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Let $A=\{f\in \operatorname{Hom}(V,V) \mid g \circ f = 0\}$. Find a basis of $A$. Here $g$ is a homomorphism of the vector space $V$ with the basis $\{e_1,e_2,e_3,e_4\}$ (canonical vectors) such that \begin{align} g(e_1+e_2)&=-e_1\\ g(e_1-e_2)&=2e_2\\ g(e_1+e_3)&=e_1+e_4\\ g(e_1-e_4)&=e_2+e_4; \end{align}

Could you find some clever way to get the solution?

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    $\begingroup$ See edits to learn how to properly format a question. $\endgroup$ – John Hughes Mar 16 at 23:26
  • $\begingroup$ What's the language used? $\endgroup$ – psidaga Mar 16 at 23:35
  • $\begingroup$ Do you mean $g(e_1 + e_2)=...$, etc. ? $\endgroup$ – zugzug Mar 17 at 0:24
  • $\begingroup$ It's called MathJax, and it's a flavor of LaTeX. Here's a quick tutorial -- the first few paragraphs give you 90% of what you need: math.meta.stackexchange.com/questions/5020/… $\endgroup$ – John Hughes Mar 17 at 0:41
  • $\begingroup$ Thank you so much. I was using the representative matrix of g with the canonical basis... The system was really annoying $\endgroup$ – psidaga Mar 17 at 6:52
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The range of $g$ is the span of $e_1,e_2,e_4$, hence its kernel is one dimensional. Find a nonzero element $v$ of the kernel, that will generate it.

Now, $g\circ f\, (x)=0$ iff $f(x)\in\ker g$, so the range (=column space) of $f$ must be contained in ${\rm span}(v)$, and you can obtain a basis $(f_i)$ of $A$ by putting $v$ in the $i$th column and $0$ otherwise in (the standard matrix of) $f_i$.

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Hint: the matrix for $g$ rel the bases $\{e_1+e_2,e_1-e_2,e_1+e_3,e_1-e_4\}$ and the standard basis is: $\begin{pmatrix}-1&0&1&0\\0&2&0&1\\0&0&0&0\\0&0&1&1\end{pmatrix}$.

So, the matrix of $f$, when multiplied by this matrix on the left, is zero, for any $f$ in $A$.

This gives a homogeneous linear system of $16$ equations in $16$ unknowns. So you can form a $16×16$ matrix and then row-reduce, to solve.

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