# Analogy between Linear Transformations & Matrices

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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Note that it is possible to have $T^2=0$ without having $T=0$. Among other things, this implies $\lambda$ could be zero. –  Gerry Myerson Dec 6 '12 at 5:22
@GerryMyerson:Would someone of you please checkout whether my approach to the given problem is right: (4) is true: $Nullity(T)+Rank(T)=n$ $\Rightarrow$ $Nullity(T) = n - k>0$ $\Rightarrow$ $Ker$ $T \neq 0$ $\Rightarrow$ (4) is true; (1) is false: Consider $T:V_n\to V_n:v\mapsto -v$$\Rightarrow T^2=(-1)T \Rightarrow (1) is false; (3) is false: Let T =$$\begin{pmatrix}1&0\\0&0\\\end{pmatrix}$$Then$T$satisfies the given conditions but here$\lambda = 1$is not the only eigenvalue of$T$; (2) is false: The above example shows that det$T$=$0\neq |1|^2\$. –  Sugata Adhya Dec 7 '12 at 10:40
Looks good to me. –  Gerry Myerson Dec 7 '12 at 11:42
Thanks @GerryMyerson –  Sugata Adhya Dec 7 '12 at 11:48