How to calculate this determinant?

Question

$$A=\begin{bmatrix}n-1&k&k&k&\ldots& k\\k&n-1&k&k&\ldots &k\\\ldots&\ldots&\ldots &&\ldots\\\\k&k&k&k&\ldots &n-1\\ \end{bmatrix}_{n\times n}$$

where $n,k\in \Bbb N$ are fixed.

I tried for $n=3$ and got the characteristic polynomial as $(x-2-k)^2(x-2+2k).$

How to find it for general $n\in \Bbb N$?

You formatting is very good. You don't need to be nervous.--) — Juniven Acapulco, Apr 05 '17 at 13:37
If you refer to the $n\times n$ matrix as $A_n$, can you write the determinant in terms of $A_{n-1}, A_{n-2}$ etc.? — user112495, Apr 05 '17 at 13:38
As new user of this site, there are also rules for you to follow. One of which can be found in here — Juniven Acapulco, Apr 05 '17 at 14:45
Just use the technique discussed in this post: http://math.stackexchange.com/questions/86644/determinant-of-a-specially-structured-matrix-as-on-the-diagonal-all-other-e?noredirect=1&lq=1. — StubbornAtom, Apr 05 '17 at 14:48

score 6 · Accepted Answer · edited Apr 06 '17 at 08:25

6

Here I've followed the same initial step as K. Miller. Instead of using a determinant identity I examine the eigenvalues $A$ and consider their product.

If $J$ denotes the $n\times n$ matrix of all $1$'s, then then eigenvalues of $J$ are $0$ with multiplicity $n-1$ and $n$ with multiplicity $1$. This can be seen by noting that $J$ has $n-1$ dimensional kernel and trace $n$.

Your matrix $A$ is exactly $kJ+(n-k-1)I$ where $I$ denotes the $n\times n$ identity matrix. The eigenvalues of $A$ are therefore $n-k-1$ with multiplicity $n-1$ and $nk+n-k-1$ with multiplicity $1$. The determinant of $A$ is then $(nk+n-k-1)(n-k-1)^{n-1}$.

edited Apr 06 '17 at 08:25

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31,062

answered Apr 05 '17 at 14:25

Zack Cramer

770

To clarify why the eigenvalues of $A$ are as given, note that all vectors are eigenvectors of $I$, so each eigenvector of $J$ is also an eigenvector of $A$. – Paul Sinclair Apr 05 '17 at 16:23
In general, if $A$ is a matrix with eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$, and $p$ is a polynomial, then the eigenvalues of $p(A)$ are $p(\lambda_1),p(\lambda_2),\ldots, p(\lambda_n)$. – Zack Cramer Apr 05 '17 at 17:30

K. Miller · Answer 2 · 2017-04-05T15:16:12.370

4

You can write $A = (n-k-1)I + kee^T$, where $e$ is the $n$-vector of all ones. Now use the matrix determinant lemma which states that

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

for $B$ an invertible square matrix. Applying this result to $A$, we find that

$$ \det(A) = (n - k - 1)^n\left(1 + \frac{kn}{n-k-1}\right) = (n-1)(k+1)(n-k-1)^{n-1}. $$

edited Apr 05 '17 at 15:16

answered Apr 05 '17 at 13:54

K. Miller

4,688

More precisely, the 'general result' is the so-called matrix determinant lemma. Additionally, it's worth noting that there's a geometric meaning to the matrix $ee^T$: it's a (unnormalized) projection matrix onto the vector $e$. The vector $e$ then generates a 1D subspace of eigenvectors of $ee^T$ while the remaining $n-1$ eigenvectors lie in the orthogonal complement. – Semiclassical Apr 05 '17 at 14:56
@Semiclassical Thanks for the additional information. I have updated my answer accordingly. – K. Miller Apr 05 '17 at 15:15

score 3 · Answer 3 · answered Apr 05 '17 at 16:51

Here is another method using only rows and columns manipulations. Less smart than previous answers but less demanding in prior knowledge.

First let's replace the first column by the sum of all colums: $$\begin{bmatrix}n-1&k&\ldots& k\\k&n-1&\ldots &k\\\ldots&\ldots&\ldots &\ldots\\\\k&k&\ldots &n-1\\ \end{bmatrix} = \left(n-1 + (n-1) \cdot k \right)\begin{bmatrix}1&k&k&\ldots &k\\ 1&n-1&k&\ldots &k\\ \ldots&\ldots&\ldots &\ldots&\ldots\\ \\1&k&k&\ldots &n-1\\ \end{bmatrix} $$ Now let's subtract the first row from all the other ones. It remains: $$ (n-1) (k+1)\begin{bmatrix}1&k&k&\ldots &k\\ 0&n-1-k&0&\ldots &0\\ \ldots&\ldots&\ldots &\ldots&\ldots\\ \\0&0&0&\ldots &n-1-k\\ \end{bmatrix} = (n-1) (k+1) (n-1-k)^{n-1} $$

How to calculate this determinant?

3 Answers3