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To give the close-form of $\det(\mathbf{G})$, where $\mathbf{G}$ is \begin{align} \mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T \end{align} in which $a$ and $b$ are constant, and $\boldsymbol{e}$ is a column vector with all elements being $1$. In addition, $(\cdot)^T$ is transposition operation. $\mathbf{G}$ is $u\times u$. We use $\mathbf{G}_u$ to underline the dimension of $\mathbf{G}$.

The question is to determine $\det(\mathbf{G})$.

As I know: We rewrite $\mathbf{G}_u$ as \begin{align} \mathbf{G}_u=\left[\begin{array}{ccc} \mathbf{G}_{u-1} & b\\ b & a+b \end{array} \right] \end{align} Using the determinant of block matrix lemma \begin{align} \det\left[\begin{array}{ccc} \mathbf{A} & \mathbf{B}\\ \mathbf{C} & \mathbf{D} \end{array} \right]=\det(\mathbf{A})\det(\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B}) \end{align} We then have \begin{align} \det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e}) \end{align} It still needs to get $\mathbf{G}_{u-1}^{-1}$ via matrix inverse lemma \begin{align} (\mathbf{A}+\mathbf{BC})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1}\mathbf{B}(\mathbf{I}+\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1} \end{align} We then have \begin{align} \mathbf{G}_{u} &=\frac{1}{a}\mathbf{I}-\frac{1}{a^2}b\boldsymbol{e}\left(1+\frac{b}{a}\boldsymbol{e}^T\boldsymbol{e}\right)^{-1}\boldsymbol{e}^T\\ &=\frac{1}{a}\mathbf{I}-\frac{b}{a(a+bu)}\boldsymbol{ee}^T \end{align} where $\boldsymbol{e}^T\boldsymbol{e}=u$ is used. Plugging it into \begin{align} &\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})\\ =&a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right) \end{align} Then \begin{align} \det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\left[{a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)}\right] \end{align} Although, the connection between $\mathbf{G}_u$ and $\mathbf{G}_{u-1}$ is found, I can't give the expression of $\mathbf{G}_u$. Please give me hand, thanks a lot!

Qiuyun.Zou
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4 Answers4

5

Assume $a\ne 0$, We know that

$$\det(A+uv^T)=(1+v^TA^{-1}u)\det(A)$$

Here $A=aI, u = be, v=e$

\begin{align}\det(aI+bee^T)&=(1+be^T(aI)^{-1}e)\det(aI) \\ &=\left(1+\frac{bu}a\right)a^u\\ &=a^{u}+bua^{u-1}\end{align}

If $a=0$ and $u>1$, then the determinant is $0$.

If $a=0$ and $u=1$, then the determinant is $b$.

Siong Thye Goh
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Note that $G$ is a circulant matrix, so we know the eigenvectors are $(1,\zeta,\zeta^2,\dots,\zeta^{u-1})$ where $\zeta^u=1$. Its eigenvalue is $$ a+b\sum_{j=0}^{u-1} \zeta^j= \begin{cases} a & \text{if }\zeta\neq 1\\ a+bu & \text{if }\zeta=1 \end{cases} $$ and so the determinant is $a^{u-1}(a+bu)$.

user10354138
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Use eigenvalues!

The matrix has $a$ as eigenvalue and the corresponding eigenspace is the space that is orthogonal to $e$: If $v^Te = 0$, then $$ Gv = (aI + bee^T)v = av + bee^Tv = av. $$ Hence the eigenvalue $a$ has multiplicity $n-1$ (if the matrix is $n\times n$). The other eigenvalue is $a+nb$ and the eigenspace is one dimensional an spanned by $e$: $$ Ge = (aI + bee^T)e = ae + bee^Te = (a+nb)e. $$ Since the determinant is the product of eigenvalues, we get $$ \det(G) = a^{n-1}(a+nb). $$

Dirk
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A More Basic Approach

Let us consider for illustration, $G_4$: $$ G=aI_{4\times4}+bee^T \\ \text{where } e = \begin{bmatrix} 1\\ 1\\ 1\\ 1 \end{bmatrix} \\ \text{Hence, } G = \begin{bmatrix} a & 0 & 0 & 0 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a \\ \end{bmatrix} + \begin{bmatrix} b & b & b & b \\ b & b & b & b \\ b & b & b & b \\ b & b & b & b \\ \end{bmatrix} = \begin{bmatrix} a + b & b & b & b \\ b & a + b & b & b \\ b & b & a + b & b \\ b & b & b & a + b \\ \end{bmatrix} $$ Taking the determinant, let us apply the transformation $\left(R_1 \rightarrow R_1 + R_2 + R_3 + R_4 \right)$, $$ \det(G) = \begin{vmatrix} a + 4b & a + 4b & a + 4b & a + 4b \\ b & a + b & b & b \\ b & b & a + b & b \\ b & b & b & a + b \\ \end{vmatrix} $$ Applying now $\left( R_1 \rightarrow \frac{1}{a+4b} R_1 \right)$ and $\prod_{k=2}^{4}{\left(R_k \rightarrow R_k - bR_1\right)}$ $$ = (a+4b) \cdot \begin{vmatrix} 1 & 1 & 1 & 1 \\ 0 & a & 0 & 0 \\ 0 & 0 & a & 0 \\ 0 & 0 & 0 & a \\ \end{vmatrix} \\ \text{So } \det{(G)} = (a+4b)\cdot a^3 $$

Similarly, for $G_n$, upon applying the transformations $\left( R_1 \rightarrow \frac{1}{a+nb} \sum_{k=1}^{n}{R_k} \right)$ and $\left( \prod_{k=2}^{n} \left( R_k \rightarrow R_k - bR_1 \right) \right)$ and then simply expanding, $$ \boxed {\det{(G_n)} = (a+nb) \cdot a^{n-1}} $$

Varad Mahashabde
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