How to prove the product of upper triangular matrices is upper triangular?

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SUMMARY

The product of two upper triangular matrices is also an upper triangular matrix. Given two upper triangular matrices A and B, where the elements of A are defined as ##a_{ij}## with ##a_{ij}=0## for ##i>j##, the resulting matrix C from the multiplication C=AB will maintain this property. Specifically, the entries of C can be expressed as dot products of the rows of A with the columns of B, confirming that ##c_{ij}=0## for ##i>j##, thus proving the upper triangular nature of the product.

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Brucezhou
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This seems easy but when I tried to do this, the best way I came up with is to list all entries and then do the multiplication work. Is there any better ,clearer and more simple way to do the proof?
 
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Brucezhou said:
This seems easy but when I tried to do this, the best way I came up with is to list all entries and then do the multiplication work. Is there any better ,clearer and more simple way to do the proof?

Did you try to express the entries in the product matrix in terms of the dot products of the row of one matrix with the corresponding column in the second matrix?
 
Suppose A is an upper triangular matrix with elements ##a_{ij}##. Then you know that ##a_{ij}=0## if ##i>j##. If C=AB where B is also upper triangular, you want to show that ##c_{ij}=0## if ##i>j##.
 

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