Matrix and its characteristic polynomial

Question

Let $A = \begin{bmatrix} a & b \\ c & d\end{bmatrix}$. Prove that the characteristic polynomial of A can be written as $p(\lambda) = \lambda^2 − trace(A)\lambda + det(A)$ and show that $A$ satisfies its own characteristic equation $p(A) = 0_{2 \times 2}$.

I had a hard time following the lecture.

Answer 1

HINTS

Let $A$ be a square matrix. The characteristic polynomial of $A$ is defined as $$ p(\lambda) = \det(A- \lambda I), \quad \forall \lambda \in \mathbb{R}. $$ It is useful, among other things, in determining eigenvalues, since $\lambda$ is an eigenvalue of $A$ if and only if it is a root of $p$.

For the specific case of a $2 \times 2$ matrix, let $$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$ then $$ p(\lambda) = \det(A- \lambda I) = \begin{vmatrix} a - \lambda & b \\ c & d - \lambda \end{vmatrix} $$
Now, you can

• compute $\det A$ and $\mathrm{tr}\ A$ in terms of $a,b,c,d$
• compute the actual expression for $p(\lambda)$
• observe that $p(\lambda) = \lambda^2 - \mathrm{tr}(A) + \det(A)$ as requested
• plug in $A$ into $p$ and compute, observing that $p(A)$ is the zero matrix.

The final observation is true not just for a $2 \times 2$ matrix, but for any square matrix, and is called the Cayley-Hamilton Theorem.