7.8 Special Determinants

A determinant of size $n$ has $n!$ terms, so a general formula for a large matrix is hopeless. Some matrices, though, have so much structure that their determinants collapse to a short product or a recurrence. This section computes the most useful of them: the Vandermonde determinant, which governs polynomial interpolation; tridiagonal determinants, which satisfy a three-term recurrence; block triangular determinants, which split into smaller ones; and the determinant of a rank-one update of the identity. The proofs share one idea: find the structure first (a polynomial with known roots, a row with two non-zero entries, a pattern of zeros) and let the defining properties of the determinant do the rest.

7.8.1. The Vandermonde determinant

Chapter 3 proved that there is exactly one polynomial of degree at most $n$ through $n + 1$ points with distinct first coordinates (Theorem 3.7.3). Behind it sat a square matrix of powers of the nodes. Its determinant turns out to be a product of all the differences of the nodes, and that product explains at a glance why distinct nodes are exactly what is needed.

For $x_1, \dots, x_n \in F$ , the Vandermonde matrix is $\mathbf{V}(x_1, \dots, x_n) \mathrel{:=} \begin{pmatrix} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} \\ \vdots & \vdots & \vdots & & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} \end{pmatrix} \in M_n(F),$ whose $(i, j)$ -entry is $x_i^{j-1}$ . Row $i$ lists the powers $1, x_i, \dots, x_i^{n-1}$ of the $i$ -th node. For $n = 1$ it is the $1 \times 1$ matrix $(1)$ , and for $n = 2$ its determinant is $x_2 - x_1$ .

Theorem 7.8.1 (Vandermonde Determinant)

Let $n \ge 1$ and $x_1, \dots, x_n \in F$ . Then $\det \mathbf{V}(x_1, \dots, x_n) = \prod_{1 \le i < j \le n} (x_j - x_i),$ where the empty product (for $n = 1$ ) is $1$ .

Idea. Induction on $n$ . Replace the last node $x_n$ by a variable $t$ . Expanding along the last row shows the determinant is a polynomial $f(t)$ of degree at most $n - 1$ , with leading coefficient the smaller Vandermonde determinant. Setting $t = x_i$ for $i < n$ makes two rows equal, so $x_1, \dots, x_{n-1}$ are roots of $f$ . A polynomial of degree at most $n - 1$ with these $n - 1$ roots is its leading coefficient times $\prod_{i<n}(t - x_i)$ . Now put $t = x_n$ back. The roots argument needs the $x_i$ distinct; if they are not, both sides are $0$ anyway.

Proof. We use induction on $n$ (Theorem 0.2.7). For $n = 1$ , $\det(1) = 1$ , the empty product.

Let $n \ge 2$ , assume the formula for $n - 1$ nodes, and write $\mathbf{V} = \mathbf{V}(x_1, \dots, x_n)$ . Let $C_{n1}, \dots, C_{nn}$ be the cofactors of $\mathbf{V}$ along row $n$ (Definition 7.6.1). Each $C_{nj}$ is computed after deleting row $n$ , so it does not involve $x_n$ . Define $f(t) \mathrel{:=} \sum_{j=1}^{n} C_{nj}\, t^{j-1} \in F[t] .$ For $c \in F$ , let $\mathbf{V}_c$ be $\mathbf{V}$ with row $n$ replaced by $(1, c, \dots, c^{n-1})$ . Then $\mathbf{V}_c$ has the same cofactors along row $n$ as $\mathbf{V}$ , so by Theorem 7.6.1 along row $n$ , $\det \mathbf{V}_c = \sum_{j=1}^{n} c^{j-1} C_{nj} = f(c) . \tag{$\ast$}$ In particular $\det \mathbf{V} = f(x_n)$ . The coefficient of $t^{n-1}$ in $f$ is $C_{nn} = (-1)^{2n}\det \mathbf{V}(x_1, \dots, x_{n-1})$ , since deleting row $n$ and column $n$ of $\mathbf{V}$ leaves the Vandermonde matrix of the first $n - 1$ nodes.

Case 1: two of $x_1, \dots, x_{n-1}$ are equal, say $x_p = x_q$ with $p < q < n$ . Then rows $p$ and $q$ of $\mathbf{V}$ are equal. By Theorem 7.4.5, $\det \mathbf{V} = \det \mathbf{V}^{\top}$ , and $\mathbf{V}^{\top}$ has two equal columns, so $\det \mathbf{V} = 0$ because the determinant is alternating (Definition 7.2.2). The product also vanishes, since it contains the factor $x_q - x_p = 0$ .

Case 2: $x_1, \dots, x_{n-1}$ are distinct. For $i < n$ , the matrix $\mathbf{V}_{x_i}$ has rows $i$ and $n$ equal, so $f(x_i) = \det \mathbf{V}_{x_i} = 0$ by $(\ast)$ and the argument of Case 1. Hence $x_1, \dots, x_{n-1}$ are $n - 1$ distinct roots of $f$ , and by Corollary 6.1.4, $g(t) \mathrel{:=} (t - x_1)(t - x_2)\cdots(t - x_{n-1}) \ \text{ divides } \ f .$ Write $f = hg$ with $h \in F[t]$ . If $h \ne 0$ , then $\deg h = \deg f - (n - 1) \le 0$ by Theorem 0.8.3, since $\deg f \le n - 1$ ; so in every case $h$ is a constant $h \in F$ . As $g$ is monic of degree $n - 1$ , the coefficient of $t^{n-1}$ in $hg$ is $h$ , so $h = C_{nn} = \det \mathbf{V}(x_1, \dots, x_{n-1})$ . Evaluating at $x_n$ (Theorem 0.8.5) and using the induction hypothesis, $\begin{aligned} \det \mathbf{V} &= f(x_n) = \det \mathbf{V}(x_1, \dots, x_{n-1}) \prod_{i=1}^{n-1}(x_n - x_i) \\ &= \prod_{1 \le i < j \le n-1}(x_j - x_i) \prod_{i=1}^{n-1}(x_n - x_i) . \end{aligned}$ The two products together run over all pairs $i < j \le n$ , the second one collecting exactly the pairs with $j = n$ . This proves the formula for $n$ , and the theorem follows by induction.

