On Linear Transformations Tsquared = T

[cjqsg]

Diagonisability of Linear Transformations Tsquared = T

Let T be a linear transformation such that T^2 = T.

i. Show that if v is not 0, then either T(v) = 0 or T(v) is an eigenvector of eigenvalue 1. (easy)

ii. Show that T is diagonalisable.

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Sorry, I misread the question just now. For part 2, I need n linearly independent eigenvectors. How can I get them?

(my thoughts; consider a standard basis. T(e1), ..., T(en). Take out all T(e_i) such that T(e_i) = 0. Clearly, all e_i are linearly independent eigenvectors with eigenvalue 0. For the remaining e_j, we have T(ej) = eigenvector with eigenvalue 1. How do we show that these T(ej) are linearly independent?)

 

[lippyka]

Answer:We can prove that T is diagonalisable by showing that it has n linearly independent eigenvectors. We begin by considering a standard basis, i.e. e_1, ..., e_n. We note that T(e_i) may be either 0 or a vector with an eigenvalue of 1. First, we collect all the vectors T(e_i) such that T(e_i) = 0. These are all eigenvectors with eigenvalue 0. Clearly, these are all linearly independent. Now consider the remainder of the T(e_j). By the first part of the question, we know that these are eigenvectors with eigenvalue 1. To show that they are linearly independent, we will take two distinct T(e_j) and consider the linear combination c_1T(e_j1) + c_2T(e_j2). Applying T to both sides, we get c_1T^2(e_j1) + c_2T^2(e_j2). Since T^2 = T, this simplifies to c_1T(e_j1) + c_2T(e_j2). Since we started with c_1T(e_j1) + c_2T(e_j2), this implies that c_1 = c_2 = 0 and therefore the two eigenvectors T(e_j1) and T(e_j2) are linearly independent. Thus, we have shown that the set of all T(e_i) such that T(e_i) = 0, plus the set of all T(e_j) with eigenvalue 1, form a set of n linearly independent eigenvectors. This means that T is diagonalisable.