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Advanced Calculus of Several Variables, 5.6:

Two vector spaces V and W are called isomorphic if and only if there exist linear mappings [itex] S : V \to W[/itex] and [itex]T : W \to V[/itex] such that [itex]S \circ T[/itex] and [itex] T \circ S [/itex] are the identity mappings of [itex]S[/itex] and [itex]W[/itex] respectively. Prove that two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

I'm having a hard time proving the problem formally. I'm also having a hard time proving sufficiency of the criterion. In terms of necessity, I have this much:

It's clear that if one space, say V, has more dimensions than the other, then V which all map to the same value of W. For instance, [itex] v \in V : <v,w> = 0 [/itex] for all [itex]w \in W [/itex] all map to 0 in W (by the positivity property of inner products). There is then no function W -> V that can reclaim the lost information.

Formally, for all [itex] S : V -> W, S(v) = 0 [/itex] when v is orthogonal to every vector in W.

Consider two vectors [itex]v_{1}[/itex] and [itex]v_{2}[/itex] which are both orthogonal to every vector in W.

Then, for example, [itex]S(v_{1}) = S(v_{2}) = 0 \in W[/itex], and we have [itex] (T \circ S) (v_{1}) = (T(S(v_{1})) = T(S(v_{2})) = T(0). [/itex]

In order for the isomorphism to exist, [itex] (T \circ S) (v_{1}) = v_{1} [/itex] and [itex] (T \circ S) (v_{2}) = v_{2} [/itex]. But this is impossible since [itex](T \circ S) (v_{2}) = (T \circ S) (v_{1})[/itex], and [itex]v_{1} ≠ v_{2}[/itex]

Though I am satisfied by this, I have the feeling that I'm supposed to show this using matrix multiplication. The formalism that is eluding me is hiding in the phrase "The identity mappings of S and W respectively". What does that mean? [itex]S \circ T = I_{w}[/itex]? What does

Two vector spaces V and W are called isomorphic if and only if there exist linear mappings [itex] S : V \to W[/itex] and [itex]T : W \to V[/itex] such that [itex]S \circ T[/itex] and [itex] T \circ S [/itex] are the identity mappings of [itex]S[/itex] and [itex]W[/itex] respectively. Prove that two finite-dimensional vector spaces are isomorphic if and only if they have the same dimension.

I'm having a hard time proving the problem formally. I'm also having a hard time proving sufficiency of the criterion. In terms of necessity, I have this much:

It's clear that if one space, say V, has more dimensions than the other, then V which all map to the same value of W. For instance, [itex] v \in V : <v,w> = 0 [/itex] for all [itex]w \in W [/itex] all map to 0 in W (by the positivity property of inner products). There is then no function W -> V that can reclaim the lost information.

Formally, for all [itex] S : V -> W, S(v) = 0 [/itex] when v is orthogonal to every vector in W.

Consider two vectors [itex]v_{1}[/itex] and [itex]v_{2}[/itex] which are both orthogonal to every vector in W.

Then, for example, [itex]S(v_{1}) = S(v_{2}) = 0 \in W[/itex], and we have [itex] (T \circ S) (v_{1}) = (T(S(v_{1})) = T(S(v_{2})) = T(0). [/itex]

In order for the isomorphism to exist, [itex] (T \circ S) (v_{1}) = v_{1} [/itex] and [itex] (T \circ S) (v_{2}) = v_{2} [/itex]. But this is impossible since [itex](T \circ S) (v_{2}) = (T \circ S) (v_{1})[/itex], and [itex]v_{1} ≠ v_{2}[/itex]

Though I am satisfied by this, I have the feeling that I'm supposed to show this using matrix multiplication. The formalism that is eluding me is hiding in the phrase "The identity mappings of S and W respectively". What does that mean? [itex]S \circ T = I_{w}[/itex]? What does

*that*mean?
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