Example (A

3 \times 3

Vandermonde Determinant)

Compute $\det \mathbf{V}(1, 2, 4)$ from the formula and by cofactor expansion.

Solution. The formula gives $(2 - 1)(4 - 1)(4 - 2) = 1 \cdot 3 \cdot 2 = 6$ . Directly, expanding $\mathbf{V}(1, 2, 4) = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 4 \\ 1 & 4 & 16 \end{pmatrix}$ along the first row gives $1\cdot(32 - 16) - 1\cdot(16 - 4) + 1\cdot(4 - 2) = 16 - 12 + 2 = 6$ .

The product vanishes exactly when a factor does, because a field has no zero divisors. So:

Corollary 7.8.2 (Distinct Nodes Give an Invertible Vandermonde Matrix)

$\mathbf{V}(x_1, \dots, x_n)$ is invertible if and only if $x_1, \dots, x_n$ are distinct.

Proof. In a field, a product is non-zero if and only if every factor is non-zero (Theorem 0.7.1 (f), applied repeatedly). So by Theorem 7.8.1, $\det \mathbf{V}(x_1, \dots, x_n) \ne 0$ if and only if $x_j \ne x_i$ for all $i < j$ . By Theorem 7.5.2, this is equivalent to invertibility.

Now the promised two-line proof of Theorem 3.7.3. Given distinct $x_0, \dots, x_n$ and values $y_0, \dots, y_n$ , a polynomial $p = c_0 + c_1x + \dots + c_nx^n$ satisfies $p(x_i) = y_i$ for all $i$ if and only if $\mathbf{V}(x_0, \dots, x_n)\,\mathbf{c} = \mathbf{y}$ , since row $i$ of this system reads $c_0 + c_1x_i + \dots + c_nx_i^n = y_i$ . By Corollary 7.8.2 the matrix is invertible, so the system has exactly one solution $\mathbf{c} = \mathbf{V}(x_0, \dots, x_n)^{-1}\mathbf{y}$ (Theorem 3.4.4), and the interpolating polynomial exists and is unique.

Remark. This is not a proof from nothing. The root bound of Chapter 6 (Corollary 6.1.4) did the real work inside Theorem 7.8.1, exactly as the root bound did in Chapter 3 (Lemma 3.7.2). What the determinant adds is an explicit number: it says how close to singular the interpolation problem is, since nodes that nearly coincide make the product small.

Warning. Watch the shape and the order. The formula is for rows $(1, x_i, \dots, x_i^{n-1})$ starting at the power $0$ , and for the factors $x_j - x_i$ with $i < j$ , “later node minus earlier node”. Dropping the column of ones changes the answer: $\det\begin{pmatrix} x_1 & x_1^2 \\ x_2 & x_2^2 \end{pmatrix} = x_1x_2(x_2 - x_1)$ , not $x_2 - x_1$ . Writing the factors as $x_i - x_j$ instead flips the sign of each of the $\binom{n}{2}$ factors, so it multiplies the answer by $(-1)^{\binom{n}{2}}$ : already for $n = 2$ and $n = 3$ this is $-1$ , a wrong sign.

Quick check. Without expanding, find $\det \mathbf{V}(1, 2, 3, 5)$ .

Solution. The pairs $i < j$ give $(2 - 1)(3 - 1)(5 - 1)(3 - 2)(5 - 2)(5 - 3) = 1 \cdot 2 \cdot 4 \cdot 1 \cdot 3 \cdot 2 = 48$ .

7.8.2. Tridiagonal determinants

Many matrices in applications have non-zero entries only on the diagonal and next to it: discretized differential equations, chains of springs, the recurrences of orthogonal polynomials. Such a matrix is tridiagonal. Its determinant is not a closed formula in general, but it obeys a recurrence in the size, because the last row has only two non-zero entries.

Let $\mathbf{T}_n \in M_n(F)$ have diagonal entries $a_1, \dots, a_n$ , entries $b_1, \dots, b_{n-1}$ just above the diagonal ( $(i, i+1)$ -entry $b_i$ ), entries $c_1, \dots, c_{n-1}$ just below it ( $(i+1, i)$ -entry $c_i$ ), and zeros elsewhere: $\mathbf{T}_n = \begin{pmatrix} a_1 & b_1 & & & \\ c_1 & a_2 & b_2 & & \\ & c_2 & \ddots & \ddots & \\ & & \ddots & a_{n-1} & b_{n-1} \\ & & & c_{n-1} & a_n \end{pmatrix}.$ For $1 \le k \le n$ , the top-left $k \times k$ block of $\mathbf{T}_n$ is the tridiagonal matrix $\mathbf{T}_k$ built from $a_1, \dots, a_k$ , $b_1, \dots, b_{k-1}$ , $c_1, \dots, c_{k-1}$ . Write $D_k = \det \mathbf{T}_k$ , and set $D_0 = 1$ .

Theorem 7.8.3 (Tridiagonal Recurrence)

With this notation, $D_1 = a_1$ , and for $2 \le k \le n$ , $D_k = a_k\,D_{k-1} - b_{k-1}c_{k-1}\,D_{k-2} .$

Proof. $D_1 = \det(a_1) = a_1$ . For $k = 2$ , $D_2 = a_1a_2 - b_1c_1 = a_2D_1 - b_1c_1D_0$ . Let $k \ge 3$ . Row $k$ of $\mathbf{T}_k$ has only two possibly non-zero entries, $c_{k-1}$ in column $k - 1$ and $a_k$ in column $k$ . By Theorem 7.6.1 along row $k$ , $D_k = c_{k-1}(-1)^{2k-1}M_{k,k-1} + a_k(-1)^{2k}M_{kk} = a_kM_{kk} - c_{k-1}M_{k,k-1},$ where $M_{kk} = \det \mathbf{T}_{k-1} = D_{k-1}$ . The minor $M_{k,k-1}$ is the determinant of the $(k - 1) \times (k - 1)$ matrix obtained by deleting row $k$ and column $k - 1$ . Its last column consists of rows $1, \dots, k - 1$ of column $k$ of $\mathbf{T}_k$ , which is $b_{k-1}$ in the last position and $0$ elsewhere. Expanding along this last column (Theorem 7.6.1), with sign $(-1)^{(k-1)+(k-1)} = 1$ , gives $M_{k,k-1} = b_{k-1}\det \mathbf{T}_{k-2} = b_{k-1}D_{k-2}$ , because deleting the last row and column leaves rows and columns $1, \dots, k - 2$ of $\mathbf{T}_k$ . Substituting proves the recurrence.

