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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
Please suggest a better one.
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
It is clear that this can happen over the complex numbers anyway.
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.
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