The determinant of a triangular matrix is the product of its diagonal entries — proof using permutations

Question

Prove that the determinant of a triangular matrix is the product of its diagonal entries.

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My attempt:

Suppose $A$ is an $n\times n$ matrix, then:

$$\det(A) = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1\sigma_1} \cdots A_{n\sigma_n}$$

The product of diagonal entries is $A_1\sigma_1 \cdots A_n\sigma_n$ when $\sigma_i = i$ for $i = 1, \ldots, n$.

I claim that when the permutation is not an identity, there exist $i$ and $j$ s.t. $\sigma_i < i$ and $\sigma_j > j$.

Let $\sigma_j \neq j$, then $\sigma_k = j$ where either $k > j$ or $k < j$.

If $k > j$, let $\sigma_\ell = k$ where $\ell > k$ or $\ell < k$.

If $\ell < k$, done. Otherwise, $\ell > k$ and suppose for $\sigma_\ell, \ldots, \sigma_n$ that $n>\cdots>\ell$ and follow the above sequence of operations incrementally.

Then, $\sigma_n = n - 1$.

This implies that $\sigma_p = n$ where $p \leq j$.

If $k <j$, let $\sigma_h = k$ where $h > k$ or $h < k$.

Likewise, if $h > k$, done. Otherwise, $h < k$ and suppose for $\sigma_1, \ldots, \sigma_h$ that $1<\ldots<h$ and follow that above sequence of operations incrementally.

Then, $\sigma_1 = 2$.

This implies that $\sigma_q = 1$ where $q \geq j$.

Thus, there is at least one $i$ and $j$ s.t. $\sigma_i < i$ and $\sigma_j > j$.

By definition, if $A$ is upper-triangular, $A_{ij} = 0$ when $i > j$ and so, $A_{i\sigma_i} = 0$. Alternatively, if $A$ is lower triangular, $A_{ij} = 0$ when $i < j$ and $A_{j\sigma_j} = 0$.

So, the determinant of a triangular matrix is indeed the product of diagonal entries.

QED.

I'm not sure if my proof is correct and I wanted someone to go over it. Please let me know if I made any mistake(s) anywhere and if so what I can and should do to correct it.

Answer 1

Let ${\rm U}_n$ be an invertible $n \times n$ upper triangular matrix. Let

$${\rm U}_{n+1} := \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix}$$

Using the Schur complement,

$$\det \left( {\rm U}_{n+1} \right) = \det \begin{bmatrix} {\rm U}_n & {\rm c}_{n+1}\\ {\rm 0}_n^\top & u_{n+1}\end{bmatrix} = \left( u_{n+1} - {\rm 0}_n^\top {\rm U}_n^{-1} {\rm c}_{n+1} \right) \det \left( {\rm U}_{n} \right) = u_{n+1} \det \left( {\rm U}_{n} \right)$$

Let ${\rm U}_{1} =: u_1$. Hence,

$$\begin{aligned} \det \left( {\rm U}_{1} \right) &= u_1\\ \det \left( {\rm U}_{2} \right) &= u_2 \, u_1\\ &\vdots\\ \det \left( {\rm U}_{n} \right) &= \color{blue}{u_n \, u_{n-1}\cdots u_2 \, u_1} \end{aligned}$$

The case where ${\rm U}_n$ is non-invertible is left as an exercise for the reader.

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matrices block-matrices schur-complement determinant

Answer 2

Suppose $A$ is upper-triangular, so that $A_{ij} = 0$ whenever $i < j$. Then it is clear that the product $A_{1 \sigma(1)} \cdots A_{n \sigma(n)}$ will be zero whenever there is an $i \in \{1, \ldots, n\}$ such that $i < \sigma(i)$. I claim that for every non-identity permutation $\sigma$ we can find an $i$ such that $i < \sigma(i)$, and hence $$ \det A = \sum_{\sigma} (\operatorname{sgn} \sigma) A_{1 \sigma(1)} \cdots A_{n \sigma(n)} = \sum_{\sigma = \operatorname{id}} (\operatorname{sgn} \sigma) A_{1 \sigma(1)} \cdots A_{n \sigma(n)} = A_{1 1} \cdots A_{n n}. $$

Proof of claim: (by contrapositive) suppose $i \geq \sigma(i)$ for all $i \in \{1, \ldots, n\}$. The value $\sigma(1)$ must be an integer between $1$ and $n$, and $1 \geq \sigma(1)$, so we must have $\sigma(1) = 1$. Then the value $\sigma(2)$ must be an integer between $1$ and $n$, but it cannot be that $\sigma(2) = 1$ (since $\sigma(1) = 1$ and $\sigma$ is a bijection), and then $2 \geq \sigma(2)$ forces $\sigma(2) = 2$. Continuing in this way, we see that $\sigma$ must be the identity permutation.