Example (Fibonacci Numbers as Determinants)

Let $\mathbf{T}_n$ have $1$ on the diagonal, $1$ above it and $-1$ below it. Find $D_n = \det \mathbf{T}_n$ .

Solution. Here $a_k = 1$ and $b_{k-1}c_{k-1} = 1 \cdot (-1) = -1$ , so Theorem 7.8.3 gives $D_k = D_{k-1} + D_{k-2}$ for $k \ge 2$ , with $D_0 = 1$ and $D_1 = 1$ . Hence $D_1, D_2, D_3, \dots = 1, 2, 3, 5, 8, 13, 21, 34, \dots,$ the Fibonacci numbers: $D_n = F_{n+1}$ , where $F_1 = F_2 = 1$ and $F_{k} = F_{k-1} + F_{k-2}$ . For instance $D_3 = \det\begin{pmatrix} 1 & 1 & 0 \\ -1 & 1 & 1 \\ 0 & -1 & 1 \end{pmatrix} = 1\cdot(1 + 1) - 1\cdot(-1 - 0) + 0 = 3 .$

Only the products $b_{k-1}c_{k-1}$ enter the recurrence, never $b_{k-1}$ and $c_{k-1}$ separately. For example, for a real number $t$ , putting $2t$ on the diagonal and $1$ on both neighboring lines gives $D_k = 2t\,D_{k-1} - D_{k-2}$ , so $D_1 = 2t$ , $D_2 = 4t^2 - 1$ , $D_3 = 8t^3 - 4t$ . These are the Chebyshev polynomials of the second kind.

7.8.3. Block triangular matrices

A matrix can be cut into blocks by a horizontal and a vertical line. When the block in the lower left is zero, the determinant should see only the two square blocks on the diagonal, just as a triangular determinant sees only diagonal entries. This result is used constantly, for instance in Chapter 8 for Schur complements.

Theorem 7.8.4 (Block Triangular Determinant)

Let $\mathbf{A} \in M_k(F)$ , $\mathbf{D} \in M_l(F)$ and $\mathbf{B} \in M_{k \times l}(F)$ , with $k, l \ge 1$ . Then $\det\begin{pmatrix} \mathbf{A} & \mathbf{B} \\ 0 & \mathbf{D} \end{pmatrix} = \det \mathbf{A} \cdot \det \mathbf{D} ,$ where $0$ is the $l \times k$ zero matrix. The same holds for $\begin{pmatrix} \mathbf{A} & 0 \\ \mathbf{C} & \mathbf{D} \end{pmatrix}$ with $\mathbf{C} \in M_{l \times k}(F)$ .

Idea. In the Leibniz formula for the whole matrix, a term picks one entry from each column. The first $k$ columns have zeros below row $k$ , so a non-zero term must pick rows $1, \dots, k$ for the first $k$ columns, and is then forced to pick rows $k+1, \dots, n$ for the last $l$ columns. Such a permutation is a permutation of each block, its inversions are the inversions inside the blocks, and the Leibniz sum factors into two Leibniz sums.

Proof. Let $n = k + l$ and $\mathbf{M} = \begin{pmatrix} \mathbf{A} & \mathbf{B} \\ 0 & \mathbf{D} \end{pmatrix} = (m_{ij})$ . By Definition 7.4.1, $\det \mathbf{M} = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{j=1}^{n} m_{\sigma(j)j}$ . Since $m_{ij} = 0$ for $i > k \ge j$ , the term of $\sigma$ is $0$ unless $\sigma(j) \le k$ for every $j \le k$ . For such $\sigma$ , the injective map $\sigma$ sends $\{1, \dots, k\}$ into itself, hence onto itself (a finite set), and so it sends $\{k+1, \dots, n\}$ onto itself. Therefore $\sigma$ determines $\alpha \in S_k$ and $\beta \in S_l$ by $\alpha(j) = \sigma(j) \quad (j \le k), \qquad \beta(q) = \sigma(k + q) - k \quad (q \le l),$ and every pair $(\alpha, \beta)$ arises from exactly one such $\sigma$ .

We compare signs through inversions (Definition 7.3.1). A pair $i < j$ with $i, j \le k$ is an inversion of $\sigma$ exactly when it is one of $\alpha$ ; a pair $k + p < k + q$ is an inversion of $\sigma$ exactly when $p < q$ is one of $\beta$ ; and a pair $i \le k < j$ is never an inversion, since $\sigma(i) \le k < \sigma(j)$ . By Definition 7.3.2, $\operatorname{sgn}(\sigma) = \operatorname{sgn}(\alpha)\operatorname{sgn}(\beta)$ . Also $m_{\sigma(j)j} = a_{\alpha(j)j}$ for $j \le k$ and $m_{\sigma(k+q),k+q} = d_{\beta(q)q}$ . Hence $\det \mathbf{M} = \sum_{\alpha \in S_k}\sum_{\beta \in S_l} \operatorname{sgn}(\alpha)\operatorname{sgn}(\beta) \prod_{j=1}^{k} a_{\alpha(j)j} \prod_{q=1}^{l} d_{\beta(q)q} = \det \mathbf{A} \cdot \det \mathbf{D},$ using Definition 7.4.1 twice. For the lower block triangular case, the transpose of $\begin{pmatrix} \mathbf{A} & 0 \\ \mathbf{C} & \mathbf{D} \end{pmatrix}$ is $\begin{pmatrix} \mathbf{A}^{\top} & \mathbf{C}^{\top} \\ 0 & \mathbf{D}^{\top} \end{pmatrix}$ , so by Theorem 7.4.5 and the first case its determinant is $\det \mathbf{A}^{\top} \det \mathbf{D}^{\top} = \det \mathbf{A} \det \mathbf{D}$ .

