Abstract Nonsense

Crushing one theorem at a time

Halmos Section 24 and 25: Tensor Product and Product Bases

Point of post: This will be the solutions to the problems in Halmos’s book corresponding to this post.


Problem: If x=(1,1) and y=(1,1,1) are vectors in \mathbb{R}^2 and \mathbb{R}^3 respectively, find the coordinates of x\otimes y in \mathbb{R}^2\otimes\mathbb{R}^3 with respect to the product basis \{x_i\otimes y_j\} where x_i=(\delta_{i,1},\delta_{i,2}) and y_j=(\delta_{1,j},\delta_{2,j},\delta_{3,j}).

Remark: For those working out of Halmos as well, I’d be surprised if you understood this on a first read. What I’ve determined it to mean is that since \left\{e_i\otimes e_j:i\in[2]\text{ and }j\in[3]\right\} is a basis for \mathbb{R}^2\otimes\mathbb{R}^3 (where the e_i,e_j are the canonical basis elements for \mathbb{R}^2,\mathbb{R}^3) then how can we express x\otimes y in terms of this basis.

Proof: We notice that x=x_1+x_2 and y=y_1+y_2+y_3 and so

\displaystyle \begin{aligned} x\otimes y &=\left(x_1+x_2\right)\otimes (y_1+y_2+y_3)\\ &= \sum_{i=1}^{2}x_i\otimes (y_1+y_2+y_3)\\ &= \sum_{i=1}^{2}\sum_{j=1}^{3}x_i\otimes y_j\end{aligned}


Problem: Let \mathbb{C}_{n,m}[x,y], \mathbb{C}_n[x], and \mathbb{C}_m[y] be defined as in my post on tensor products. Prove there exists an isomorphism


such that

f:p(x)\otimes q(y)\mapsto p(x)q(y)

Proof: Note that \mathcal{B}_1=\left\{x^j\otimes y^i:j\in[n-1]\text{ and }i\in[m-1]\right\} and \mathcal{B}_2=\{x^jy^i:j\in[n-1]\text{ and }i\in[m-1]\} are bases for \mathbb{C}_n[x]\otimes \mathbb{C}_m[y] and \mathbb{C}_{n,m}[x,y] respectively. So, define

:\mathcal{B}_1\to\mathcal{B}_2:x^j\otimes y^i\mapsto x^jy^i

Clearly this is a linear bijection and so we may extend f to


by linearity. By a previous post we may conclude that \tilde{f} is an isomorphism. Also, we note that if \displaystyle p(x)=\sum_{j=0}^{n-1}\alpha_j x^j and \displaystyle q(y)=\sum_{i=0}^{m-1}\beta_i y^i we see that

\displaystyle \begin{aligned}f\left(\left(\sum_{j=0}^{n-1}\alpha_j x^j\right)\otimes\left(\sum_{i=0}^{m-1}\beta_i y^i\right)\right) &= f\left(\sum_{j=0}^{n-1}\sum_{i=0}^{m-1}\alpha_j \beta_i f\left(x^j\otimes y^i\right)\right)\\ &= \sum_{j=0}^{n-1}\sum_{i=0}^{m-1}\alpha_j \beta_i x^jy^i\\ &= \left(\sum_{j=0}^{n-1}\alpha_j x^j\right)\left(\sum_{i=0}^{m-1}\beta_i y^i\right)\\ &= p(x)q(y)\end{aligned}


Problem: To what extent is the formation of tensor products commutative and associative? What about the distributive laws?


Proof: It’s true that \mathscr{U}\otimes\mathscr{V}\cong\mathscr{V}\otimes \mathscr{U}, besides a clear dimension argument one immediately sees that if \{x_1,\cdots,x_n\} and \{y_1,\cdots,y_n\} are bases for \mathscr{U},\mathscr{V} respectively then the canonical map determined by

f:x_i\otimes y_j\mapsto y_j\otimes x_i

is evidently an isomorphism. Similarly, \mathscr{U}\otimes\left(\mathscr{V}\otimes\mathscr{W}\right)\cong\left(\mathscr{U}\otimes\mathscr{V}\right)\otimes\mathscr{W} via the natural isomorphism which is determined by

g(x_i\otimes (y_j\otimes z_k)\mapsto (x_i\otimes y_j)\otimes z_k

Lastly, the above distributivity property holds vai the isomorphism determined by

h:x_i\otimes(y_j\otimes z_k)\mapsto \left(x_i\otimes y_j,x_i\otimes z_k\right)

Regardless, all of these are trivially true by a dimension argument.


