Prove that $\det(X), \det(X+I), \det(X-I)$ are linearly independent functions 
  
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*Let $X \in M_3(\mathbb{C})$. Prove that functions $\det(X), \det(X+I), \det(X-I)$ are linearly independent.
  
*Prove that there is such $m \in \mathbb{N}$, that functions
  $$\det(X-mI),\ \det(X-(m-1)I),\ \det(X-(m-2)I),\ ...,\ \det(X+mI)$$
  are linearly dependent.
  

I have a question about both parts of the task.
Let denote by $V$ the vector space of complex polynomials of $9$ variables with degree $\leq 3$ (btw, is there some standard notation for such vector spaces?).


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*I have no better idea of how to solve this except by writing down determinants explicitly and find some coordinates of these vectors (as vectors in $V$) showing that they are not linearly dependent. For example coordinates of free term are $0,1,-1$, coordinates of $x_{11}x_{22}x_{33}$ are $1,1,1$ and coordinates of $x_{11}$ are $0,1,1$ respectively for $\det X, \det (X+I), \det (X-I)$. And as vectors $(0,1,-1), (1,1,1), (0,1,1)$ are linearly independent, our functions are independent too. But this way of proof is too cumbersome. Is there a way to show independence in a more elegant way?

*In the second part, is it enough to say, that we can just pick $m$ large enough so that the amount of our functions, which grows as $2m+1$, will outgrow the dimensionality of $V$, which is finite and fixed?
 A: The first one is fairly straightforward. Indeed, if there are constants $a,b,c$ such that $a \det X  + b \det (X-I) + c \det (X+I) = 0$ for all $X$ in the domain, then we can choose three such $X$ which produce equations in $a,b,c$ which are unsatisfiable unless they are all zero.
For example, take $X = 0$, the zero matrix, then the equation gives $-b +c = 0$.
Then take $X = I$, so we get $a+8c = 0$. Finally, take $X = 2I$, this gives $8a +b + 27c = 0$. 
You can see that $a=b=c = 0$ is forced from here.
Note : a similar approach is used to show linear independence of functions over $\mathbb R$ or the complex numbers (for example, of $e^x,e^{2x}$ and $e^{3x}$, say). The idea of substituting appropriate values to get equations relating the coefficients, is therefore worth remembering as a technique to establish linear independence in this case.
The answer to the second one is correct : indeed, any such expression $\det(X + nI)$ is a polynomial of degree atmost $3$ over nine variables (A nice notation is suggested in the comments), so any set of such functions is a subset of a finite dimensional vector space. Thus, ensuring there are more functions in the set that the dimension of the vector space (you can find this explicitly if you like), makes sure that such a set is linearly dependent.
EDIT : When you noticed that the determinant is a polynomial of bounded degree, you did yourself a great favour. Indeed, the vector space of all functions, or of all polynomials, is infinite dimensional, so in that case such an argument could not have worked out.