The proof uses only sums and products of entries, so, like the results flagged in §4–§6, it holds for matrices over any commutative ring; §10 uses this over $F[x]$ . By induction on the number of blocks, a matrix with square blocks $\mathbf{A}_1, \dots, \mathbf{A}_r$ on the diagonal and zeros below them has determinant $\det \mathbf{A}_1 \cdots \det \mathbf{A}_r$ ; Theorem 7.4.6 is the case where every block is $1 \times 1$ .

For example, $\det\begin{pmatrix} 2 & 1 & 5 & -3 \\ 1 & 1 & 0 & 7 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 4 & 2 \end{pmatrix} = \det\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\det\begin{pmatrix} 3 & 1 \\ 4 & 2 \end{pmatrix} = 1 \cdot 2 = 2,$ and the entries $5, -3, 0, 7$ of the upper right block play no role.

Warning. There is no “block $ad - bc$ ” rule. For a general block matrix, $\det\begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{pmatrix} \ne \det \mathbf{A}\det \mathbf{D} - \det \mathbf{B}\det \mathbf{C}$ . Take $\mathbf{A} = \mathbf{D} = 0$ and $\mathbf{B} = \mathbf{C} = \mathbf{I}_2$ : the $4 \times 4$ matrix $\begin{pmatrix} 0 & \mathbf{I}_2 \\ \mathbf{I}_2 & 0 \end{pmatrix}$ is the permutation matrix of $(1\ 3)(2\ 4)$ , a product of two transpositions, so its determinant is $+1$ ; the rule would give $0 \cdot 0 - 1 \cdot 1 = -1$ . The zero block in the corner is what the theorem needs. The correct formulas for general blocks use Schur complements, in Chapter 8.

Quick check. Is $\det\begin{pmatrix} \mathbf{A} & \mathbf{B} \\ 0 & \mathbf{D} \end{pmatrix} = \det \mathbf{A} \det \mathbf{D}$ meaningful when $\mathbf{A}$ is $2 \times 3$ and $\mathbf{D}$ is $3 \times 2$ (so the whole matrix is $5 \times 5$ )?

Solution. No. The theorem requires $\mathbf{A}$ and $\mathbf{D}$ square; here $\det \mathbf{A}$ and $\det \mathbf{D}$ are undefined. In fact the whole determinant is $0$ whatever the entries: the first $3$ columns have their non-zero entries only in the first $2$ rows, so they are linearly dependent in a $2$ -dimensional space, and a determinant with dependent columns is $0$ (Theorem 7.2.1).

7.8.4. Rank-one updates

Changing a matrix by a rank-one matrix $\mathbf{u}\mathbf{v}^{\top}$ is the smallest possible change in the sense of rank, and it happens often: updating one observation in statistics, modifying one row of a system, or adding the same number to every entry. The determinant of the updated identity turns out to be a single dot product.

Theorem 7.8.5 (Matrix Determinant Lemma)

Let $\mathbf{u}, \mathbf{v} \in F^n$ . Then $\det(\mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top}) = 1 + \mathbf{v}^{\top}\mathbf{u} .$

Idea. Border $\mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top}$ by an extra column $\mathbf{u}$ and an extra row $(\mathbf{0}^{\top}, 1)$ , giving a block triangular matrix $\mathbf{M}$ with the same determinant. Then clear the update by elimination with blocks. Rows: add $\mathbf{v}^{\top}$ times the top $n$ rows to the new bottom row (multiply by $\mathbf{L}$ on the left). Columns: subtract the last column times $\mathbf{v}^{\top}$ from the first $n$ columns (multiply by $\mathbf{L}'$ on the right). The factors $\mathbf{L}$ , $\mathbf{L}'$ have determinant $1$ , and the result is block triangular with $\mathbf{I}_n$ and the $1 \times 1$ block $1 + \mathbf{v}^{\top}\mathbf{u}$ on the diagonal.

Proof. Consider the $(n+1) \times (n+1)$ block matrices $\mathbf{L} = \begin{pmatrix} \mathbf{I}_n & \mathbf{0} \\ \mathbf{v}^{\top} & 1 \end{pmatrix}, \qquad \mathbf{M} = \begin{pmatrix} \mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top} & \mathbf{u} \\ \mathbf{0}^{\top} & 1 \end{pmatrix}, \qquad \mathbf{L}' = \begin{pmatrix} \mathbf{I}_n & \mathbf{0} \\ -\mathbf{v}^{\top} & 1 \end{pmatrix}.$ Splitting the sum $\sum_{j}$ in Theorem 0.9.2 (1) into the indices $j \le n$ and $j = n + 1$ shows that matrices partitioned in this way multiply block by block, like $2 \times 2$ matrices whose entries are blocks. Hence $\mathbf{L}\mathbf{M} = \begin{pmatrix} \mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top} & \mathbf{u} \\ \mathbf{v}^{\top} + (\mathbf{v}^{\top}\mathbf{u})\mathbf{v}^{\top} & \mathbf{v}^{\top}\mathbf{u} + 1 \end{pmatrix},$ where we used $\mathbf{v}^{\top}(\mathbf{u}\mathbf{v}^{\top}) = (\mathbf{v}^{\top}\mathbf{u})\mathbf{v}^{\top}$ by associativity, $\mathbf{v}^{\top}\mathbf{u}$ being a $1 \times 1$ matrix, that is, a scalar. Multiplying on the right by $\mathbf{L}'$ , $\mathbf{L}\mathbf{M}\mathbf{L}' = \begin{pmatrix} \mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top} - \mathbf{u}\mathbf{v}^{\top} & \mathbf{u} \\ \mathbf{v}^{\top} + (\mathbf{v}^{\top}\mathbf{u})\mathbf{v}^{\top} - (1 + \mathbf{v}^{\top}\mathbf{u})\mathbf{v}^{\top} & 1 + \mathbf{v}^{\top}\mathbf{u} \end{pmatrix} = \begin{pmatrix} \mathbf{I}_n & \mathbf{u} \\ \mathbf{0}^{\top} & 1 + \mathbf{v}^{\top}\mathbf{u} \end{pmatrix}.$ By Theorem 7.8.4, $\det \mathbf{L} = \det \mathbf{I}_n \cdot 1 = 1$ , $\det \mathbf{L}' = 1$ , $\det \mathbf{M} = \det(\mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top}) \cdot 1$ , and $\det(\mathbf{L}\mathbf{M}\mathbf{L}') = \det \mathbf{I}_n\,(1 + \mathbf{v}^{\top}\mathbf{u}) = 1 + \mathbf{v}^{\top}\mathbf{u}$ . By Theorem 7.5.4, $\det(\mathbf{L}\mathbf{M}\mathbf{L}') = \det \mathbf{L}\det \mathbf{M}\det \mathbf{L}' = \det(\mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top})$ . This proves the lemma.

