Smart method for computing the determinant of the matrix defined by $A_{ii}=a+b$ and $A_{ij}=a$?

Question

We are given the following $ntimes n$ matrix, whose terms are:

$a+b$ for the terms in the diagonal ($i=j$).
$a$ for the rest of the elements.

I know we can simply compute this and try to find a pattern, but is there a faster way? Maybe using some properties of the determinant of a matrix?

Zhanxiong · Accepted Answer

begin{align*}
& begin{vmatrix}
a + b & a     & a     & cdots & a \
a     & a + b & a     & cdots & a \
a     & a     & a + b & cdots & a\
vdots & vdots & vdots & ddots & vdots \
a     & a     & a     & cdots &a + b 
end{vmatrix} \
= & begin{vmatrix}
na + b & na + b   & na + b     & cdots & na + b \
a     & a + b & a     & cdots & a \
a     & a     & a + b & cdots & a\
vdots & vdots & vdots & ddots & vdots \
a     & a     & a     & cdots &a + b 
end{vmatrix} \
= & (na + b)begin{vmatrix}
1 & 1   & 1     & cdots & 1 \
a     & a + b & a     & cdots & a \
a     & a     & a + b & cdots & a\
vdots & vdots & vdots & ddots & vdots \
a     & a     & a     & cdots &a + b 
end{vmatrix} \
= & (na + b)begin{vmatrix}
1 & 1   & 1     & cdots & 1 \
0     & b & 0    & cdots & 0 \
0     & 0    & b & cdots & 0 \
vdots & vdots & vdots & ddots & vdots \
0     & 0    & 0   & cdots & b 
end{vmatrix} \
=& (na + b)b^{n - 1}
end{align*}
Can you figure out what operations I used in each step?

Hagen von Eitzen · Answer

Let  $mathbf e_i$ denote the $i$th standard base vector and $mathbf j=summathbf e_i$ denote the "all-ones" vector.
Clearly, $Amathbf j=(b+na)mathbf j$, so $mathbf j$ is an eigenvector of eigenvalue $b+na$.
Also $Amathbf e_i=bmathbf e_i+amathbf j$ so that for $2le ile n$, $A(mathbf e_i-mathbf e_1)=b(mathbf e_i-mathbf e_1)$, i.e., we know $n-1$ linearly independent eigenvectors of eigenvalue $b$.
All in all, we have found a basis consisting of eigenvectors an know their eigenvalues, and can
conclude that $$det A=(b+na)cdot b^{n-1}.$$

leonbloy · Answer

You can use the matrix determinant lemma.
You have
$$C = b I + a {bf u}  {bf u}^t$$
where ${ bf u}$ a all-ones column vector.
Then $$det(C) = (1 + a {bf u}^t({b}I)^{-1}  {bf u}) det(b I) $$
Can you go on from here?

Parcly Taxel · Answer

The given matrix may be written as $bI_n+mathbf amathbf1^T$ where $mathbf x$ denotes a vector consisting of $n$ copies of $mathbf x$. The matrix determinant lemma gives the required determinant as
$$(1+mathbf 1^T(I_n/b)mathbf a)det bI_n=(1+na/b)b^n=b^n+nab^{n-1}$$

Smart method for computing the determinant of the matrix defined by $A_{ii}=a+b$ and $A_{ij}=a$?

4 Answers

Add your own answers!

Ask a Question