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I'm wondering whether can we define an explicit Kronecker product of Hilbert–Schmidt integral operators. $T(f)=\int k(x,y)f(y)dy$.

Since integral operator is an extension of matrix operation, I'm curious whether there's a corresponding concept of Kronecker product.

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The Kronecker product of two matrices is just the matrix corresponding to the tensor product of the two linear maps that the matrices. For any two linear maps of vector spaces you always get a linear map between the tensor products of the domains and the tensor products of the codomains. So yes you can definitely do that, it's not necessarily clear that it will have a nice interpretation, but let's see what happens.

Let's say for example that the domain you are looking at is the unit interval $[0,1]$, then these operators are linear maps from $L^2([0,1])$ to itself. Hence the tensor product of two such maps should be a linear map from $L^2([0,1]) \otimes L^2([0,1])$ to itself. Explicitly the map just sends $f_1 \otimes f_2$ to $T_1(f_1) \otimes T_2(f_2)$.

Now let's identify $L^2([0,1]) \otimes L^2([0,1])$ with $L^2([0,1]^2)$ in the natural way. That is we send $f_1 \otimes f_2$ to the function $f_1(x)f_2(y)$ on $[0,1]^2$, the span of such functions is dense in $L^2([0,1]^2)$ so this identification makes sense when we use the typical completed tensor product for Hilbert spaces. Now by working it out explicity for functions $f(x,y) = f_1(x)f_2(y)$ and extending linearly we get that our map is just: $$T_1 \otimes T_2(f)(x,y) = \int \int k_1(x,t)k_2(y,z)f(x,y)dtdz$$ which we can recognize as another Hilbert-Schmidt integral operator, but this time it's on $L^2([0,1]^2)$.

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  • $\begingroup$ Thank you for providing this example. I'm not familiar with Tensor product, it seems there may be many constructions. The double integral haunts me a little bit. Maybe Kronecker product does not have an exact analog at all. I'll accept your answer if no one comes up with a even nicer one. Thanks again. $\endgroup$ – iridium Mar 17 '15 at 19:22
  • $\begingroup$ Might I ask, how are you interpreting the what the Kronecker product is without tensor products? $\endgroup$ – Nate Mar 17 '15 at 19:37
  • $\begingroup$ I'm seeing it as something similar to Hadamard product. I know it's conceptually incorrect, yet they are somewhat similat in some applications. $\endgroup$ – iridium Mar 17 '15 at 21:16

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