Prove that for a real matrix $A$, $\ker (A) = \ker (A^TA)$
Is it possible to solve this supposing that inner product spaces haven't been covered in the class yet?
$\ker (A^TA) = \ker (A)$ - Mathematics Stack Exchange
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.
linear algebra - Prove that $\ker (AB) = \ker (A) + \ker (B ...
It is not right: take $A=B$, then $\ker (AB)=\ker (A^2) = V$, but $ker (A) + ker (B) = ker (A)$.
If Ker (A)=$\ {\vec {0}\}$ and Ker (B)=$\ {\vec {0}\}$ Ker (AB)=?
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 …
How to find $ker (A)$ - Mathematics Stack Exchange
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 …
Can $RanA=KerA^T$ for a real matrix $A$? And for complex $A$?
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.
Prove that $\operatorname {im}\left (A^ {\top}\right) = \ker (A ...
I need help with showing that $\ker\left (A\right)^ {\perp}\subseteq Im\left (A^ {T}\right)$, I couldn't figure it out.
linear algebra - proof of KerA = ImB implies ImA^T = KerB^T ...
Sep 3, 2019 · proof of KerA = ImB implies ImA^T = KerB^T Ask Question Asked 6 years, 3 months ago Modified 6 years, 3 months ago
Prove that if $\ker (A) = \ker (A^2 )$, then $ \ker (A^k ) = \ker (A ...
Feb 6, 2015 · Let $b$ be a vector such that $Ab =0$, and $b$ is that kernel. Let's call that kernel $A$. Then $b$ is also the same as $\\ker(A^2)$. Any hints would be appreciated ...
How to prove strong duality by Slater's condition without the ...
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 …