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I have a matrix $A$ whose rows sum to zero, such that $\sum_j A_{ij} = 0, \forall i$. If I multiply it by any matrix, $B$, can it be proven that the resulting matrix, $C = BA$, must also have zero sum rows? I find that they are empirically.

Is such a proof available as a reference in any text book?

user664303
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3 Answers3

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Note that, by matrix multiplication rule, the rows of the matrix

$$C = BA$$

are linear combination of the rows of the matrix $A$ and thus also $\sum_j C_{ij} = 0, \forall i$.

As an alternative since the vector $x=(1,1,...,1)\in N(A)$ we have that

$$Cx=BAx=B\cdot 0=0$$

user
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  • I like the alternative. The first proof feels like an the assertion; I don't see how it follows. – user664303 Jun 21 '18 at 20:43
  • @user664303 I agree the second method is more fine and elegant but also the first one is very important. Indeed in general given $v=(c_1,c_2,...,c_n)$ we have that $w=Av$ is a column vector obtained by linear combination of the columns of A by the coefficients $c_1,c_2,...,c_n$ and $v^TB$ is a row vector obtained by linear combination of the rows of B by the coefficients $c_1,c_2,...,c_n$. – user Jun 21 '18 at 20:49
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For any $i = 1, \ldots, n$ we have

$$\sum_{j=1}^n C_{ij} = \sum_{j=1}^n (BA)_{ij} = \sum_{j=1}^n \sum_{k=1}^n B_{ik}A_{kj} = \sum_{k=1}^n\left(B_{ik}\cdot\sum_{j=1}^n A_{kj}\right) = \sum_{k=1}^n B_{ik} \cdot 0 = 0$$

Hence $C$ also has zero mean rows.

mechanodroid
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1

The answer is yes.

Note that the sum of the $i_{th}$ row of $C$ is $$C_{i1} + C_{i2}+...+ C_{in}$$

$$ C_{i1} = B_{i1}A_{11}+B_{i2}A_{21}+...+B_{im}A_{m1}$$ $$ C_{i2} = B_{i1}A_{12}+B_{i2}A_{22}+...+B_{im}A_{m2}$$

$$.\\.\\.\\ C_{in} = B_{i1}A_{1n}+B_{i2}A_{2n}+...+B_{im}A_{mn}$$

Upon adding and factorization we get

$$ C_{i1} + C_{i2}+...+C_{in} = B_{i1} ( A_{11}+A_{12}+...+A_{1n}) + B_{i2} (A_{21}+A_{22}+...+A_{2n}) + ... +B_{im} (A_{m1}+A_{m2}+...+A_{mn}) =0$$

Mohammad Riazi-Kermani
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