Exercise 4.3.24

Question

Let $A \in M_{n 	imes n} $ have the form 
$$\begin{pmatrix} 
0 & 0 & 0 & ... & 0 & a_0\ -1 & 0 & 0 & ... & 0 &a_1 \ 0 & -1 & 0 & ... &0 & a_2 \\vdots &\vdots &\vdots &...&\vdots&\vdots \ 0 & 0 & 0 & . . . & -1 & a_{n-1}
\end{pmatrix}$$
Compute $\det(A + tI)$, where $I$ is the $n 	imes n$ identity matrix.

Jephian C.-H. Lin · Accepted Answer

We use induction to claim that 
$$\det(A+tI)=t^n+\sum_{i=n-1}^{0}{a_it^i}.$$
For $n=1$, it's easy to see that $\det(A+tI)=t+a_0$. Suppose that the statement holds for $n=k-1$, consider the case for $n=k$. We can expand the matrix and get 
$$\det\begin{pmatrix}t&0&0&\cdots &0&a_0\-1&t&0&\cdots &0&a_1\0&-1&t&\cdots &0&a_2\\vdots &\vdots &\vdots & &\vdots &\vdots \0&0&0&\cdots &-1&a_{k-1}+t\end{pmatrix} $$
$$=t\det\begin{pmatrix}t&0&\cdots &0&a_1\-1&t&\cdots &0&a_2\\vdots &\vdots &\vdots & &\vdots &\vdots \0&0&\cdots &-1&a_{k-1}\end{pmatrix}+(-1)^{1+k}a_0\det\begin{pmatrix}-1&t&\cdots &0\0&t&\cdots &0\\vdots &\vdots &\vdots &\ddots &\vdots \0&0&\cdots &t\end{pmatrix}$$
$$=t(t^{k-1}+\sum_{i=k-2}^{0}{a_{i+1}t^i})+(-1)^{1+k}a_0(-1)^{k-1}$$
$$=t^n+\sum_{i=n-1}^{0}{a_it^i}.$$

Exercise 4.3.24

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Exercise 4.3.24

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