Can an idempotent matrix be complex?
A matrix $A$ is called idempotent if $A^2 = A$. I am just wondering if such matrix can be complex. Anyone can help give an example or proof that it has to be real? Thanks!
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$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$Any matrix $A = \pmatrix{a&b\\c&1-a}$ will be idempotent provided that $a^2+bc=a$
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