What happens when you multiply the dirac delta function, $\delta(t)$, by some scalar k? From my understanding, $\delta(t)$=0 for t≠0, and $\delta(t)$=∞ for t=0. Based on this, does k*$\delta(t)$ simply equal to $\delta(t)$ since it doesn't change the function definition?
For example, let's say I have 2$\delta(t)$ - $\delta(t)$, will this result in 0 since the 2 simply gets "absorbed" into $\delta(t)$, basically making it $\delta(t)$ - $\delta(t)$?
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The dirac delta function isn't really a function, strictly speaking. It's a distribution. With the definition you've supplied, it is common to place the additional constraint that $\int_{-\infty}^{\infty} \delta(x)dx = 1$, which allows to avoid the logical paradox that you've outlined.
In particular, $k \delta(x)$ is the "function" satisfying $k \delta(0) = +\infty$ and is $0$ otherwise but now with the constraint $\int_{-\infty}^{\infty} k\delta(x)dx = k.$
One of the properties of Dirac Delta generalised Function is:
$$\delta(ax) = \dfrac{1}{|a|}\delta(x) \qquad \qquad a\in\mathbb{R}\backslash\{0\}$$
In your case, writing $2\delta(x)$ just means $\delta\left(\dfrac{x}{2}\right)$, which is zero when $x \neq 0$.