# Proving (non-)invertibility of (nearly) tri-diagonal matrix?

When discussing a transformation for cryptographic keys over on Crypto.SE I noticed that the transformation could be described using a matrix, so I wrote it up and ran some basic online-tools against it to confirm whether the function is a bijection or not and found (empirically) the following conjecture for which I'd like to ask for help in proving.

Let $\mathbb F_2^n$ be the $n$-dimensional vector space built on-top of $\mathbb F_2$, the field with the elements $0,1$ and the addition and multiplication modulo 2. Let $b\in\mathbb F_2$ (in the practical scenario the value of $b$ indicates whether we use a cyclical or a logical shift).

My conjecture is now that the following matrix is invertible iff $b=0$: \begin{pmatrix} 1&0&0&\cdots&0&b\\ 1&1&0&\cdots&0&0\\ 0&1&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&1&1 \end{pmatrix}

Textually the matrix is the all-zero matrix with the top-right entry holding the value $b$, the diagonal being all-ones and each entry below the diagonal (if it exists) being $1$ as well.

My question is now, can this above conjecture be (easily?) proven and if so how?

I have empirically verified the above conjecture for $n=2,3,4$ but have no idea how to prove it in all generality.

• Is it not possible for you to compute the determinant inductively? – Pedro Tamaroff Mar 1 '18 at 20:09
• @PedroTamaroff while I see the idea, I think I completely lack the mathematical toolkit to make a proof using induction over $n$ as this would seem to bring myriads of other difficult (?) matrices when using laplace's theorem (?) to compute the new determinant in the induction step :( (And Laplace's theorem (?) is the only one I know that would remotely support something like this) – SEJPM Mar 1 '18 at 20:43

Following up on the determinant comment, note that by expanding through the first row, where I omit signs since you are in $\mathbb F_2$:

$$\left| \begin{matrix} 1&0&0&\cdots&0&b\\ 1&1&0&\cdots&0&0\\ 0&1&1&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&1&1 \end{matrix} \right|= \left| \begin{matrix} 1&0&\cdots&0&0\\ 1&1&\cdots&0&0\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&1 \end{matrix}\right|+b \left|\begin{matrix} 1&1&0&\cdots&0\\ 0&1&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots&1 \end{matrix}\right| =1+b$$ since the two matrices to the right are upper and lower triangular. This means that your matrix has $0$ determinant if and only if $b=1$, as you wanted.

• Wow, this was way simpler than I expected. Particularly because I forgot that the $0$ will kill all but two terms in the laplace expansion. – SEJPM Mar 1 '18 at 21:08

When $b=1 \; \; : \; \;$Over the rational or real field, there is an eigenvalue of $2$ with eigenvector all entries equal to $1.$ However, over the field of two elements, the same eigenvector has eigenvalue $0.$

if you subtract off the identity matrix, you have a "companion matrix" for $x^n - b.$ Now that I think of it, $x^n - b$ is also the minimal polynomial for this revised matrix; whatever $b,$ we have a single Jordan block.

If you stick with the field of two elements, when $b=1$ the only eigenvalue is $1,$ so when you add $I$ back in the only eigenvalue is $0$

• An alternative (maybe simpler) formulation would be to call the above matrix $A$ and say that the characteristic polynomial of $A-I$ is $x^n-b$ per the linked article. Now this implies $\det((A-I)-xI)=\det(A-(x+1)I)=x^n-b$ and plugging in $x=-1$ yields $\det(A)=(-1)^n-b$ which is equal to $1-b$ over $\mathbb F_2$ and which is $0$ iff $b=1$ and $\det A=0$ iff A is not invertible. [This may be easier to understand than the more fancy way using eigenvalues] – SEJPM Mar 1 '18 at 20:40
• @SEJPM added: in field $\mathbb Z / 2 \mathbb Z$ there is an eigenvector with all entries equal to $1$ but eigenvalue $0.$ – Will Jagy Mar 1 '18 at 20:48