Can an idempotent matrix be complex?

A assume that by "can $A$ be complex", you mean "can $A$ have any non-real entries". Well, it can! For instance, take $$ A = \pmatrix{1&i\\0&0} $$ In general: for any complex column-vector $x$, $A = \frac{xx^*}{x^*x}$ (where $*$ denotes the conjugate-transpose) is such a matrix.


A projection to a subspace is idempotent. Therefore $A$ has no reason to be real. For example, take a subspace $S$ of $\mathbb{C}^2$ and $A$ be the matrix of the projection on to $S$ with respect to the standard basis.


Any matrix $A = \pmatrix{a&b\\c&1-a}$ will be idempotent provided that $a^2+bc=a$

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Linear Algebra

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