Take $a=b$, then $\ker (ab)=\ker (a^2) = v$, but $ker (a) + ker (b) = ker (a)$. Convex optimization, strong duality is proved under the assumption that ker(a^t)={0} for the linear map describing the equality constraint, though it is rema. Is it possible to solve this supposing that inner product spaces haven't been covered in the class yet?
I am sorry i really had no clue which title to choose In section 5.3.2 of boyd, vandenberghe I thought about that matrix multiplication and since it does not change the result it is idempotent
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 with my keyboard We have ker (a)= {x∈v:a⋅x=0} this means if a vector x when applied to our system of equations (matrix) are takin to the zero vector 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 i know that ker (a)=ker (rref (a)) for any matrix a, i am not sure if i can say that ker (rref (a) * rref (b))=ker (ab) Is this statement true? just out of my curiosity? It does address complex matrices in the comments as well
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 need help with showing that $\ker\left (a\right)^ {\perp}\subseteq im\left (a^ {t}\right)$, i couldn't figure it out. You'll need to complete a few actions and gain 15 reputation points before being able to upvote
Upvoting indicates when questions and answers are useful What's reputation and how do i get it Instead, you can save this post to reference later.
