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Apr
30
comment Prove that a product of two complex numbers has zero imaginary part
Thanks a lot for posting your solution. I'm going to accept Catalin Zara's anther though because he was also trying to be more educational.
Apr
30
accepted Prove that a product of two complex numbers has zero imaginary part
Apr
30
comment Prove that a product of two complex numbers has zero imaginary part
This is a very nice way to prove it. Thank you! But I'll accept Catalin Zara's answer because it was trying to be more educational.
Apr
30
revised Prove that a product of two complex numbers has zero imaginary part
added 1 character in body
Apr
30
comment Prove that a product of two complex numbers has zero imaginary part
When I try doing this, first by multiplying the numerator and denominator by denominator's conjugate, I get an expression with several dozens terms, and if I try to expand it further, I eventually might get cubes or even higher powers etc. But it never really simplifies to something manageable. There must be some "trick" I'm missing in simplification...
Apr
30
asked Prove that a product of two complex numbers has zero imaginary part
Apr
3
accepted Closed formula for finite product series
Apr
3
comment Closed formula for finite product series
Thank you! Would you mind saying what errors? Otherwise, this is a nice way to think about it.
Apr
3
comment What happens to dimension of kernel when squaring non-invertible matrix?
But your explanation leaves the possibility that $\ker u^k = \ker^k$ even when $k < n$. But in your original post you say that kernels must be proper subsets, i.e. there is no way that $\ker A = \ker A^2$. Did you perhaps mean that kernels of subsequent powers are at most of the same dimension (i.e. could be equal)?
Apr
3
accepted What happens to dimension of kernel when squaring non-invertible matrix?
Apr
3
comment What happens to dimension of kernel when squaring non-invertible matrix?
In your second paragraph, did you want to show that $\ker A \neq \ker A^2$? This still leaves room for the possibility that $\ker A = \ker A^2$. You say there are two possibilities, but mention only one in the third paragraph, but there would be just two, if you showed that $\ker A \neq \ker A^2$. But the last paragraph answers my original question, so I'm going to accept your answer. Thanks again!
Apr
3
comment What happens to dimension of kernel when squaring non-invertible matrix?
Thank you! Can you briefly explain why this is the case?
Apr
2
comment What happens to dimension of kernel when squaring non-invertible matrix?
@WilliamKrinsman thanks for the pointer! I'll need to read on it, when I'm awake :)
Apr
2
asked What happens to dimension of kernel when squaring non-invertible matrix?
Apr
2
comment Closed formula for finite product series
@AndrásSalamon I think I figured the solution: instead of lumping the $(k^i + 2^i)$ terms together, I could simplify the $(k^i+2^i)(k^i-2^i)$ terms to get in the end something that roughly resembles $O(n^4)$. Though, admittedly, I'm too scared to verify this once again, because if this falls into pieces, then oh woe... :)
Apr
2
asked Closed formula for finite product series
Apr
2
asked Recurrence: how to compute the base case when $n$ is its root on each step?
Apr
2
accepted Combination of invertible linear transformation
Apr
1
comment Combination of invertible linear transformation
@Jonas oh, indeed, that would work!
Apr
1
asked Combination of invertible linear transformation