Determinant of $n \times n$ matrix of a sort of skew symmetric matrix plus some diagonal

Question

Given, a matrix:

$$\begin{pmatrix} a & b & \ldots & b & b \\ -b & a & \ldots & b & b \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -b & -b & \ldots & a & b \\ -b & -b & \ldots & -b & a \end{pmatrix}.$$

I need to find a determinant. So initially what I did, was I added the first column to other ones:

$$\begin{pmatrix} a & b+a & \ldots & b+a & b+a \\ -b & a-b & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -b & -2b & \ldots & a-b & 0 \\ -b & -2b & \ldots & -2b & a-b \end{pmatrix},$$

then added the last row to the first one

$$\begin{pmatrix} a-b & a-b & \ldots & a-b & 2a \\ -b & a-b & \ldots & 0 & 0 \\ \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -b & -2b & \ldots & a-b & 0 \\ -b & -2b & \ldots & -2b & a-b\end{pmatrix},$$

then multiplied the first column by 2 and subtracted the second one

$$\frac{1}{2}\cdot\begin{pmatrix} a-b & a-b & \ldots & a-b & 2a \\ -a-b & a-b & \ldots & 0 & 0 \\ \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & -2b & \ldots & a-b & 0 \\ 0 & -2b & \ldots & -2b & a-b\end{pmatrix},$$

then expanded it using the first column and got:

$$\frac{1}{2}\cdot(a-b)^n + (a+b)\cdot \begin{pmatrix} a-b & a-b & \ldots & a-b & 2a \\ -2b & a-b & \ldots & 0 & 0 \\ \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -2b & -2b & \ldots & a-b & 0 \\ -2b & -2b & \ldots & -2b & a-b\end{pmatrix}_{(n-1)\times(n-1)}.$$

Then I conjecture that $(n-1)\times(n-1)$ determinant in the latter expression is equal to $(a+b)^(n-1)$, because:

$$ B_{n-1} := \begin{pmatrix} a-b & a-b & \ldots & a-b & 2a \\ -2b & a-b & \ldots & 0 & 0 \\ \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -2b & -2b & \ldots & a-b & 0 \\ -2b & -2b & \ldots & -2b & a-b\end{pmatrix}_{(n-1)\times(n-1)} (R_{1} - R_{2})$$

$$\begin{pmatrix} a+b & 0 & \ldots & a-b & 2a \\ -2b & a-b & \ldots & 0 & 0 \\ \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -2b & -2b & \ldots & a-b & 0 \\ -2b & -2b & \ldots & -2b & a-b\end{pmatrix}_{(n-1)\times(n-1)} (C_{1} - C_{2})$$

$$\begin{pmatrix} a+b & 0 & \ldots & a-b & 2a \\ -(a+b) & a-b & \ldots & 0 & 0 \\ \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & -2b & \ldots & a-b & 0 \\ 0 & -2b & \ldots & -2b & a-b\end{pmatrix}_{(n-1)\times(n-1)} (R_{1}+R_{2})$$

$$\begin{pmatrix} 0 & 0 & \ldots & a-b & 2a \\ -(a+b) & a-b & \ldots & 0 & 0 \\ \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & -2b & \ldots & a-b & 0 \\ 0 & -2b & \ldots & -2b & a-b\end{pmatrix}_{(n-1)\times(n-1)}.$$

So we have $B_{n-1} = (a+b)\cdot B_{n-2} = (a+b)^{n-1}$. So far it looks okay, but I don't see any mistake, cause the results do not correct when I start plugging in numbers.

Answer 1

Let $U$ be the upper triangular matrix of ones. The matrix in question can be expressed as $$ A_n=aI_n+b(U-U^T)=UB_nU^T $$ where $$ B_n=aU^{-1}(U^{-1})^T+b\left[(U^{-1})^T-U^{-1}\right]. $$ Clearly, we have $\det A_n=\det B_n$ because $\det U=1$. It is straightforward to verify that $$ U^{-1}=\pmatrix{1&-1\\ &\ddots&\ddots\\ &&\ddots&\ddots\\ &&&\ddots&-1\\ &&&&1\\},\quad B_n=\pmatrix{2a&b-a\\ -b-a&\ddots&\ddots\\ &\ddots&\ddots&\ddots\\ &&\ddots&2a&b-a\\ &&&-b-a&a}. $$ Being a tridiagonal matrix, the determinant of $B_n$ satisfies the recurrence relation $\det B_n=2a\det B_{n-1}+(b^2-a^2)\det B_{n-2}$ with initial conditions $\det B_0=1$ and $\det B_1=a$. (If you don’t see this, try Laplace expansion along the first row of $B_n$.) By mathematical induction, it can be proved that $\det B_n=\frac12[(a+b)^n+(a-b)^n]$.

Answer 2

We can write $$A=\begin{pmatrix} a & b & \ldots & b & b \\ -b & a & \ldots & b & b \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -b & -b & \ldots & a & b \\ -b & -b & \ldots & -b & a \end{pmatrix}=aI_n+bX$$,

where $$X=\begin{pmatrix} 0 & 1 & \ldots & 1 & 1 \\ -1 & 0 & \ldots & 1 & 1 \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ -1 & -1 & \ldots & 0 & 1 \\ -1 & -1 & \ldots & -1 & 0 \end{pmatrix}$$

Now note that, if $\alpha$ is an eigenvalue of $X$, then $a+b\alpha$ is an eigenvalue of $A$.

Can you take it forward?

Answer 3

Let the matrix be $A_n$. We prove by induction the following observation made by @Pythagoras in the comment:

$$\det A_n =\frac{(a+b)^n + (a-b)^n}{2}$$

When $n=1, 2$, this is obvious.