Example (

a

on the Diagonal,

1

Elsewhere)

Let $a \in \mathbb{R}$ and $n \ge 1$ , and let $\mathbf{A}_n \in M_n(\mathbb{R})$ have every diagonal entry $a$ and every other entry $1$ . Find $\det \mathbf{A}_n$ .

Solution. Let $\mathbf{1} = (1, \dots, 1) \in \mathbb{R}^n$ . Every entry of $\mathbf{1}\mathbf{1}^{\top}$ is $1$ , so $\mathbf{A}_n = (a - 1)\mathbf{I}_n + \mathbf{1}\mathbf{1}^{\top}$ .

Case $a \ne 1$ . Then $\mathbf{A}_n = (a - 1)\big(\mathbf{I}_n + \mathbf{u}\mathbf{1}^{\top}\big)$ with $\mathbf{u} = \frac{1}{a-1}\mathbf{1}$ . Scaling an $n \times n$ matrix by $c$ multiplies its determinant by $c^n$ (multilinearity in each of the $n$ columns), so by Theorem 7.8.5 $\det \mathbf{A}_n = (a - 1)^n\Big(1 + \frac{n}{a - 1}\Big) = (a - 1)^{n-1}(a - 1 + n),$ since $\mathbf{1}^{\top}\mathbf{u} = \frac{n}{a-1}$ .

Case $a = 1$ . For $n \ge 2$ , all columns are equal, so $\det \mathbf{A}_n = 0$ , which the formula also gives. For $n = 1$ , $\det \mathbf{A}_1 = 1$ , and the formula gives $(a - 1)^0 \cdot 1 = 1$ .

So $\det \mathbf{A}_n = (a - 1)^{n-1}(a + n - 1)$ for every $a$ . For $n = 3$ this is $(a - 1)^2(a + 2)$ ; with $a = 3$ it gives $4 \cdot 5 = 20$ .

Warning. The lemma is about $\mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top}$ , not $\mathbf{u}\mathbf{v}^{\top}$ or $\mathbf{u}^{\top}\mathbf{v}$ . The rank-one matrix $\mathbf{u}\mathbf{v}^{\top}$ is $n \times n$ and, for $n \ge 2$ , has determinant $0$ ; the scalar $\mathbf{v}^{\top}\mathbf{u}$ is its trace. Do not confuse $\det(\mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top}) = 1 + \mathbf{v}^{\top}\mathbf{u}$ with $\det(\mathbf{I}_n) + \det(\mathbf{u}\mathbf{v}^{\top}) = 1$ (for $n \ge 2$ ), which is false whenever $\mathbf{v}^{\top}\mathbf{u} \ne 0$ : for $\mathbf{u} = \mathbf{v} = \mathbf{e}_1$ , $\mathbf{I}_n + \mathbf{e}_1\mathbf{e}_1^{\top} = \operatorname{diag}(2, 1, \dots, 1)$ has determinant $2$ .

7.8.5. Exercises

A. Check your understanding

Exercise (A1)

(a) State the Vandermonde determinant formula, including the order of the factors.
(b) True or false: $\mathbf{V}(x_1, \dots, x_n)$ is invertible whenever $x_1, \dots, x_n$ are non-zero. Justify your answer.
(c) Write down the recurrence for the determinant of a tridiagonal matrix, and say which row the proof expands along.
(d) True or false: $\det\begin{pmatrix} \mathbf{A} & \mathbf{B} \\ \mathbf{C} & \mathbf{D} \end{pmatrix} = \det \mathbf{A}\det \mathbf{D}$ whenever $\mathbf{A}$ and $\mathbf{D}$ are square and $\mathbf{B} = 0$ . Justify your answer.
(e) True or false: $\det(\mathbf{I}_n + \mathbf{u}\mathbf{v}^{\top}) = \det(\mathbf{I}_n + \mathbf{v}\mathbf{u}^{\top})$ for all $\mathbf{u}, \mathbf{v} \in F^n$ . Justify your answer.

Solution.

(a) $\det \mathbf{V}(x_1, \dots, x_n) = \prod_{1 \le i < j \le n}(x_j - x_i)$ , where $\mathbf{V}$ has $(i, j)$ -entry $x_i^{j-1}$ (Theorem 7.8.1).
(b) False. $\mathbf{V}(1, 1) = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$ has non-zero nodes and determinant $0$ . The condition is that the nodes are distinct (Corollary 7.8.2); $0$ is allowed as a node.
(c) $D_k = a_kD_{k-1} - b_{k-1}c_{k-1}D_{k-2}$ , with $D_0 = 1$ , $D_1 = a_1$ (Theorem 7.8.3). The proof expands along the last row, which has only two non-zero entries.
(d) True. With $\mathbf{B} = 0$ the matrix is lower block triangular, and Theorem 7.8.4 applies.
(e) True. Both equal $1 + \mathbf{v}^{\top}\mathbf{u} = 1 + \mathbf{u}^{\top}\mathbf{v}$ by Theorem 7.8.5, since $\mathbf{v}^{\top}\mathbf{u} = \sum_i v_iu_i = \mathbf{u}^{\top}\mathbf{v}$ .

