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Suppose we have a square upper triangular matrix $R$, I want to show that it is singular if and only if one of its diagonal elements is zero. I know a matrix is singular if and only if (or is it "if" and not "iff"?) its determinant is zero, right? According to Wikipedia, the determinant of a square matrix ($n \times n$) can be defined by the Leibniz formula or the Laplace formula which I don't understand very well. At the end of that article it says for a lower (upper) triangular matrix the determinant equals the product of the diagonal entries. If it is so, the prove is quite straight forward. But could someone explain to me the reason of the above bold statement?

P.S: I've also read the following questions, but none of them answers my question specifically:

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I guess it is easier to see this with the Laplace formula. If you choose the first column to apply it, as the only non-zero element is the first you have $$\det A=a_{1,1}M_{1,1}$$ But again, $M_{1,1}$ is the determinant of an upper triangular, so if you apply the formula again to the first column, you will have the product of the first and second elements of the diagonal with the determinant of a new upper triangular matrix. Applying it $n$ times, you get the desired result.

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The rank of a matrix can be the number of independent rows it has. Now after reduction to Echoleon Form, if its rank is 0, which means less than $n$, it must have a single row that will have all its elements 0.Now as each row has one diagonal matrix. Hence proved.

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  • $\begingroup$ Thank you for your answer. If the rank is $0$, it means the number of independent rows is zero. How did you conclude that it must have a single row that will have all elements 0? $\endgroup$ – Gigili Dec 6 '13 at 14:27
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    $\begingroup$ When you calculate rank through Echleon Form, you calculate the rank by counting the rows that have at least one non-zero element. $\endgroup$ – Isomorphic Dec 6 '13 at 14:31
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If $A=(a_{ij})$ is an upper triangle matrix in $\mathcal M_n(\mathbb F)$ then by definition $$\det A=\sum_{\sigma\in S_n}\epsilon(\sigma)\prod_{k=1}^n a_{k\sigma(k)}$$ or if $\sigma\ne \mathrm{id}$ then there's $k\in\{1,\ldots,n\}$ such that $a_{k\sigma(k)}=0$ hence we have $$\det A=\prod_{k=1}^na_{k\mathrm{id}(k)}=\prod_{k=1}^na_{kk}$$

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  • $\begingroup$ Very nicely done, Sami! +1 $\endgroup$ – Namaste Dec 7 '13 at 15:20
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From the definition of determinant, it is equal to the signed summation of products of elements lying in $n$ different columns and also $n$ different rows. Hence for a lower (upper) triangular matrix, the unique product that is not zero is the one that all factors lie in the diagonal of the matrix. That forces the determinant to be the product of diagnoal elements.

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Compute the determinant of the following matrix using the formula you learnt in kindergarten by expanding along the first column: $$ \begin{bmatrix} a_{11} & b\\ 0 & A_{22} \end{bmatrix} $$ where $a_{11} \in \mathbb{R}$, $A_{22} \in \mathbb{R}^{N \times N}$ and $b \in \mathbb{R}^{1,N}$. Is it now clear why the determinant of a triangular matrix is just the product of its diagonal entries?

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    $\begingroup$ We didn't have mathematics course in kindergarten, perhaps it was restricted to wiser students like you. $\endgroup$ – Gigili Dec 6 '13 at 14:23

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