Is it possible to solve this supposing that inner product spaces haven't been covered in the class yet?

So before I answer this we have to be clear with what objects we are working with here. Also, this is my first answer and I cant figure out how to actually insert any kind of equations, besides what I can type …

Sep 9, 2018 · I am sorry i really had no clue which title to choose. I thought about that matrix multiplication and since it does not change the result it is idempotent? Please suggest a better one.

It is not right: take $A=B$, then $\ker (AB)=\ker (A^2) = V$, but $ker (A) + ker (B) = ker (A)$.

I think that the correct path is to show that ImB ⊆ KerA which is when AB=0 and KerA⊆ImB at the same time implies KerA=ImB. But I don't know when that is true. I also need to prove that A cannot have …

Thank you Arturo (and everyone else). I managed to work out this solution after completing the assigned readings actually, it makes sense and was pretty obvious. Could you please comment on "Also, while …

Jan 27, 2021 · It does address complex matrices in the comments as well. It is clear that this can happen over the complex numbers anyway.

I need help with showing that $\ker\left (A\right)^ {\perp}\subseteq Im\left (A^ {T}\right)$, I couldn't figure it out.

Dec 20, 2019 · Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant …

Feb 11, 2021 · In Section 5.3.2 of Boyd, Vandenberghe: Convex Optimization, strong duality is proved under the assumption that ker(A^T)={0} for the linear map describing the equality constraint, though …