Deriving a simpler expression for $\left({I}-T\otimes T\right)^{-1}\left(T\otimes T\right)$ using Kronecker product properties

Question

Let $T\in\mathbb{R}^{n\times n}$.

Is there a simpler expression for $$\left(I_{n^{2}}-T\otimes T\right)^{-1}\left(T\otimes T\right)~?$$

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When $T$ is symmetric PSD, we could do use the unitary decomposition as in this question, but I am interested in a general square $T$.

Answer 1

In general, if $M$ is invertible:

$$(I-M)^{-1}M=(M(M^{-1}-I))^{-1}M=(M^{-1}-I)^{-1}M^{-1}M=(M^{-1}-I)^{-1}$$

giving here the alternative form, in the case $T$ (therefore $T \otimes T$) is invertible:

$$\left(\left(T\otimes T\right)^{-1} -I_{n^{2}}\right)^{-1}$$

with $\left(T\otimes T\right)^{-1}=T^{-1} \otimes T^{-1}.$

Is it really "simpler" ? It certainly depends on the use you will have of this expression.