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Is it possible to solve this supposing that inner product spaces haven't been covered in the class yet? Proof of kera = imb implies ima^t = kerb^t ask question asked 6 years ago modified 6 years ago Take $a=b$, then $\ker (ab)=\ker (a^2) = v$, but $ker (a) + ker (b) = ker (a)$.
I am sorry i really had no clue which title to choose I'm working with the following kernel I thought about that matrix multiplication and since it does not change the result it is idempotent
Please suggest a better one.
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. 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? 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 We actually got this example from the book, where it used projection on w to prove that dimensions of w + w perp are equal to n, but i don't think it mentioned orthogonal projection, though i could be wrong (maybe we are just assumed not to do any other projections at our level, or maybe it was assumed it was a perpendicular projection, which i guess is the same thing).
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