Welcome to our community

Be a part of something great, join today!

Formula of determinant

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
Hey!!

Let $\mathbb{K}$ be a field and let $1\leq n\in \mathbb{N}$. Let $a_0, \ldots , a_{n-1}\in \mathbb{K}$ and let $m_n\in M_n(\mathbb{K})$ be given by \begin{equation*}m_n:=\begin{pmatrix}0 & 0 & \ldots & 0 & -a_0 \\ 1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & 0 & \vdots \\ 0 & \ldots & 0 & 1 & -a_{n-1}\end{pmatrix}\end{equation*}

I want to show that for $\lambda\in \mathbb{K}$ it holds that $\displaystyle{\det (\lambda u_n-m_n) =\lambda^n+\sum_{i=0}^{n-1}a_i \lambda^i}$.


First I applied the Laplace theorem to show that the formula bust be this one:
\begin{equation*}\det (\lambda u_n-m_n)=\begin{vmatrix}\lambda & 0 & \ldots & 0 & a_0 \\ -1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \lambda & \vdots \\ 0 & \ldots & 0 & -1 & \lambda+a_{n-1}\end{vmatrix}\end{equation*}
Laplace as for the first row:
\begin{equation*}\det (\lambda u_n-m_n)=\begin{vmatrix}\lambda & 0 & \ldots & 0 & a_0 \\ -1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \lambda & \vdots \\ 0 & \ldots & 0 & -1 & \lambda+a_{n-1}\end{vmatrix}=\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^{n+1}a_0\begin{vmatrix}-1 & \lambda & \ddots & \vdots \\ 0 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \lambda \\ 0 & \ldots & 0 & -1 \end{vmatrix}\end{equation*}
The matrix $\begin{pmatrix}-1 & \lambda & \ddots & \vdots \\ 0 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \lambda \\ 0 & \ldots & 0 & -1 \end{pmatrix}$ is a triangular matrix, so the determiannt is the product of the diagonal elements, i.e. $(-1)^{n-1}$.

So we get \begin{align*}\det (\lambda u_n-m_n)&=\begin{vmatrix}\lambda & 0 & \ldots & 0 & a_0 \\ -1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \lambda & \vdots \\ 0 & \ldots & 0 & -1 & \lambda+a_{n-1}\end{vmatrix} =\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^{n+1}a_0(-1)^{n-1}\\ & =\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^{2n}a_0 =\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+a_0\end{align*}
We expand again as for the first row:
\begin{align*}\det (\lambda u_n-m_n)&=\lambda \left (\lambda \begin{vmatrix} \lambda & 0 & a_2 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^na_1\begin{vmatrix} -1 & \ddots & 0 \\ \ddots & \ddots & \lambda \\ \ldots & 0 & -1 \end{vmatrix}\right )+a_0\\ & =\lambda \left (\lambda \begin{vmatrix} \lambda & 0 & a_2 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^na_1(-1)^{n-2}\right )+a_0 \\ & =\lambda \left (\lambda \begin{vmatrix} \lambda & 0 & a_2 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^{2n-2}a_1\right )+a_0 \\ & =\lambda^2 \begin{vmatrix} \lambda & 0 & a_2 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+ a_1 \lambda+a_0\end{align*}
We expand again as for the first row:
\begin{align*}\det (\lambda u_n-m_n)&=\lambda^2 \left (\lambda \begin{vmatrix} \lambda & 0 & a_3 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^{n-1}a_2\begin{vmatrix} -1 & \ddots & 0 \\ \ddots & \ddots & \lambda \\ \ldots & 0 & -1 \end{vmatrix}\right )+a_1\lambda +a_0\\ & =\lambda^2 \left (\lambda \begin{vmatrix} \lambda & 0 & a_3 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^{n-1}a_2(-1)^{n-3}\right )+a_1\lambda +a_0 \\ & =\lambda^2 \left (\lambda \begin{vmatrix} \lambda & 0 & a_3 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+(-1)^{2n-4}a_2\right )+a_1\lambda +a_0 \\ & =\lambda^3 \begin{vmatrix} \lambda & 0 & a_3 \\ \ddots & \lambda & \vdots \\ 0 & -1 & \lambda+a_{n-1}\end{vmatrix}+a_2\lambda^2+a_1\lambda +a_0\end{align*}
If we aply the Laplace explansion several times, we see that the formula is:
\begin{equation*}\det (\lambda u_n-m_n) =\lambda^n+\sum_{i=0}^{n-1}a_i \lambda^i\end{equation*}

