# Need help proving the determinant of a particular sum of matrices.

I'm just learning how to use Mathematica and I was screwing around with it and I noticed that the following expression holds for a bunch of numbers that I threw into it.

I was wondering if someone could help me prove/disprove this?

$det(cJ_n+I)=cn+1$

Where $c\in \mathbb{R}$ is some constant

$J_n$ is the $n\times n$ matrix of ones

$I$ is the $n\times n$ identity matrix

I feel like there might be something simple here, though I'm not really sure how to approach determinants of sums. I'm a first year undergrad, FWIW.

Hints/help is much appreciated!

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The easiest way to see this is by changing the basis and noting that this leaves the determinant unchanged. I'll do it for $c=1$, you can do it analogously for other $c$. By inspection the eigenvalues are $n$ with multiplicity $1$, and $0$ with multiplicity $n-1$, with corresponding eigenvectors $(1,1,...,1)$ and $(1,-1,0,...,0)$, $(1,0,-1,0,...0)$,...,$(1,0,...,0,-1)$, respectively. Note that these eigenvectors form a basis for the vector space. The identity matrix looks the same in any basis, so diagonalize $J_n$, so that it has $n$ in the top-left corner, and is zero everywhere else. Then $$J_n+I=\begin{bmatrix} n+1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0 \\ \vdots & 0 & \ddots &\vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}$$ and so the determinant is $n+1$.

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Thanks JLA, this is a nice solution, and best of all, it made me learn something new! I knew what eigenvalues and eigenvectors were, but now I know what it means to diagonalise a matrix. – kurtgirdle Jun 6 '13 at 5:54

This statement is true!

We will prove this using expansion by minors, inducting on $n$.

For $n = 1$, $\det(c + 1) = c+1$ is trivial.

Now for $n+1$, take $\det ( c J_{n+1} + I )$ and use expansion by minors across the first row.

The first number is $c+1$, while the rest of the numbers are $c$. So:

$$\det(c J_{n+1} + I) = ( c+1 ) \det( c J_n + I ) + c \sum_{i=2}^{n+1} (-1)^{i+1} \det(A_{i-1})$$

$A_i$ is an $n \times n$ matrix with its $i$-th row consisting entirely of $c$ while all other rows look like:

$$\left( \begin{array}{ccccc} c & c & c+1 & \dots & c \end{array} \right)$$

By subtracting the first row from all others, we get a matrix that looks like:

$$\left( \begin{array}{ccc} c & c+1 & c \\ c & c & c \\ c & c & c+1 \end{array} \right) \to \left( \begin{array}{ccc} 0 & 1 & 0 \\ c & c & c \\ 0 & 0 & 1 \end{array} \right)$$

Factor out $c$ and make $A$ into a permutation matrix by subtracting the rows consisting of one $1$ from the $i$-th row (addition and subtraction of rows does not change the determinant). Our permutation matrix will correspond to the permutation $( 123 \dots i)$. So $\det(A_i) = (-1)^i c$ (follows by induction and the fact that swapping two rows reverses the sign on the determinant).

So our determinant is

$$( c + 1 ) ( cn + 1 ) + c \sum_{i=2}^{n+1} (-1)^{i+1} (-1)^i c = c^2n + c( n+1 ) + 1 - c^2n = c(n+1) + 1$$

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This is an excellent question. To me, the best approach seems to be induction. Taking the base case as $n = 2$, this is clearly true for a $2 \times 2$ matrix, as we can see:

$$\det\begin{bmatrix} c+1 & c\\ c & c+1 \end{bmatrix} = (c+1)^{2} - c^{2} = 2c +1$$

However, we can write this more suggestively. Note that applying elementary row operations to any matrix does not change its determinant! Hence, we can see, subtracting the bottom row from the top:

$$\det\begin{bmatrix} c+1 & c\\ c & c+1 \end{bmatrix} = \det\begin{bmatrix} 1 & -1\\ c & c+1 \end{bmatrix} = (c+1)\cdot 1 - c\cdot(-1) = 2c +1$$

Now assume that this is true for an $n \times n$ matrix. We want to show it is true for an $(n+1) \times (n+1)$ matrix. Writing it out, our determinant of our matrix looks like

$$\det\begin{bmatrix} c+1 & c & \cdots & c\\ c & c+1 & \cdots & c \\ \vdots & c & \ddots &\vdots \\ c & c & \cdots & c+1 \end{bmatrix} = \det\begin{bmatrix} 1 & -1 & 0 & \cdots & 0\\ c & c+1 & c & \cdots & c \\ 0 & -1 & 1 & \cdots & 0 \\ \vdots & \vdots & 0 & \ddots &\vdots \\ 0 & -1 & \cdots & 0 & 1 \end{bmatrix}$$

We're so close! With a little work and our induction hypothesis, we can show that this is indeed $$(cn+1)\cdot 1 - c\cdot(-1) = c(n+1) +1$$

To do so, we need to see that:

$$\det\begin{bmatrix} c & c & c & \cdots & c \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ \end{bmatrix} = c$$

But this isn't hard to show with expansion along a smart row choice.

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Ahh, fantastic. I was going down this route but got a bit drowned in the details. Just to clarify, in the step where you say 'We're so close!', do you get $(cn+1)$ because the matrix you obtain by removing row 1 and column 1 is our $n\times n$ matrix after some row operations? – kurtgirdle Jun 6 '13 at 5:16
@Rife168 Absolutely! You're 100% correct. :) – Alex Wertheim Jun 6 '13 at 5:17