Idempotent linear transformations are precisely subspace projections

Let V be an n-dimensional vector space over a field F. Let \varphi : V \rightarrow V be a linear transformation, and suppose \varphi^2 = \varphi.

  1. Prove that \mathsf{im}\ \varphi \cap \mathsf{ker}\ \varphi = 0.
  2. Prove that V = \mathsf{im}\ \varphi \oplus \mathsf{ker}\ \varphi.
  3. Prove that there is a basis B of V such that M^B_B(\varphi) is diagonal, with only 1 and 0 as diagonal entries.

  1. Suppose x \in \mathsf{im}\ \varphi \cap \mathsf{ker}\ \varphi; say x = \varphi(y). Now 0 = \varphi(x) = \varphi^2(y) = \varphi(y) = x.
  2. Let v \in V. Now \varphi(v - \varphi(v)) = \varphi(v) = \varphi^2(v) = \varphi(v) - \varphi(v) = 0, so that v - \varphi(v) \in \mathsf{ker}\ \varphi. That is, v = \varphi(v) + (v - \varphi(v)) \in \mathsf{im}\ \varphi + \mathsf{ker}\ \varphi. Using part (1), we have V = \mathsf{im}\ \varphi \oplus \mathsf{ker}\ \varphi.
  3. Choose a basis E_I for \mathsf{im}\ \varphi and a basis E_K for \mathsf{ker}\ \varphi. Using this previous exercise, B = E_I \cup E_K is an ordered basis for V, and moreover M^B_B(\varphi) = \left[ \begin{array}{c|c} M^{E_I}_{E_I}(\varphi) & 0 \\ \hline 0 & M^{E_K}_{E_K}(\varphi) \end{array} \right]. (Where we restrict \varphi as necessary). Clearly this matrix is diagonal, and moreover, the upper left block is I while the lower right block is 0.
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