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

Sep 9, 2018 · " " Assuming that Ax = 0 A x = 0, we obtain that ATAx = A0 = 0 A T A x = A 0 = 0. " " Assuming ATAx = 0 A T A x = 0 xTATAx = 0 x T A T A x = 0 (Ax)T(Ax) = 0 Ax = 0 (A x) T (A …

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

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 …

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 …

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.

May 14, 2020 · In fact, the dimensions are always equal. Because we generally have dim ker(AB) = dim ker B + dim(im(B) ∩ ker A) dim ker (A B) = dim ker B + dim (im (B) ∩ ker A), it suffices to …

Oct 11, 2023 · I'm trying to investigate what $\\ker(A) \\cap \\operatorname{im}(A)$ implies for $\\operatorname{rank}(A)$ and $\\operatorname{null}(A)$. Specifically, if we have ...

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 …