Proof: Let A & B be Matrices; Show BA has a Row of Zeros

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SUMMARY

In the discussion, the proof demonstrates that if matrix A is an m x n matrix with at least one row of zeros, and matrix B is an n x p matrix, then the product BA will also contain a row of zeros. The proof relies on the definition of matrix multiplication, specifically showing that the entries in the resulting matrix BA will be zero due to the zero entries in A. The conclusion is that the presence of a zero row in A guarantees a corresponding zero row in the product BA.

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Homework Statement


28. Let A be an m x n matrix with a row consisting entirely of zeros. Show that if B is an n x p matrix, then BA has a row of zeros.

Homework Equations


N/A

The Attempt at a Solution


A = (aij)_{mxn} and B = (bij)_{nxp}. Assuming that the entries for jth column of A are all zeros, I infer that the jth column of BA is also a row of zeros. Let cij be an entry in jth column. By the definition of multiplication:
cij = b_{i1}a_{j1} + b_{i2}a_{j2} + ...+ b_{in}a_{nj} = \sum_{k=1}^n b_{ik}a_{kj}

Since the jth column of A is zero, then there is:
a_{1j} = a_{2j} = ... = a_{nj} = 0. Hence, cij = 0 and, therefore, the jth column of BA is a column of zeros.

Did I do this proof correctly? Thanks in advance.
 
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Retribution said:

Homework Statement


28. Let A be an m x n matrix with a row consisting entirely of zeros. Show that if B is an n x p matrix, then BA has a row of zeros.

Homework Equations


N/A


The Attempt at a Solution


A = (aij)_{mxn} and B = (bij)_{nxp}. Assuming that the entries for jth column of A are all zeros,
You're given that one row of A consists of zeros. You can't also assume that one column (column j) is all zeros.
Retribution said:
I infer that the jth column of BA is also a row of zeros. Let cij be an entry in jth column. By the definition of multiplication:
cij = b_{i1}a_{j1} + b_{i2}a_{j2} + ...+ b_{in}a_{nj} = \sum_{k=1}^n b_{ik}a_{kj}

Since the jth column of A is zero, then there is:
a_{1j} = a_{2j} = ... = a_{nj} = 0. Hence, cij = 0 and, therefore, the jth column of BA is a column of zeros.

Did I do this proof correctly? Thanks in advance.
 

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