Does this solution guarantee $det(A)=0$ where $A\in M(R)$? [closed]

Question

Suppose $R$ is a commutative ring with identity $1$ and the following matrix equation holds:

$\begin{pmatrix} a_n & & \\ \vdots & \ddots & \\ a_1 & \cdots & a_n \end{pmatrix} \begin{pmatrix} b_n \\ \vdots \\ b_1 \end{pmatrix} =0$ where $a_i,b_j\in R$ and $b_1 = 1$.

Can we deduce that $det(A) = a_n^n = 0 $ ?

Note that $b_1 = 1$, and this assertion is true when $n=2$:

$\begin{pmatrix} a_2 & & \\ a_1 & a_2 \end{pmatrix} \begin{pmatrix} b_2 \\ 1 \end{pmatrix} =0$ gives

$a_2 = -a_1 b_2 \Rightarrow 0=a_2 b_2=-a_1b_2^2 \Rightarrow a_2^2=a_1 a_1b_2^2=a_1 0=0$ .

I can't prove or disprove whether it is true for all $n$.

Answer 1

You have $n$ equations

$$a_nb_n=0$$$$a_{n-1}b_n+a_nb_{n-1}=0$$$$a_{n-2}b_n+a_{n-1}b_{n-1}+a_nb_{n-2}=0$$ of which the $j+1$st is (for $0 \le j \le n-1$) $$a_{n-j}b_n+a_{n-j+1}b_{n-1}+\cdots a_nb_{n-j}=0 .$$

Prove by induction on $k\ge0$ that also $$a_n^{k+1}b_{n-k}=0.$$The case $k=n-1$ is what you want and the case $k=0$ is the first equation.

Suppose this has been shown up to $a_n^jb_{n-j+1}=0.$ Multiply the $j+1$st equation by $a_n^{j}$ to get $$a_{n-j}a_n^{j}b_n+a_{n-j+1}a_n^{j}b_{n-1}+\cdots +a_{n-1}a_n^jb_{n-j+1}+a_na_n^{j}b_{n-j}=0 .$$ All but the last term on the left hand side have already been shown to be zero leaving only $$a_n^{j+1}b_{n-j}=0.$$