Trace and determinant encode eigenvalue sum and product

For a $2\times 2$ matrix $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the characteristic polynomial is quadratic. Expanding $\det(A-\lambda I)=0$ gives a monic polynomial whose coefficients you can read from the matrix without hunting for roots .

Two symmetric facts drive the shortcut in this chapter. The sum of the eigenvalues $\lambda_1+\lambda_2$ equals the trace $a+d$. The product $\lambda_1\lambda_2$ equals the determinant $ad-bc$. These are the Vieta relations for a quadratic, packaged in linear-algebra language.

Example: $\begin{bmatrix}8&?\\?&6\end{bmatrix}$ has trace $14$, so the eigenvalue mean is $7$. If $\det=48-8=40$, the eigenvalues multiply to $40$. You have not solved yet, but you already know the pair must average to $7$ and multiply to $40$.

Geometric intuition: eigenvalues measure how much $A$ stretches along eigen-directions; trace tracks combined stretching along the coordinate axes in this $2\times 2$ case, while determinant tracks area scaling. That geometric story motivates why trace and determinant appear together in the characteristic polynomial .

Check your understanding. The tasks below rest on these ideas: Correct: $\lambda_1 + \lambda_2 = \mathrm{tr}(A)$. Not quite: the determinant is the product, and rank or a diagonal product are not the sum. Correct: $\lambda_1\lambda_2 = \det(A)$. Not quite: the trace is the sum, and the max entry or entry-sum are unrelated. Correct: a scalar matrix $3I$ scales every direction by $3$, so both eigenvalues are $3$. Not quite: $0$ is not an eigenvalue (the matrix is invertible), and $9,1$ ignore the diagonal. Correct: the monic quadratic has $-(\text{trace})$ as the linear coefficient and the determinant as the constant. Not quite: the sign of $t$, and swapping the roles of $t$ and $p$, give the wrong polynomial.