B. Practice

Exercise (B1: A

4 \times 4

Vandermonde determinant)

Compute $\det\begin{pmatrix} 1 & -1 & 1 & -1 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 4 & 8 \end{pmatrix}$ . Hence decide whether there is exactly one polynomial $p \in \mathbb{R}[x]_{\le 3}$ with $p(-1) = 3$ , $p(0) = 1$ , $p(1) = 0$ , $p(2) = 5$ .

Solution. Row $i$ is $(1, x_i, x_i^2, x_i^3)$ with nodes $-1, 0, 1, 2$ , so this is $\mathbf{V}(-1, 0, 1, 2)$ . By Theorem 7.8.1 its determinant is $(0 - (-1))(1 - (-1))(2 - (-1))(1 - 0)(2 - 0)(2 - 1) = 1 \cdot 2 \cdot 3 \cdot 1 \cdot 2 \cdot 1 = 12 .$ (Check by expanding along row $2$ , which is $(1, 0, 0, 0)$ : the determinant is $(-1)^{2+1}\det\begin{pmatrix} -1 & 1 & -1 \\ 1 & 1 & 1 \\ 2 & 4 & 8 \end{pmatrix}$ , that is $-\big({-1}(8 - 4) - 1(8 - 2) - 1(4 - 2)\big) = -(-4 - 6 - 2) = 12$ .) Hence the coefficient matrix of the interpolation system is invertible, and there is exactly one $p \in \mathbb{R}[x]_{\le 3}$ with the given values.

Exercise (B2: The second-difference matrix)

Let $\mathbf{K}_n \in M_n(\mathbb{R})$ have $2$ on the diagonal, $-1$ directly above and below it, and $0$ elsewhere. Prove that $\det \mathbf{K}_n = n + 1$ . Hence show that $\mathbf{K}_n$ is invertible for every $n$ .

Solution. Here $a_k = 2$ and $b_{k-1}c_{k-1} = (-1)(-1) = 1$ . By Theorem 7.8.3, $D_k = 2D_{k-1} - D_{k-2}$ for $k \ge 2$ , with $D_0 = 1$ , $D_1 = 2$ . We show $D_k = k + 1$ by strong induction on $k$ (Theorem 0.2.10). It holds for $k = 0, 1$ . If it holds for $k - 1$ and $k - 2$ , then $D_k = 2k - (k - 1) = k + 1$ . Hence $\det \mathbf{K}_n = D_n = n + 1 \ne 0$ , and $\mathbf{K}_n$ is invertible by Theorem 7.5.2.

Exercise (B3: A rank-one update)

Let $\mathbf{u} = (1, 2, 3)$ and $\mathbf{v} = (1, -1, 2)$ in $\mathbb{Q}^3$ .

(a) Use Theorem 7.8.5 to find $\det(\mathbf{I}_3 + \mathbf{u}\mathbf{v}^{\top})$ .
(b) Write out $\mathbf{I}_3 + \mathbf{u}\mathbf{v}^{\top}$ and check your answer by elimination.
(c) Find all $t \in \mathbb{Q}$ for which $\mathbf{I}_3 + t\,\mathbf{u}\mathbf{v}^{\top}$ is not invertible.

Solution.

(a) $\mathbf{v}^{\top}\mathbf{u} = 1 - 2 + 6 = 5$ , so $\det(\mathbf{I}_3 + \mathbf{u}\mathbf{v}^{\top}) = 6$ .
(b) $\mathbf{u}\mathbf{v}^{\top} = \begin{pmatrix} 1 & -1 & 2 \\ 2 & -2 & 4 \\ 3 & -3 & 6 \end{pmatrix}$ , so $\mathbf{I}_3 + \mathbf{u}\mathbf{v}^{\top} = \begin{pmatrix} 2 & -1 & 2 \\ 2 & -1 & 4 \\ 3 & -3 & 7 \end{pmatrix}$ . Subtract row $1$ from row $2$ , which does not change the determinant (Theorem 7.5.1), to get row $2$ equal to $(0, 0, 2)$ . Expanding along that row, $\det = 2 \cdot (-1)^{2+3}\det\begin{pmatrix} 2 & -1 \\ 3 & -3 \end{pmatrix} = -2(-6 + 3) = 6 .$
(c) $\mathbf{I}_3 + t\,\mathbf{u}\mathbf{v}^{\top} = \mathbf{I}_3 + (t\mathbf{u})\mathbf{v}^{\top}$ , whose determinant is $1 + t\,\mathbf{v}^{\top}\mathbf{u} = 1 + 5t$ . It is zero exactly for $t = -\frac15$ , and by Theorem 7.5.2 that is the only $t$ for which the matrix is not invertible.

