Let a be a real matrix of order n, and prove that if AA '= 0, then a = 0 How can I prove it? I don't know what matrix a 'is,

A 'is the transpose of a, right
According to the definition of matrix multiplication, the element of row I and column J of AA 'is equal to the product of row I of a and column J of a' (that is, transpose of row J of a). So the element of diagonal line I of AA 'is the result of point multiplication of row I vector of a and transpose itself, that is, its own square. Assuming that the vector is V, then VV' = 0. Since only the square of 0 vector is 0, so V must be 0 vector
So all row vectors of matrix A are 0, so matrix A = 0

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