# Would linear algeba be considered idempotent?

Sorry but I'm not a mathematician, so please bear with me. I understand what idempotence is from a communications point of view, and am wondering if it is correct to include linear algebraic formulas as being idempotent. After all, if y= 2x, then for a given input X, you always get the same result. Is y=2x idempotent?

Cheers, Nap

• Formulas are not idempotent. Operators can be idempotent. Nov 11, 2015 at 7:46
• please explain what is it that makes the difference. I know that Absolute value of is idempotent, as is multiplying by one. But I don't see the general aspect of the rule.
– gone
Nov 11, 2015 at 7:56
• An operator is idempotent if applying it twice is the same as applying it once. In particular it has to be something you can apply. Nov 11, 2015 at 8:24

Any function always gives the same value on the same input: $f(a) = f(b)$ whenever $a = b$. Hence, the property you've observed is really just the fact that expressions like "$2x$" can be thought of as a function of $x$.
Idempotence of $f$ is something different. It means that for all $x$, $f(f(x)) = f(x)$.
Some linear operators are idempotent, e.g. the identity operator $I(x) = x$, or the one that always returns the zero vector. More generally, an operator that "projects" a vector onto a subspace, while keeping the coordinates on that subspace's basis unchanged and zeroing the coordinates on the other dimensions, is an idempotent linear operator.
But most operators are not, as in this very example where $f(x) = 2x$. We then have $f(f(x)) = 4x$, so $f$ is not idempotent unless we're in the trivial zero-dimensional vector space with only the zero vector present.