C. Going deeper

Exercise (C1: The Cauchy determinant)

Let $x_1, \dots, x_n, y_1, \dots, y_n \in F$ with $x_i + y_j \ne 0$ for all $i, j$ , and let $\mathbf{Q}_n \in M_n(F)$ have $(i, j)$ -entry $\dfrac{1}{x_i + y_j}$ . The goal is $\det \mathbf{Q}_n = \frac{\prod_{1 \le i < j \le n}(x_j - x_i)(y_j - y_i)}{\prod_{i=1}^{n}\prod_{j=1}^{n}(x_i + y_j)} .$

(a) For $i < n$ , subtract row $n$ from row $i$ . Show that the new $(i, j)$ -entry is $\dfrac{x_n - x_i}{(x_i + y_j)(x_n + y_j)}$ , and deduce that $\det \mathbf{Q}_n = \dfrac{\prod_{i<n}(x_n - x_i)}{\prod_{j}(x_n + y_j)}\det \mathbf{Q}'$ , where $\mathbf{Q}'$ agrees with $\mathbf{Q}_n$ in rows $1, \dots, n-1$ and has last row $(1, \dots, 1)$ .
(b) In $\mathbf{Q}'$ , for $j < n$ , subtract column $n$ from column $j$ . Deduce that $\det \mathbf{Q}' = \dfrac{\prod_{j<n}(y_n - y_j)}{\prod_{i<n}(x_i + y_n)}\det \mathbf{Q}_{n-1}$ , where $\mathbf{Q}_{n-1}$ is built from $x_1, \dots, x_{n-1}$ and $y_1, \dots, y_{n-1}$ .
(c) Prove the formula by induction on $n$ .
(d) The Hilbert matrix $\mathbf{H}_n$ has $(i, j)$ -entry $\frac{1}{i + j - 1}$ . Show that it is a Cauchy matrix, and use (c) to compute $\det \mathbf{H}_3$ .

Hint: in (a) and (b), factor scalars out of whole rows and columns using multilinearity.

Solution.

(a) For $i < n$ , $\frac{1}{x_i + y_j} - \frac{1}{x_n + y_j} = \frac{(x_n + y_j) - (x_i + y_j)}{(x_i + y_j)(x_n + y_j)} = \frac{x_n - x_i}{(x_i + y_j)(x_n + y_j)}$ . These row operations do not change the determinant (Theorem 7.5.1). In the new matrix, row $i < n$ has the factor $x_n - x_i$ in every entry, and column $j$ has the factor $\frac{1}{x_n + y_j}$ in every entry (in row $n$ , $\frac{1}{x_n + y_j} = \frac{1}{x_n + y_j} \cdot 1$ ). Pulling these scalars out of each row and column by multilinearity (Theorem 7.5.1, in its row and column forms) leaves the matrix $\mathbf{Q}'$ with entries $\frac{1}{x_i + y_j}$ in rows $i < n$ and $1$ in row $n$ . This gives the stated formula.
(b) For $j < n$ , subtracting column $n$ from column $j$ leaves the determinant unchanged, makes the last row $(0, \dots, 0, 1)$ , and for $i < n$ gives $\frac{1}{x_i + y_j} - \frac{1}{x_i + y_n} = \frac{y_n - y_j}{(x_i + y_j)(x_i + y_n)}$ . Pull $y_n - y_j$ out of column $j$ ( $j < n$ ) and $\frac{1}{x_i + y_n}$ out of row $i$ ( $i < n$ ); the last row $(0, \dots, 0, 1)$ is unaffected by the column factors because its first $n - 1$ entries are $0$ , and row factors do not touch row $n$ . What remains has $\frac{1}{x_i + y_j}$ in positions $i, j < n$ , entries $(x_i + y_n)\cdot\frac{1}{x_i + y_n} = 1$ in column $n$ above the corner, and last row $(0, \dots, 0, 1)$ . Expanding along the last row (Theorem 7.6.1) gives $\det \mathbf{Q}_{n-1}$ , with sign $(-1)^{2n} = 1$ .
(c) For $n = 1$ , $\det \mathbf{Q}_1 = \frac{1}{x_1 + y_1}$ , which is the formula with empty products upstairs. For $n \ge 2$ , combining (a) and (b), $\det \mathbf{Q}_n = \frac{\prod_{i<n}(x_n - x_i)(y_n - y_i)}{\prod_{j \le n}(x_n + y_j)\prod_{i<n}(x_i + y_n)}\det \mathbf{Q}_{n-1} .$ The numerator supplies exactly the factors of $\prod_{i<j}(x_j - x_i)(y_j - y_i)$ with $j = n$ , and the denominator supplies exactly the factors $x_i + y_j$ with $\max(i, j) = n$ , each once. By the induction hypothesis, $\det \mathbf{Q}_{n-1}$ supplies all the remaining factors. This proves the formula.
(d) With $x_i = i$ and $y_j = j - 1$ , $x_i + y_j = i + j - 1 \ge 1$ , so $\mathbf{H}_n$ is a Cauchy matrix over $\mathbb{Q}$ . For $n = 3$ , the numerator is $\big((2-1)(3-1)(3-2)\big)^2 = 4$ , since $y_j - y_i = j - i = x_j - x_i$ . The denominator is the product of all $i + j - 1$ : the rows give $1 \cdot 2 \cdot 3 = 6$ , $2 \cdot 3 \cdot 4 = 24$ , $3 \cdot 4 \cdot 5 = 60$ , in total $8640$ . Hence $\det \mathbf{H}_3 = \frac{4}{8640} = \frac{1}{2160}$ . (In general $\det \mathbf{H}_n = c_n^4 / c_{2n}$ with $c_n = \prod_{k=1}^{n-1} k!$ ; we state this without proof. These determinants shrink extremely fast, which is why Hilbert matrices are a standard test of numerical methods.)

Exercise (C2: A circulant determinant over

\mathbb{C}

)

Let $a, b, c \in \mathbb{C}$ , let $\omega = e^{2\pi i/3}$ , and let $\mathbf{Z} = \begin{pmatrix} a & b & c \\ c & a & b \\ b & c & a \end{pmatrix}.$

(a) For $k = 0, 1, 2$ , let $\mathbf{w}_k = (1, \omega^k, \omega^{2k})$ and $\lambda_k = a + b\omega^k + c\omega^{2k}$ . Show that $\mathbf{Z}\mathbf{w}_k = \lambda_k\mathbf{w}_k$ .
(b) Let $\mathbf{W}$ be the matrix with columns $\mathbf{w}_0, \mathbf{w}_1, \mathbf{w}_2$ . Show that $\mathbf{Z}\mathbf{W} = \mathbf{W}\operatorname{diag}(\lambda_0, \lambda_1, \lambda_2)$ and that $\det \mathbf{W} \ne 0$ . Deduce $\det \mathbf{Z} = \lambda_0\lambda_1\lambda_2$ .
(c) Using $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$ , deduce $\det \mathbf{Z} = a^3 + b^3 + c^3 - 3abc$ , and check this against the Leibniz formula.