Now we show that this formula is true using induction. (Is the way I do that correct?)

At the base case we have $n=1$, or not? But how is the matrix for $n=1$ ? Do we have that \begin{equation*}\det (\lambda u_1-m_1)=\begin{vmatrix}\lambda \end{vmatrix}=\lambda\end{equation*} ? But then the formula wouldn't hold, would it?

:unsure:
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,679
At the base case we have $n=1$, or not? But how is the matrix for $n=1$ ? Do we have that \begin{equation*}\det (\lambda u_1-m_1)=\begin{vmatrix}\lambda \end{vmatrix}=\lambda\end{equation*} ? But then the formula wouldn't hold, would it?
The question says "let $1\leq n\in \mathbb{N}$", but the matrix $m_n$ does not appear to be defined when $n=1$. In fact, $a_1$ has only one entry, which has to be equal to both $0$ and $-a_0$. I think that the question should have said "let $2\leq n\in \mathbb{N}$" and that you should take the base case to be $n=2$.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
The question says "let $1\leq n\in \mathbb{N}$", but the matrix $m_n$ does not appear to be defined when $n=1$. In fact, $a_1$ has only one entry, which has to be equal to both $0$ and $-a_0$. I think that the question should have said "let $2\leq n\in \mathbb{N}$" and that you should take the base case to be $n=2$.
Ahh ok! So we have the following:

Base Case: For $n=2$ we have that \begin{equation*}\det (\lambda u_2-m_2)=\begin{vmatrix}\lambda & a_0\\ -1 & \lambda +a_1 \end{vmatrix}=\lambda(\lambda+a_1)-(-1)a_0=\lambda^2+a_1 \lambda+a_0=\lambda^2+\sum_{i=0}^{1}a_i \lambda^i\end{equation*} and so the formula holds. .

Inductive hypothesis: We assume that the formula holds for $n=k$, i.e. \begin{equation*}\det (\lambda u_k-m_k) =\lambda^k+\sum_{i=0}^{k-1}a_i \lambda^i, \ \ \ \ \ (IH)\end{equation*}

Inductive step: We want to show that the formula hodls also for $n=k+1$, i.e. \begin{equation*}\det (\lambda u_{k+1}-m_{k+1}) =\lambda^{k+1}+\sum_{i=0}^{(k+1)-1}a_i=\lambda^{k+1}+\sum_{i=0}^{k}a_i \lambda^i\end{equation*}
We have that
\begin{equation*}\det (\lambda u_{k+1}-m_{k+1})=\begin{vmatrix}\lambda & 0 & \ldots & 0 & a_0 \\ -1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \lambda & \vdots \\ 0 & \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}\end{equation*}
Do we expand as for the first row? Then we would get the following:
\begin{equation*}\det (\lambda u_{k+1}-m_{k+1})=\begin{vmatrix}\lambda & 0 & \ldots & 0 & a_0 \\ -1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \lambda & \vdots \\ 0 & \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}=\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}+(-1)^{k+2}a_0\begin{vmatrix}-1 & \lambda & \ddots & \vdots \\ 0 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \lambda \\ 0 & \ldots & 0 & -1 \end{vmatrix}\end{equation*}

Now the matrix $\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}$ is a $k\times k$ matrix, but is it of the form$\lambda u_{k}-m_{k}$. so that we can use $(IH)$ ? I ask because at the first row we have now $a_1$ instead of $a_0$. Or do we have to amend the formula of (IH) accordingly?

