Please don’t ignore I really need some help on this
2 answers:
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Answer: Although A and AT have the same eigenvalues, they may actually have different eigenvectors for that particular eigenvalue. Therefore A and AT have at least one eigenvector that aren't in common.
Step-by-step explanation:
These two attachment explains the provings and counterexamples briefly and simply
u(x) = -2x², v(x)= 1/x
(u o v)(x) = u(v(x)) = -2(1/x)²= -2/x²
We can see that domain for x is going to bee all real numbers except 0. From the equation above we can see that graph of the function (u o v)(x) = -2/x² has horizontal asymptote y=0, because degree of numerator is less than degree of denominator ( (u o v)(x) = -2x⁰/x² ).
x² is always going to be positive, so range is going to be all negative numbers.
y<0, or y∈(0,-∞)