Circulant matrices of every size are treated in Chapter 12.

Solution.

(a) By Example (Roots of Unity), $\omega^3 = 1$ , so $\omega^{-k} = \omega^{2k}$ and $\omega^{-2k} = \omega^{k}$ . Row $1$ of $\mathbf{Z}\mathbf{w}_k$ is $a + b\omega^k + c\omega^{2k} = \lambda_k$ . Row $2$ is $c + a\omega^k + b\omega^{2k} = \omega^k(c\omega^{-k} + a + b\omega^k) = \omega^k(a + b\omega^k + c\omega^{2k}) = \omega^k\lambda_k$ . Row $3$ is $b + c\omega^k + a\omega^{2k} = \omega^{2k}(b\omega^{-2k} + c\omega^{-k} + a) = \omega^{2k}\lambda_k$ . Hence $\mathbf{Z}\mathbf{w}_k = \lambda_k\mathbf{w}_k$ .
(b) By Theorem 0.9.2, column $k$ of $\mathbf{Z}\mathbf{W}$ is $\mathbf{Z}\mathbf{w}_k = \lambda_k\mathbf{w}_k$ , which is column $k$ of $\mathbf{W}\operatorname{diag}(\lambda_0, \lambda_1, \lambda_2)$ . The $(i, k)$ -entry of $\mathbf{W}$ is $(\omega^k)^{i-1}$ , so $\mathbf{W} = \mathbf{V}(1, \omega, \omega^2)^{\top}$ . The nodes $1, \omega, \omega^2$ are distinct (they are $e^{2\pi ik/3}$ for $k = 0, 1, 2$ ), so $\det \mathbf{W} = \det \mathbf{V}(1, \omega, \omega^2) \ne 0$ by Theorem 7.4.5 and Corollary 7.8.2. By Theorem 7.5.4 and Theorem 7.4.6, $\det \mathbf{Z}\det \mathbf{W} = \det \mathbf{W}\,\lambda_0\lambda_1\lambda_2$ , and canceling $\det \mathbf{W} \ne 0$ gives $\det \mathbf{Z} = \lambda_0\lambda_1\lambda_2$ .
(c) Multiply out $\lambda_1\lambda_2 = (a + b\omega + c\omega^2)(a + b\omega^2 + c\omega)$ . The nine terms are $a^2$ , $ab\omega^2 + ab\omega$ , $ac\omega + ac\omega^2$ , $b^2\omega^3$ , $bc\omega^2$ , $bc\omega^4 = bc\omega$ , $c^2\omega^3$ . Using $\omega^3 = 1$ and $\omega + \omega^2 = -1$ , $\lambda_1\lambda_2 = a^2 + b^2 + c^2 - ab - ac - bc .$ Then $\det \mathbf{Z} = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = a^3 + b^3 + c^3 - 3abc$ , expanding and canceling. The Leibniz formula for $\mathbf{Z}$ gives the three even terms $a \cdot a \cdot a$ , $c \cdot c \cdot c$ (from $\sigma = (1\ 2\ 3)$ ) and $b \cdot b \cdot b$ , and the three odd terms, each equal to $abc$ with sign $-$ : the same answer.

Exercise (C3: Rank-one updates of an invertible matrix)

Let $\mathbf{A} \in M_n(F)$ be invertible and $\mathbf{u}, \mathbf{v} \in F^n$ .

(a) Prove that $\det(\mathbf{A} + \mathbf{u}\mathbf{v}^{\top}) = \det \mathbf{A}\,\big(1 + \mathbf{v}^{\top} \mathbf{A}^{-1}\mathbf{u}\big)$ .
(b) Deduce that $\mathbf{A} + \mathbf{u}\mathbf{v}^{\top}$ is invertible if and only if $\mathbf{v}^{\top} \mathbf{A}^{-1}\mathbf{u} \ne -1$ .
(c) Give an example over $\mathbb{Q}$ with $n = 2$ where $\mathbf{A}$ is invertible, $\mathbf{u}\mathbf{v}^{\top} \ne 0$ , and $\mathbf{A} + \mathbf{u}\mathbf{v}^{\top}$ is not invertible.

Solution.

(a) Since $\mathbf{A}$ is invertible, $\mathbf{A} + \mathbf{u}\mathbf{v}^{\top} = \mathbf{A}(\mathbf{I}_n + \mathbf{A}^{-1}\mathbf{u}\mathbf{v}^{\top}) = \mathbf{A}\big(\mathbf{I}_n + (\mathbf{A}^{-1}\mathbf{u})\mathbf{v}^{\top}\big)$ by Theorem 0.9.4. By Theorem 7.5.4 and Theorem 7.8.5 applied to the vectors $\mathbf{A}^{-1}\mathbf{u}$ and $\mathbf{v}$ , $\det(\mathbf{A} + \mathbf{u}\mathbf{v}^{\top}) = \det \mathbf{A}\,(1 + \mathbf{v}^{\top} \mathbf{A}^{-1}\mathbf{u})$ .
(b) $\det \mathbf{A} \ne 0$ by Theorem 7.5.2, so $\det(\mathbf{A} + \mathbf{u}\mathbf{v}^{\top}) \ne 0$ exactly when $1 + \mathbf{v}^{\top} \mathbf{A}^{-1}\mathbf{u} \ne 0$ . Apply Theorem 7.5.2 again.
(c) Take $\mathbf{A} = \mathbf{I}_2$ , $\mathbf{u} = (-1, 0)$ , $\mathbf{v} = (1, 0)$ . Then $\mathbf{v}^{\top}\mathbf{u} = -1$ , and indeed $\mathbf{A} + \mathbf{u}\mathbf{v}^{\top} = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ is not invertible.