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In Strassen's algorithm, why does padding the matrices with zeros, in order to multiply matrices that are not powers of 2, not affect the asymptopic complexity?

Hi, I was reading this question but I do not follow Yuval Filmus's answer completely.

He offers two ways of padding the matrix with zeroes, I am interested in his first suggestion, padding the entire matrix with zeroes at the start such that the new matrix has dimensions $N\times N$ where $N = 2^c$.

He says "$N < 2n$ so this doesn't affect the asymptotic complexity."

Could someone please elaborate as I do not follow, I understand calculating time complexity with the master theorem and $T(n) = aT(n/b) + f(n)$ so if someone could explain how this works in reference to that it would be greatly appreciated.

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Suppose that if $N = 2^c$ then you can multiply two $N \times N$ matrices in time $O(N^{\log_2 7})$. For concreteness, let us say that two such matrices can be multiplied in time at most $CN^{\log_27}$.

Now suppose that we are given two $n\times n$ matrices. We pad them to $N \times N$ matrices, where $N = 2^{\lceil \log_2 n \rceil} < 2n$. We can extract the product of the original matrices from the product of the new matrices. The two new matrices can be multiplied using at most this many operations: $$ CN^{\log_2 7} < C (2n)^{\log_2 7} = 7C n^{\log_2 7} = O(n^{\log_2 7}). $$

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