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 $n\times n$ 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?

Hint. Do you know what happens to the determinant when you subtract one row from another? — Ethan Bolker, Oct 28 '20 at 18:58
Correct. That and expansion by minors on a column that's mostly $0$ should help you. — Ethan Bolker, Oct 28 '20 at 19:02
Does this answer your question? Determining eigenvalues, eigenvectors of $A\in \mathbb{R}^{n\times n}(n\geq 2)$. — Jean Marie, Oct 28 '20 at 19:04
I still do not know what eigenvalues are, neither eigenvectors. — Uri Toti, Oct 28 '20 at 19:07

score 1 · Answer 1 · answered Oct 28 '20 at 21:15

Let $\mathbf e_i$ denote the $i$th standard base vector and $\mathbf j=\sum\mathbf e_i$ denote the "all-ones" vector.

Clearly, $A\mathbf j=(b+na)\mathbf j$, so $\mathbf j$ is an eigenvector of eigenvalue $b+na$.

Also $A\mathbf e_i=b\mathbf e_i+a\mathbf j$ so that for $2\le i\le n$, $A(\mathbf e_i-\mathbf e_1)=b(\mathbf e_i-\mathbf e_1)$, 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 $$\det A=(b+na)\cdot b^{n-1}.$$

Thank you! Awesome, although I still do not know what eigen-stuff is. — Uri Toti, Oct 29 '20 at 18:58

score 0 · Answer 2 · answered Oct 28 '20 at 19:14

The given matrix may be written as $bI_n+\mathbf a\mathbf1^T$ where $\mathbf x$ denotes a vector consisting of $n$ copies of $\mathbf x$. The matrix determinant lemma gives the required determinant as $$(1+\mathbf 1^T(I_n/b)\mathbf a)\det bI_n=(1+na/b)b^n=b^n+nab^{n-1}$$

score 0 · Answer 3 · answered Oct 28 '20 at 19:15

0

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?

answered Oct 28 '20 at 19:15

leonbloy

63,430

The problem is supposed to be solved by only using determinant properties... not that advanced lemma, although it makes things a lot easier. – Uri Toti Oct 28 '20 at 19:26

score 0 · Accepted Answer · answered Oct 28 '20 at 19:55

\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?

Thank you! It's almost what I did! Thank you very much. – Uri Toti Oct 29 '20 at 18:58 — Uri Toti, Oct 29 '20 at 18:58

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

4 Answers4