Problem: If \mathscr{V} is a finite-dimensional vector space, and if x,y\in\mathscr{V} is it true that x\otimes y=y\otimes x? Not always, take \mathscr{V}=\mathbb{C}^2 then notice that

\left(\begin{bmatrix} 1 \\ 1 \end{bmatrix}\otimes\begin{bmatrix} 1\\ -1 \end{bmatrix}\right)\left(\det\right)=-2\ne 2=\left(\begin{bmatrix}1\\ -1\end{bmatrix}\otimes\begin{bmatrix} 1\\ 1\end{bmatrix}\right)\left(\det\right)



a) Suppose that \mathscr{V} is a finite-dimensional real vector space, and  \mathbb{C} of all complex numbers regarded as a real vector space. Form the tensor product \mathscr{V}^{\mathbb{C}}=\mathbb{C}\otimes \mathscr{V}. Prove that there is a way of defining products of complex numbers with elements of \mathscr{V}^{\mathbb{C}} so that \alpha(x\otimes y)=(\alpha x)\otimes y whenever \alpha,x\in\mathbb{C}.

b) Prove that with respect to vector addition, and with respect to scalar multiplication as defined in a) the space \mathscr{V}^{\mathbb{C}} is a complex vector space.

c) Find the dimension of the complex vector space \mathscr{V}^{\mathbb{C}} in terms of the dimension of the real vector space \mathscr{V}.

d) Prove that the vector space \mathscr{V} is isomorphic to a subspace in \mathscr{V}^{\mathbb{C}}.


a) For \alpha,z\in\mathbb{C} and v\in\mathscr{V} define

\alpha\cdot (z\otimes v)=(\alpha z)\otimes v

Let \{x_1,\cdots,x_n\} be a basis for \mathscr{V} and let z_1=1,z_2=i and so  extend this to a map



\displaystyle \alpha\left(\alpha\otimes v\right)=\alpha\sum_{i=1}^{2}\sum_{j=1}^{n}\alpha_i\beta_j (z_i\otimes x_j)=\sum_{i=1}^{2}\sum_{j=1}^{n}\alpha_i\beta_j \left(\alpha\cdot(z_i\otimes x_j)\right)

This map is evidently associative and 1\cdot(z\otimes v)=z\otimes v. To prove distributivity it suffices to prove that (\alpha+\beta)\cdot (z\otimes v)=\alpha\cdot(z\otimes v)+\beta\cdot(z\otimes v) and \alpha((z_1\otimes v_1)+(z_2\otimes v_2))=\alpha\cdot (z_1\otimes v_1)+\alpha(z_2\otimes v_2). The first one is proven as follows:

\begin{aligned}(\alpha+\beta)\cdot(z\otimes v) &= \left((\alpha+\beta)z\right)\otimes v\\ &= \left(\alpha z+\beta z\right)\otimes v\\ &= \left((\alpha z)\otimes v\right)+\left((\beta z)\otimes v\right)\\ &=\left(\alpha\cdot(z\otimes v)\right)+\left(\beta\cdot(z\otimes v)\right)\end{aligned}

The second one follows by definition of the extension.

b) This is all pretty standard stuff. In other words, all of the axioms are just rote definitions.

c) We know that when we consider \mathbb{C}\otimes\mathscr{V} to be a real vector space that \{z\otimes x_j:z=1,i\text{ and }j\in[n]\} is a basis. We claim that \{1\otimes x_j:j\in[n]\} is a basis for \mathscr{V}^{\mathbb{C}}. To see this suppose that

\displaystyle \sum_{j=1}^{n}(\alpha_j+\beta_j i)\cdot (1\otimes x_j)=\sum_{j=1}^{n}(\alpha_j+\beta_j i)\otimes x_j0=\sum_{j=1}^{n}\alpha_j\otimes x_j+\sum_{j=1}^{n}(\beta_j i)\otimes x_j=\bold{0}

but, since \{z\otimes x_j:z=1,i\text{ and }j\in[n]\} is linearly independent (over \mathbb{R}) it follows that \alpha_1=\cdots=\alpha_n=\beta_1=\cdots=\beta_n=0. Also, noting (again as above) that

\displaystyle \sum_{j=1}^{n}(\alpha_j\otimes \beta_j i)\cdot(1\otimes x_j)=\sum_{j=1}^{n}\alpha_j\otimes x_j+\sum_{j=1}^{n}(\beta_j i)\otimes x_j

and \text{span} _{\mathbb{R}}\text{ }\{z\otimes x_j:z=1,i\text{ and }j\in[n]\}=\mathbb{C}\otimes\mathscr{V} it follows that \text{span}_{\mathbb{C}}\text{ }\{1\otimes x_j:j\in[n]\}=\mathbb{C}\otimes\mathscr{V}. Thus, \{1\otimes x_j:j\in[n]\} is a basis for \mathbb{C}\otimes\mathscr{V} over \mathbb{C} and we get the pleasant result that


c) We can clearly see that f:\mathscr{V}\to\mathbb{C}\otimes\mathscr{V}:v\mapsto 1\otimes v is an embedding whether we consider \mathbb{C}\otimes\mathscr{V} to be a real or complex vector space, but in the latter case it’s an isomorphism.



1. Halmos, Paul R. “Bilinear Forms.” Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print


November 1, 2010 - Posted by | Uncategorized | , , , ,

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