^\dagger$.

Etymologically, it's worth noting that isomorphism derives from the Ancient Greek words "ίσος" meaning "the same" and "μορφή" basically meaning shape or form. This suggests that the concatenated word "iso+morphism" might mean having the same shape or form. Indeed, this is the case.

$\dagger:$ To be more precise, this map preserves addition of vectors and multiplication of vectors by scalars. For instance, if $T(v_1)=w_1$ and $T(v_2)=w_2$ and we wish to compute $w_1+w_2$, then we can compute $v_1+v_2$ and apply $T$. By linearity, $T(v_1+v_2)=T(v_1)+T(v_2)=w_1+w_2$.

Alekos Robotis
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  • "That is, to compute the sum of two vectors in W we can compute the product in V" Do you mean the sum in V? – dagny Feb 01 '17 at 08:33
  • Let me edit for clarity. – Alekos Robotis Feb 01 '17 at 08:34
  • Thanks for the edit, I think I was misinterpreting when you said "product" to mean "multiplication." – dagny Feb 01 '17 at 08:47
  • I'm still a little confused because it seems like your answer is just saying that because linear transformations preserve addition and scalar multiplication, the vector spaces are therefore kind of the same. That makes sense, but that's just the definition of a linear transformation, right? But not every linear transformation is an isomorphism. Where does the transformation having to be one-to-one and onto come into play? – dagny Feb 01 '17 at 08:47
  • That's a great question. So, I think we agree that the transformation preserves the "structure" of the space (this makes it called a homomorphism of vector spaces by the way). Now, that's nice, but we'd also like to be able to think of each element in $V$ as an element of $W$, and that's where the bijection comes in to play. The fact that $T$ is a bijection ensures that we can represent each element of $V$ with exactly one in $W$ and vice versa. So, as sets we can think of them as being the same, and their additional structure is also the same. – Alekos Robotis Feb 01 '17 at 08:50
  • Okay, that makes a little more sense. Thanks – dagny Feb 01 '17 at 08:56
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Suppose we have addition and multiplication tables for the vector space. For the addition table, all the vectors in the space appear in the first column, and likewise for the first row. Then, to determine $\mathbf{v} + \mathbf{w}$ for some vectors $\mathbf{v}, \mathbf{w} \in V$, we simply look at the entry associated with row $\mathbf{v}$ and column $\mathbf{w}$.

For the multiplication table, the first column will list the elements of the scalar field, and first row will list the vectors. To find out what $c \mathbf{v}$ is for some $c \in F$ and $\mathbf{v} \in V$, we simply go down to the row labeled with $c$ and across to the column labeled with $\mathbf{v}$ and read that entry.

We say two vector spaces are isomorphic if their operation tables are the same up to relabeling of the elements. In other words, suppose $V$ is a vector space over $F$ and $V'$ a vector space over $F'$. If we can find bijections $V \rightarrow V'$ and $F \rightarrow F'$ such that applying these to the operation tables for $V$ yields the operation tables for $V'$, then $V \cong V'$.

More generally, linear transformations between vector spaces (that aren't necessarily isomorphisms) still "transport" structure from one vector space to another. Algebra, by and large, is the study of algebraic structures and the transformations / homomorphisms between them; there is quite a bit of analogy to be made between vector space transformations, group homomorphisms, ring homomorphisms, and so forth. My answer here provides more detail in the context of ring homomorphisms, and the ideas are more-or-less applicable to any other structure -- vector spaces, modules, groups -- with only minor changes to the notation / wording.

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Kaj Hansen
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