Show that $Tr(A)=(-1)^{n-1} a_{n-1}$

Question

Let $A$ be an $n \times n$ matrix with characteristic polynomial: $$\phi_A(t)=(-1)^n t^n + a_{n-1} t^{n-1} + \cdot \cdot \cdot +a_1 t + a_0$$

Show that $Tr(A)=(-1)^n a_{n-1}$

I think I have seen similar problems like this one solved here but they are not focused on the trace of $A$ but on other properties of the characteristic polynomial of $A$

My attempt:

By the Cayley-Hamilton Theorem we have: $\phi_A(A)=0$, so:

$$(-1)^n A^n + a_{n-1} A^{n-1} + \cdot \cdot \cdot +a_1 A + a_0 I = 0$$

If we get the trace of both sides of the last expression and use the linearity of the trace then:

$$(-1)^n Tr(A^n) + a_{n-1} Tr(A^{n-1}) + \cdot \cdot \cdot +a_1 Tr(A) + n a_0 = 0$$

I'm not sure how to proceed next. Could someone give me a hint?

Answer 1

Triangularize your matrix in a field extension $\mathbb{L}$, you will easily find that $\text{tr} = (-1)^n a_{n-1}$ since neither of the characteristic polynomial and the trace depend on the field extension.

Answer 2

I suppose that the characteristic polynomial of $A$ is defined to be the determinant of $A - tI$, where $t$ is the indeterminate. I suppose that the trace of $A$ is defined to be the sum of the diagonal entries of $A$.

I do not know how you define the determinant of a square matrix. However, I will give a solution that is independent of the definitions.

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Before solving your question, I need a formula, which you might have seen.

Let $A$ be an $n \times n$ matrix. Suppose that the $(i, j)$-entry of $A$ is $[A]_{i,j}$. Suppose that column $j$ of the $n \times n$ identity matrix $I$ is $e_j$. Suppose that column $j$ of $A$ is $\alpha_j$ (which means that $A = [\alpha_1, \alpha_2, \dots, \alpha_n]$.) One can write $$ \alpha_{k} = [A]_{1,k} e_{1} + [A]_{2,k} e_{2} + \dots + [A]_{n,k} e_{n} = \sum_{i_k = 1}^{n} {[A]_{i_k,k} e_{i_k}}. $$ Hence, by the multilinear property, $$ \begin{align*} \det {(A)} = {} & \det {[\alpha_1, \alpha_2, \dots, \alpha_n]} \\ = {} & \det {\left[ \sum_{i_1 = 1}^{n} {[A]_{i_1,1} e_{i_1}, \alpha_2, \dots, \alpha_n} \right]} \\ = {} & \sum_{i_1 = 1}^{n} {[A]_{i_1,1}\, \det {[ e_{i_1}, \alpha_2, \dots, \alpha_n ]}} \\ = {} & \sum_{i_1 = 1}^{n} {[A]_{i_1,1}\, \det {\left[ e_{i_1}, \sum_{i_2 = 1}^{n} [A]_{i_2,2} e_{i_2}, \dots, \alpha_n \right]}} \\ = {} & \sum_{i_1 = 1}^{n} { \sum_{i_2 = 1}^{n} {[A]_{i_1,1} [A]_{i_2,2}\, \det {[ e_{i_1}, e_{i_2}, \dots, \alpha_n ]}}} \\ = {} & \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \\ = {} & \sum_{i_1 = 1}^{n} { \sum_{i_2 = 1}^{n} { \dots \sum_{i_n = 1}^{n} { [A]_{i_1,1} [A]_{i_2,2} \dots [A]_{i_n,n} \det {[e_{i_1}, e_{i_2}, \dots, e_{i_n}]}}}}. \end{align*} $$ Because of the alternating property, one can write $$ \boxed{ \det {(A)} = \sum_{\substack{ 1 \leq i_1, i_2, \dots, i_n \leq n \\ i_1, i_2, \dots, i_n\,\text{are distinct} }} {[A]_{i_1,1} [A]_{i_2,2} \dots [A]_{i_n,n} \det {[e_{i_1}, e_{i_2}, \dots, e_{i_n}]} }. } $$

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Okay. I am going to solve your question.

The characteristic polynomial of $A$ is $$ \det {(A - tI)} = \sum_{\substack{ 1 \leq i_1, i_2, \dots, i_n \leq n \\ i_1, i_2, \dots, i_n\,\text{are distinct} }} {[A - tI]_{i_1,1} [A - tI]_{i_2,2} \dots [A - tI]_{i_n,n} \det {[e_{i_1}, e_{i_2}, \dots, e_{i_n}]} }. $$ Your question amounts to finding the coefficient of $t^{n-1}$ (of the expansion). You are invited to prove, that among the $n!$ (unexpanded) terms, there is exactly one term which contains $t^{n-1}$, which is $$ \begin{align*} & [A - tI]_{1,1} [A - tI]_{2,2} \dots [A - tI]_{n,n} \det {[e_{1}, e_{2}, \dots, e_{n}]} \\ = {} & ([A]_{1,1} - t) ([A]_{2,2} - t) \dots ([A]_{n,n} - t) \cdot 1. \end{align*} $$ Hence the coefficient of $t^{n-1}$ is $$ (-1)^{n-1} ([A]_{1,1} + [A]_{2,2} + \dots + [A]_{n,n}). $$