Show that last term of the characteristic polynomial of $A$ is $(-1)^n \det A$, and that the term of degree $(n-1)$ is $Tr (A)$.

Question

I was given the following problem:

Let $A \in \mathbb{C}^{n\times n}$. Show the last term of $p$ the characteristic polynomial of $A$ is $(-1)^n \det A$, and that the term of degree $(n-1)$ is $Tr (A)$.

I attempted a proof by induction. Particularly, we know this holds for $A \in \mathbb{C}^{2 \times 2}$. Assume it does for $A \in \mathbb{C}^{k-1 \times k-1}$. Now let $M = xI - A$ for $A \in \mathbb{C}^{k \times k}$. Then the characteristic polynomial of $A$ is given by

\begin{align*} \det M &= (x - a_{11}) \det \big( M(1\mid 1) \big) - a_{22} \det \big( M(2 \mid 1) \big) + \ldots + (-1)^{k + 1} a_{k1} \det\big( M (k \mid 1)\big) \end{align*}

Since $\mathbb{C}$ is algebraically closed, the characteristic polynomial of each $M(j\mid 1) \big), 1 \leq j \leq k$, exists and is of the form

\begin{align*} (x - c_1)\ldots(x-c_{k-1}) \end{align*}

where there might be repeated eigenvalues. Let $P_j$ denote the polynomial, of the above expressed form, pertaining to $M(j\mid 1)$. Then what we have is

\begin{align*} \det M = (x - a_{11})P_{1} - a_{22}P_2 + \ldots + (-1)^{k + 1}P_k \end{align*}

I was unable to continue from here. Other desperate approaches lead nowhere. How does one tackle this proof? Thanks in advance.

Answer 1

There are better methods; see, for example, this one, in which the characteristic polynomial is defined a little bit differently.

However, you can of course prove one of your claim by induction.

The main idea is to expand more.

I will give you some hints.

Suppose that $A$ is an $n \times n$ matrix. I will denote the $(n-2) \times (n-2)$ matrix obtained from $A$ by removing rows $i_1$, $i_2$ and columns $j_1$, $j_2$ by $A({i_1,i_2}|{j_1,j_2})$.

Remember that one can compute the determinant of a square matrix by expansion about column $1$; remember also that one can compute the determinant of a square matrix by expansion about row $1$. Hence $$ \begin{align*} & \det {(A)} \\ = {} & [A]_{1,1} \det {(A(1|1))} + \sum_{i = 2}^{n} { (-1)^{i+1} [A]_{i,1} \det {(A(i|1))} } \\ = {} & [A]_{1,1} \det {(A(1|1))} + \sum_{i = 2}^{n} { (-1)^{i+1} [A]_{i,1} \sum_{j = 2}^{n} { (-1)^{1+j-1} [A]_{1,j} \det {(A(i,1|1,j))} } } \\ = {} & [A]_{1,1} \det {(A(1|1))} + \sum_{i = 2}^{n} { \sum_{j = 2}^{n} { (-1)^{i+1} [A]_{i,1} (-1)^{1+j-1} [A]_{1,j} \det {(A(i,1|1,j))} } } \\ = {} & [A]_{1,1} \det {(A(1|1))} + \sum_{i, j = 2}^{n} { (-1)^{i+j-1} [A]_{i,1} [A]_{1,j} \det {(A(1,i|1,j))} }, \end{align*} $$ in which $[A]_{i,j}$ is the $(i, j)$-entry of $A$.

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Let $I_n$ be the $n \times n$ identity matrix. Then $$ \begin{align*} & \det {(xI_n - A)} \\ = {} & \hphantom{{} + {}} [xI_n - A]_{1,1} \det {(\color{red}{(xI_n - A)}(1|1))} & \\ & + \sum_{i, j = 2}^{n} { (-1)^{i+j-1} [xI_n - A]_{i,1} [xI_n - A]_{1,j} \det {(\color{red}{(xI_n - A)} (1,i|1,j))} } \\ = {} & \hphantom{{} + {}} (x - [A]_{1,1}) \det {(xI_{\color{red}{n-1}} - A(1|1))} & \\ & + \sum_{i, j = 2}^{n} { (-1)^{i+j-1} [A]_{i,1} [A]_{1,j} \det {(\color{red}{(xI_n - A)} (1,i|1,j))} }. \end{align*} $$ It is not hard to see that the determinant of $\color{red}{(xI_n - A)} (1,i|1,j) = \color{red}{(xI_{n-1} - A(1|1))}(i-1|j-1)$ is either zero or a polynomial (in $x$) of degree less than $n-1$ (it seems that you will have to prove this claim by induction before prove what the coefficient of $x^{n-1}$ is).

By the hypothesis, the coefficient of $x^{n-2}$ of $\det {(xI_{\color{red}{n-1}} - A(1|1))}$ is $\color{red}{-}([A]_{2,2} + \dots + [A]_{n,n})$ (you missed a minus sign). After doing some multiplication, you should be able to show that the coefficient of $x^{n-1}$ of $\det {(xI_n - A)}$ is $\color{red}{-}([A]_{1,1} + [A]_{2,2} + \dots + [A]_{n,n})$.

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Note that the constant term of $f(x) = a_0 + a_1 x + \dots + a_n x^n$ is $a_0 = f(0)$. Hence the constant term of $\det {(xI_n - A)}$ is $\det {(-A)}$, which is $(-1)^n \det {(A)}$.