Determinant of rank-$1$ update of multiple of identity matrix

Question

I've got to calculate determinant for such matrix:

$$ \begin{bmatrix} a_1+b & a_2 & \cdots & a_n\\ a_1 & a_2+b & \cdots & a_n\\ \vdots & \vdots & \ddots & \vdots\\ a_1 & a_2 & \cdots & a_n+b\\ \end{bmatrix} $$

Please give me some tips how to calculate this.

Answer 1

The structure of this matrix aloud to write this equation which does not generally hold $$det(A+Ib)=b^{n-1}(tr(A)+b)$$

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I guess that's related with the fact that $$A^2=tr(A)A$$ but just now I don't see how... If someone can see just edit in the comments please

Answer 2

Set $A=\sum\limits_{i=1}^na_i$. By multilinearity,

\begin{align*} &\begin{vmatrix} a_1+b &a_2&\dots&a_n\\ a_1&b+a_2 &\dots&a_n \\ \vdots&\vdots&&\vdots\\ a_1 a_2&\dots &a_n+b \end{vmatrix}= \begin{vmatrix} A+b &a_2&\dots&a_n\\ A+b&b+a_2 &\dots&a_n \\ \vdots&\vdots&&\vdots\\ A+b & a_2&\dots &b+a_n \end{vmatrix}\\[1ex] &=(A+b)\begin{vmatrix} 1 &a_2&\dots&a_n\\ 1&b+a_2 &\dots&a_n \\ \vdots&\vdots&&\vdots\\ 1 & a_2&\dots &b+a_n \end{vmatrix}=(A+b)\begin{vmatrix} 1 &a_2&\dots&a_n\\ 0&b &\dots& 0 \\ \vdots&\vdots&&\vdots\\ 0 & 0&\dots &b \end{vmatrix}\\[1ex] &=\color{red}{(A+b)b^{n-1}}. \end{align*}

Answer 3

Write the matrix as $A+bI.$ Here, all the rows of $A$ are the same, and so $A$ is rank $1$, and therefore the kernel is of dimension $n-1$ and there is only one non-trivial eigenvalue, $\operatorname{tr}(A)$. Therefore the characteristic polynomial of $A$ is $p_t(A)=\det(tI-A)=t^{n-1}(t-\operatorname{tr}(A))$.

It is now straightforward to calculate $\det(A+bI)$ from $\det(tI-A)$.

Answer 4

The matrix can be written as $A+bI$, where $A$ is the matrix with all rows equal to $a_1,a_2,\dots,a_n$. The determinant in question is $(-1)^n\chi_A(-b)$, where $\chi_A$ is the characteristic polynomial of $A$.

Since $A$ has rank $1$, we have $\chi_A(x)=x^n-tr(A)x^{n-1}$ (see this for instance).

Finally, the determinant in question is $(-1)^n\chi_A(-b)=(-1)^n(-b)^n-a(-b)^{n-1}=b^n+ab^{n-1}$.