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# Vieta's formula

## Problem

Let

$$
A = \begin{bmatrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
7 & 8 & 9
\end{bmatrix}
$$

and $\lambda_1, \lambda_2, \lambda_3$ its eigenvalues. Find the values of

$$
\begin{aligned}
a &= \lambda_1 + \lambda_2 + \lambda_3, \\
b &= \lambda_1\lambda_2 + \lambda_1\lambda_3 + \lambda_2\lambda_3, \\
c &= \lambda_1\lambda_2\lambda_3.
\end{aligned}
$$

## Thought

For any $n\times n$ matrix $A$, its eigenvalues $\lambda_1, \ldots, \lambda_n$ are the roots of its characteristic polynomial $p_A(x) = \det(A - xI)$. Therefore, we have

$$
\begin{aligned}
p_A(x) &= (-x)^n + s_1(-x)^{n-1} + \cdots + s_n \\
&= (\lambda_1 - x)(\lambda_2 - x) \cdots (\lambda_n - x),
\end{aligned}
$$

where

$$
s_k = \sum_{\substack{\alpha \subseteq [n] \\ |\alpha| = k}}\det(A[\alpha]).
$$

Note that it is $(\lambda_i - x)$ instead of $(x - \lambda_i)$ since the leading term of $p_A(x)$ is $(-x)^n$.

By comparing the coefficients (also known as the [Vieta's formula](https://en.wikipedia.org/wiki/Vieta%27s_formulas), we have

$$
\begin{aligned}
s_1 &= \lambda_1 + \lambda_2 + \cdots + \lambda_n, \\
s_2 &= \lambda_1\lambda_2 + \lambda_1\lambda_3 + \cdots + \lambda_{n-1}\lambda_n, \\
\vdots &= \\
s_n &= \lambda_1\lambda_2\cdot\cdots\cdot\lambda_n.
\end{aligned}
$$

## Sample answer

By comparing the coefficients, we know that $a = \lambda_1 + \lambda_2 + \lambda_3 = s_1$, which is the trace of the matrix. Thus, $a = \tr(A) = 1 + 5 + 9 = 15$.

Next, $b = \lambda_1\lambda_2 + \lambda_1\lambda_3 + \lambda_2\lambda_3 = s_2$. All the $\alpha\subseteq [3]$ of order $2$ are $\{1,2\}$, $\{1,3\}$, and $\{2,3\}$. The corresponding $A[\alpha]$ are

$$
\begin{bmatrix}
1 & 2 \\
4 & 5
\end{bmatrix},
\begin{bmatrix}
1 & 3 \\
7 & 9
\end{bmatrix},
\begin{bmatrix}
5 & 6 \\
8 & 9
\end{bmatrix}
$$

with their determinans $-3$, $-12$, and $-3$. Thus, $b = s_2 = -3 -12 -3 = -18$.

Lastly, $c = \lambda_1\lambda_2\lambda_3 = s_3$, which is the determinant of $A$. Since $A$ is singular, $c = \det(A) = 0$.

*This note can be found at Course website > Learning resources.*