Keeping in mind that the set of all n $\times$ n matrices over a field F forms a vector space isomorphic to L(V, V) where dim V = n & all the result of eigen values of a square matrix has its analogous outcomes for corresponding linear operator, can we transform any problems involving the eigen values of square matrix to a analogous problem for operator. Let us take an example: Let T ($\neq$ $O$) be a linear operator on a n-dimensional real vector space V such that Rank (T) = k < n. Suppose for some real $\lambda$, $T^2$=$\lambda$$T$. Then which of followings are true: 1. $\lambda$=1, 2. det T = $|\lambda|^n$, 3. $\lambda$ is the only eigen value of T, 4. $\exists$ a non-trivial subspace $V_0$$\subset V$ sucht that $Tx=0$ $\forall$ $x \in V_0$. If 2 & 4 are the correct alternatives of the problem then can we say 2 & 4 are also the correct alternatives of the corresponding problem obtained by replacing the term 'linear operator' with 'martix' & vice-versa.
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Yes, in your example. In general, some care must be taken. For example, any matrix is similar to its Jordan form, and this is useful; but there are properties of operators that are not preserved by similarity, like positivity and any orthogonal-related property.