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Is \begin{bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 4 & 4 \\ 0 & 0 & 4 & 4 \end{bmatrix}

an upper triangular matrix? My linear algebra teacher says that the main diagonals must have exclusively pivots or zeros, but I thought that the only requirement for upper triangular form is to have zeros below the main diagonal. Online sources like proofwiki seem to agree with me.

Whose definition is correct?

William Chang
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    Wikipedia only defines triangular matrices for square matrices. – Git Gud May 19 '14 at 17:09
  • Non-square matrices don't really have anything you can call a "main" diagonal; imo, this gives the result that no non-square matrix can be triangular. – Pockets May 19 '14 at 17:17
  • @SamuelLijin Wikipedia defines main diagonal for any matrix. – Git Gud May 19 '14 at 17:18
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    I don't see anything harmful about defining a matrix $A\in M_{m\times n}$ to be upper-triangular if $[A]_{ij}=0$ for $i>j$. On the other hand, I'm not sure if this is particularly useful. – Brian Fitzpatrick May 19 '14 at 17:21
  • Actually, maybe the term upper-trapezoidal would be more appropriate. – Brian Fitzpatrick May 19 '14 at 17:23
  • @GitGud, the source for that entry only defines main diagonals for square matrices. I should also note that OP's linear algebra teacher's position that "main diagonals must have exclusively pivots" is a bit nonsensical; consider for example the following upper triangular matrix: \begin{bmatrix}1&0&1&1\0&0&1&5\0&0&0&2\0&0&0&0\end{bmatrix} – Pockets May 19 '14 at 17:27
  • @SamuelLijin sorry, that wasn't what I meant; edited my post – William Chang May 20 '14 at 15:56
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    @Mathster, that sounds more to be like the definition of a reduced row-echelon form upper triangular matrix. Although there isn't really any particularly clear rationale about whether a matrix $A\in\mathbf{M}{i,j}:i\neq j$ can be upper triangular (or as Brian labeled it, _upper-trapezoidal), the one consensus here that you can see is that the criteria is dependent on exclusively whether certain elements of a matrix are zero or not, not the elements that do not have to be zero. – Pockets May 20 '14 at 16:06
  • In what way do you teacher's definition and yours differ ?? And what's the meaning of "must have exclusively pivots" ? –  Jun 09 '16 at 09:53
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    You said that: "Online sources like proofwiki seem to agree with me." Perhaps it would be a good idea to add the link to the ProofWiki article which agrees with you. – Martin Sleziak Jun 09 '16 at 11:02

2 Answers2

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One source that I have has a definition (kind of hidden away in the questions): "An $m\times n$ matrix $A$ is called upper triangular if all entries lying below the diagonal entries are zero, that is, if $A_{ij}=0$ whenever $i>j$." (p.21 Friedberg et al, Linear Algebra 4th edition)

I have yet to find a source that explicitly contradicts this definition (so deliberately states that $m \times n$ matrices cannot be upper triangular), thereby limiting upper triangular matrices to square matrices only. But in all my other sources we have something similar to "...$A \in M_{n \times n}(K)$...upper triangular iff...". The other sources I could consult here was p.37 Cullen (Matrices and linear transformations) and p.149 Golan (The linear algebra a beginning graduate student ought to know).

Christiaan Hattingh
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  • In other words, as another commenter also said, for most sources it's syntactically invalid (since not defined) to ask whether a non-square matrix is upper triangular. After all, a wide rectangular matrix cannot have an upper triangle above the first diagonal. – user21820 May 27 '16 at 06:16
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You can have LU factorization of a non-square matrix, where the U is a non-square matrix. see How can LU factorization be used in non-square matrix?

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