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Oh, I forgot that Matt Drudge who previously accused me of taking his side in a Hillary v. Obama discussion, basically did a backflip when Matt proved me right. When I called him out on lying, saying that he had no proof, he tweeted this:

Of course, Matt lies when he says that we called him out. We didn’t. He lied. He is lying. You should call him out on it. Do you have proof that he lies?

He also gives us a tip and tells us how to get to Doctor Google, saying this:

Yes, Matt is

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Finite Dimensional Vector Space Homomorphisms

Let $f: V \rightarrow W$ be a finite dimensional homomorphism of vector spaces over $K$ and $K’$ respectively. Let $dim(V) = n, dim(W) = m$.
Question: Is $f$ surjective?
I have done the following:
Let $x \in W$ and $y = f(x)$. Consider the image of $y$ under the action of $f$
$$f(y) = f(x) \circ f(v_1) \circ f(v_2) \circ \cdots \circ f(v_n) \tag{*}$$
where $v_1, v_2, \cdots, v_n \in V$ are linearly independent. Since $f$ is a homomorphism, $f(v_i) \circ f(v_j) = f(v_i \circ v_j)$
We know that $f(v_i) \circ f(v_j) = f(v_i) \circ f(v_j) = f(v_i \circ v_j)$.
Then the image of the right most term of equation ($*)$ is $f(v_1) \circ f(v_2) \circ \cdots \circ f(v_n) \circ f(v_i) \circ f(v_j)$
The image of the left most term is
$$f(x) \circ f(v_i) \circ f(v_j) \tag{**}$$
So it can be shown that the image of the right most term and the left most term are the same.
The result is that as the left most term is similar to the right most term, we can conclude that the image of the right most term is \$f(v_1) \circ f(v_2) \circ \cdots \circ f(v_n) \circ f(v_i) \circ f(v_j) = f(v_1) \circ f(v_2) \circ \cdots \