Endomorphisms, Tensors in Disguise |18 August 2020|
`tags: math.LA`

Unfortunately the radio silence remained, the good news is that even more cool stuff is on its way*!

For now, we briefly mention a very beautiful interpretation of (1,1) -tensors. Given some vectorspace mathcal{V} , let mathrm{End}(mathcal{V}) be the set of linear endomorphisms from to . Indeed, for finite-dimensional vector spaces one can just think of these maps as being parametrized by matrices. Now, we recall that (1,1) -tensors are multilinear maps which take an element of and its dual space $mathcal{V}^{star}$ and map it to some real number, that is, $T^1_1:mathcal{V}times mathcal{V}^{star}to mathbf{R}$ .

The claim is that for finite-dimensional mathcal{V} the set mathrm{End}(mathcal{V}) is isomorphic to the set of (1,1) -tensors $mathcal{T}_1^1(mathcal{V})$ . This is rather intuitive to see, let vin mathcal{V} and $v^{star}in mathcal{V}^{star}$ , then, we can write $langle v^{star},T_1^1 vrangle$ and think of $T_1^1in mathcal{T}^1_1(mathcal{V})$ as a matrix. This can be formalized, as is done here by showing there is a bijective map $phi:mathrm{End}(V)to mathcal{T}^1_1(mathcal{V})$ . So, we can identify any T_1^1 with some $Ain mathbf{R}^{ntimes n}$ , that is $T_1^1(v,v^{star})=v^{star}(Av)=langle v^{star},Av rangle$ . Similarly, we can fix $v^{star}$ and identify T_1^1 with a linear map in the dual space, that is $T_1^1(v^{star})(v)=(A^{star}v^{star})(v)=langle A^{star}v^{star},vrangle$ . Regardless, both represent the same tensor evaluation, hence

$langle v^{star},Av rangle = langle A^{star}v^{star},v rangle,$

which is indeed the standard dual map definition.

Now we use this relation, recall that discrete-time linear systems are nothing more than maps vmapsto Av , that is, linear endomorphims, usually on $mathbf{R}^n$ . Using the tensor interpretation we recover what is sometimes called the ‘‘dual system’’ to , namely $v^{star}mapsto A^{top}v^{star}$ (suggestive notation indeed). Using more traditional notation for the primal: $x_{k+1}=Ax_k$ and dual: $z_{k+1}=A^{top}z_k$ system, we recover that $langle z_{k+1},x_k rangle = langle z_k, x_{k+1}rangle forall kin mathbf{Z}$ . Then, let z_0=w and x_0=v for w_j and v_i being a left- and right-eigenvector of , respectively. We see that this implies that mu_j^k langle w_j, v_i rangle = lambda_i^k langle w_j, v_i rangle , for mu_j the eigenvalue corresponding to w_j and the eigenvalue corresponding to v_i . Hence, in general, the eigenspaces of and $A^{top}$ are generically orthogonal.