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 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.
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 Proof of kera = imb implies ima^t = kerb^t ask question asked 6 years, 3 months ago modified 6 years, 3 months ago 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)
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? 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). I'm working with the following kernel
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