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I want to show that if $L:\mathbb{R}^d\rightarrow\mathbb{R}^d$ is a linear isomorphism and $V_1$ are linear subspaces of $\mathbb{R}^d$ then:

$$||(L|V_1)^{-1}||^{-d_1}\leq|det(L|V_1)|\leq||(L|V_1)||^{d_1}$$ where $d_1=\text{dim}{V_1}$ and $(L|V_1)$ is the restriction of $L$ to $V_1.$

I haven't any idea. Can you help me?

Here there is a formal definition of det$(L|V_1)$:

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    The notation $(L|V)$ seems nonstandard; perhaps you can define it? – 2017-01-03
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    Maybe the restriction? Another question... Where is V_2 ? – 2017-01-03
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    I have add some more details in the question. Now it could be clear. Can you help me?? – 2017-01-03
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    For $\det(L\lvert_{V_1})$ to be defined you need $L\lvert_{V_1}$ to have as image $V_1$ again, which is a restriction on $L$. – 2017-01-03
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    What norm are you using to measure $\| L \|$? Is it the Hilbert-Schmidt norm? – 2017-01-03

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For everything to be well defined you need $L\lvert_{V_1}$ to be a map $V_1\to V_1$, otherwise the determinant does't exist. But with this there is no need to restrict to a subspace at all, if the statement holds for any map $L:\mathbb R^d\to\mathbb R^d$ then it also holds for the restriction to any subspace, as you will be in the same situation.

In finite dimensional spaces $\|L\|$ is the magnitude value of the largest (in absolute value) eigenvalue of $L$. The determinant is the product over all eigenvalues with multiplicity corresponding to the algebraic multiplicity of that eigenvalue. If $L$ is invertible then the eigenvalues of $L^{-1}$ are the inverses of the eigenvalues of $L$.

Put everything together: $$\|L^{-1}\|^{-1}=(\sup_{\lambda \text{ EV of }L}|\lambda^{-1}|)^{-1}=\inf_{\lambda\text{ EV of }L}|\lambda|\\ \det(L)=\prod_{\lambda\text{ EV of }L}\lambda\implies |\det(L)|=\prod_{\lambda \text{ EV of }L}|\lambda|\\ \|L\|=\sup_{\lambda\text{ EV of }L}|\lambda|$$ One has for any eigenvalue $\lambda$ that $0≤\|L^{-1}\|^{-1}≤|\lambda|≤\|L\|$, taking the product over all eigenvalues gives then: $$(\|L^{-1}\|^{-1})^d≤|\det(L)|≤\|L\|^d$$

In the case that $L\lvert_{V_1}$ is not a map $V_1\to V_1$ but a map $V_1\to V_2$ for some $V_2\subset V$ we do something similar. Since $L$ is invertible $V_1$, $V_2$ are isomorphic finite dimensional subspaces of $V=\mathbb R^n$. They inherit the inner product of $\mathbb R^n$ and are actually unitarily isomorphic as Hilbert spaces, so let $U: V_2\to V_1$ be such a unitary isomorphism. You have:

$$\|UL\lvert_{V_1}\|=\|L\lvert_{V_1}\|\qquad \|L\lvert_{V_1}^{-1}U^{-1}\|=\|L\lvert_{V_1}^{-1}\|\\ \mathrm{vol}(UL\lvert_{V_1}v_1,..,UL\lvert_{V_1}v_n)=\mathrm{vol}(L\lvert_{V_1}v_1,..,L\lvert_{V_1}v_n)\implies \det(UL\lvert_{V_1})=\det(L\lvert_{V_1})$$ all from the virtue of $U$ being unitary. This means we have reduced to the situation considered before, here $UL\lvert_{V_1}$ is our map $V_1\to V_1$ for which the inequality holds.

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    I add a formal defition of the "restricted determinant" – 2017-01-03
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    @Mr.Brown this makes things different, now the only condition you have is that $L\lvert_{V_1}$ is invertible as a linear map $V_1\to L(V_1)$. – 2017-01-03
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    Do you have any idea to adapt your proof? – 2017-01-03
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    Not really, if we are in a setting where $V$ is a Hilbert space then $V_1$ and $L(V_1)$ will be unitarily isomorphic and it will work with some identifications. If $V$ is not Hilbert you can have problems where $V_1$ and $L(V_1)$ are not isometric and eigenvalues under some identifications will not directly correspond to the norm. – 2017-01-03
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    We are working in $\mathbb{R}^d$ – 2017-01-03
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    Ok it should work, tonight its too late for me to do anything anymore I will look at this again tomorrow and see if what I thought has any value. – 2017-01-03
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    @Mr.Brown sorry for being late, I have written up how it reduces to the situation from before. – 2017-01-06