In general, starting with the first row until the second last row, let each row subtract the one below it. We get the matrix:

$$\begin{pmatrix} a+b & b-a & 0 & 0 & \cdots \\ & a+b & b-a & 0 & \cdots \\\vdots & \vdots & \vdots & \vdots & \vdots \\ & & &a+b & b-a \\ -b & \cdots & -b& -b & a\end{pmatrix}$$

We expand the determinant based on the first column. If the first row and column are eliminated, we end up with a matrix that is obtained by doing the same row operations as above from the $A_{n-1}$. If the last row and the first column are eliminated, we end up with a lower triangular matrix whose determinant is clearly $(b-a)^{n-1}$. , therefore $$\det A_n = (a+b)\det A_{n-1} + (-1)^{n+1}(-b)(b-a)^{n-1}$$

Apply the induction hypothesis $\det A_{n-1}=\frac{(a+b)^{n-1}+(a-b)^{n-1}}{2}$, we are done.

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Edit: $\DeclareMathOperator{sgn}{sgn}$

Here is another approach that is slightly more difficult than I had expected but nonetheless interesting. From the result $\det A_n = \frac{(a+b)^n + (a-b)^n}{2}$, it's clear that it's just the binomial expansion of $(a+b)^n$ with the terms $a^ib^j$ dropped iff $j$ is odd. By the definition of determinant, we have $$\det A_n = \sum_{\sigma\in S_n}\sgn(\sigma)A_{1\sigma(1)}\cdots A_{n\sigma(n)}$$

For each $\sigma$, define $p(\sigma):=|\{i:\sigma(i)<i\}|, q(\sigma):=|\{i:\sigma(i)>i\}, r(\sigma):=|\{i:\sigma(i)=i\}|$. Then we have $$\det A_n =\sum_{\sigma\in S_n}\sgn(\sigma) (-1)^{p(\sigma)}b^{p+q}a^r=\sum_{\sigma\in S_n}\sgn(\sigma) (-1)^{q(\sigma)}b^{p+q}a^r$$

If $p+q$ is odd, then $\sigma^{-1}\not=\sigma$, and we have $$\sgn(\sigma^{-1})(-1)^{p(\sigma^{-1})}=\sgn(\sigma)(-1)^{q(\sigma)}=-\sgn(\sigma)(-1)^{p(\sigma)}$$ hence the term associated with $\sigma$ will cancel the one with $\sigma^{-1}$.

Now it suffices to show that if $m=n-r$ is even, then $$\sum_{\substack{\sigma\in S_m \\ \forall i, \sigma(i)\not=i}}\sgn(\sigma)(-1)^{p(\sigma)}=1$$

Becasue if so, the coefficients of $b^{n-r}a^r$ in $\det A_n$ is clearly ${n \choose r}$.

(This is probably known in combinatorics but a bit surprising to me, as I didn't even know the number of derangement is always odd if there are in total even number of positions.)

Group the rest of the $\sigma$'s by:

• If $(12)$ is part of the cycle decomposition of $\sigma$, then sum up all of such $\sigma$'s, we have $\sum_{\tau\in S_{m-2}}-\sgn(\tau)(-1)^{p(\tau)+1} = 1$ by induction hypothesis.

• If $(12)$ is connected in the cycle decomposition of $\sigma$, i.e. there is a cycle of the form $(12x)$ or $(21x)$. It's clear that such pair of permutations have the same sign but their $p$-values differ by $1$.

• If $1$ and $2$ are not connected the cycle decomposition of $\sigma$. If they are in the same cycle $(1x2y)$ (where $x, y$ represent a non-empty list of numbers), then if paired with $\sigma\circ(12)=(1y)(2x)\cdots$, the terms cancel. Because $1, 2$ are the minimal numbers, the increasing in $(1x2y)$ is the same as in $(1y)(2x)$. (Similarly if they are in distinct cycles $(1x), (2y)$ then $\sigma\circ (12) = (1y2x)$, again the terms cancel.)

Answer 4

\begin{equation} A = \begin{pmatrix} a & b & \cdots & b & b \\ -b & a & \cdots & b & b \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ -b & -b & \cdots & a & b \\ -b & -b & \cdots & -b & a \\ \end{pmatrix} \end{equation} \begin{equation} \begin{pmatrix} a & b+a & \cdots & b+a & b+a \\ -b & a-b & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ -b & -2b & \cdots & a-b & 0 \\ -b & -2b & \cdots & -2b & a-b \\ \end{pmatrix} \end{equation} \begin{equation} \begin{pmatrix} a-b & a-b & \cdots & a-b & 2a \\ -b & a-b & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ -b & -2b & \cdots & a-b & 0 \\ -b & -2b & \cdots & -2b & a-b \\ \end{pmatrix} \end{equation} We conjecture that the \begin{equation} (n - 1) \times (n - 1) \end{equation} determinant in the latter expression is equal to \begin{equation} (a + b)(n - 1), \end{equation} based on the pattern observed from expanding the determinant.

The calculation error in the original attempt seems to come from step 5 or step 6, where the pattern needs to be recognized correctly to deduce the final determinant.

By applying these operations correctly, you will be able to find the determinant of the matrix. Remember that expanding a determinant along a row or column involves taking the elements of that row or column, multiplying by the determinant of the corresponding minor (the matrix that remains after removing the row and column of that element), and alternating the signs.

To confirm the pattern for the determinant of the \begin{equation} (n - 1) \times (n - 1) \end{equation} matrix, you may want to evaluate it explicitly for small values of \begin{equation} n \end{equation} and then generalize. The resulting expression should be valid for all \begin{equation} n, \end{equation} and you can use it to calculate the determinant for any size matrix of this form.