:unsure:
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,679
The inductive hypothesis should say that the formula holds for every element $(a_0,\ldots,a_{n-1})\in\Bbb{K}^n$. In the inductive argument you start with a $(k+1)\times(k+1)$ matrix whose last column is $(a_0,\ldots,a_{k})\in\Bbb{K}^{k+1}$, and you end up with a $k\times k$ matrix whose last column is $(a_1,\ldots,a_k)$, which is an element of $\Bbb{K}^k$. That is in accordance with the inductive hypothesis. It doesn't matter how the coordinates of that element are labelled.
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,008
The inductive hypothesis should say that the formula holds for every element $(a_0,\ldots,a_{n-1})\in\Bbb{K}^n$. In the inductive argument you start with a $(k+1)\times(k+1)$ matrix whose last column is $(a_0,\ldots,a_{k})\in\Bbb{K}^{k+1}$, and you end up with a $k\times k$ matrix whose last column is $(a_1,\ldots,a_k)$, which is an element of $\Bbb{K}^k$. That is in accordance with the inductive hypothesis. It doesn't matter how the coordinates of that element are labelled.
Ah ok! So at the inductive step we have the following:
\begin{equation*}\det (\lambda u_{k+1}-m_{k+1})=\begin{vmatrix}\lambda & 0 & \ldots & 0 & a_0 \\ -1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \lambda & \vdots \\ 0 & \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}\end{equation*}
We expand as for the first row:
\begin{equation*}\det (\lambda u_{k+1}-m_{k+1})=\begin{vmatrix}\lambda & 0 & \ldots & 0 & a_0 \\ -1 & \ddots & \ddots & \vdots & \vdots \\ 0 & \ddots & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \lambda & \vdots \\ 0 & \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}=\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}+(-1)^{k+2}a_0\begin{vmatrix}-1 & \lambda & \ddots & \vdots \\ 0 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \lambda \\ 0 & \ldots & 0 & -1 \end{vmatrix}\end{equation*}
The matrix $\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}$ is a $k\times k$ matrix of the form $\lambda u_{k}-m_{k}$. From the $(IH)$ we get that determiannt is equal to $\displaystyle{\lambda^k+\sum_{i=0}^{k-1}a_{i+1} \lambda^i}$.

The matrix $\begin{vmatrix}-1 & \lambda & \ddots & \vdots \\ 0 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \lambda \\ 0 & \ldots & 0 & -1 \end{vmatrix}$ is triangular so the determinant is equal to the product of the diagnal elements, i.e. $(-1)^k$.

So we get:
\begin{align*}\det (\lambda u_{k+1}-m_{k+1})&=\lambda \begin{vmatrix} \lambda & 0 & \vdots & a_1 \\ -1 & \ddots & 0 & \vdots \\ \ddots & \ddots & \lambda & \vdots \\ \ldots & 0 & -1 & \lambda+a_{k}\end{vmatrix}+(-1)^{k+2}a_0\begin{vmatrix}-1 & \lambda & \ddots & \vdots \\ 0 & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \lambda \\ 0 & \ldots & 0 & -1 \end{vmatrix} \\ & =\lambda \left (\lambda^k+\sum_{i=0}^{k-1}a_{i+1} \lambda^i\right )+(-1)^{k+2}a_0(-1)^k=\lambda^{k+1}+\lambda\sum_{i=0}^{k-1}a_{i+1} \lambda^i+(-1)^{2k+2}a_0\\ & =\lambda^{k+1}+\lambda\sum_{i=0}^{k-1}a_{i+1} \lambda^i+a_0=\lambda^{k+1}+\sum_{i=0}^{k-1}a_{i+1} \lambda^{i+1}+a_0=\lambda^{k+1}+\sum_{i=1}^{k}a_{i} \lambda^{i}+a_0 \\ & =\lambda^{k+1}+\sum_{i=0}^{k}a_{i} \lambda^{i} \end{align*}


Is everything correct? Could we improve something? :unsure: