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Let $f: R^2 \times R^2 \rightarrow R^{2 \times 2}$ s.t. $ f((x_1, y_1 ),(x_1, y_1 )) = \begin{bmatrix} x_{1} y_1 & x_{1} y_2 \\ x_{2} y_1 & x_{2} y_2 \end{bmatrix} $

f is bilinear, but not surjective. I am told that that (f, $R^{2 \times 2}$) has the universal property, i.e. that for any bilinear map g from $R^2\times R^2$ to some vector space H, there exists a unique linear map G from $R^{2 \times 2}$ to H s.t. G $\circ$ f = g. However, I do not think this is possible, since f is not subjective. (We can consider $\begin{bmatrix} 0 & 1 \\ 1& 0 \end{bmatrix} $ for example.)

Does (f, $R^{2 \times 2}$) have the universal property? I know $R^{2 \times 2}$ is isomorphic as a tensor product to $R \otimes R$, but I can't seem to find an isomorphism. I think this might be the first step, any help would be much appreciated.

fdzsfhaS
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1 Answers1

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This $R^4$ and $f$ do indeed have the universal property.

If the first copy of $R^2$ has the basis $(e_1, e_2)$ and the second copy $(f_1,f_2)$, you know that $R^2 \otimes R^2$ has as a basis $(e_1 \otimes f_1,e_1 \otimes f_2,e_2 \otimes f_1,e_2 \otimes f_2)$, and the mapping $f$ you describe is simply the transcription of the mapping $(v,w) \mapsto v \otimes w$ in the corresponding coordinate system.

Of course, this is a post hoc justification once the theory of tensor products is somewhat established. I won't write a direct proof here from first principles.

You say that because $f$ is not surjective, this can't be the case. I don't follow this argument. It may be that you are falsely generalizing the following property of sets. If $f \colon E \to F$ and $g \colon F \to G$, and $f$ is not surjective, then the mapping $g$ need not be uniquely determined by $g \circ f$. This is because $g$ can be defined arbitrarily on $F - f(E)$. (Assume $\operatorname{Card} (G) > 1$.)

However, if $F$ and $G$ are vector spaces and $g$ is assumed linear, then $g$ is uniquely determined as soon as the set $f(E)$ generates $F$ as a vector space. And this does in fact occur in the case at hand, since the set of all quadruples $(x_1 y_1,x_1 y_2,x_2 y_1,x_2 y_2)$ generates $R^4$.

Dave
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