# homomorphisms and vector spaces

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?

• See edits to learn how to properly format a question. – John Hughes Mar 16 at 23:26
• What's the language used? – psidaga Mar 16 at 23:35
• Do you mean $g(e_1 + e_2)=...$, etc. ? – zugzug Mar 17 at 0:24
• 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/… – John Hughes Mar 17 at 0:41
• Thank you so much. I was using the representative matrix of g with the canonical basis... The system was really annoying – psidaga Mar 17 at 6:52

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) $$f_i$$.
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.