5. More on the characteristic polynomial

For the $2 \times 2$ case, let's see what happens if we ``precompute'' the characteristic polynomial using letters, by writing $A = \matp{cc}{a&c\\ b&d}$:

$A - \lambda I = \matp{cc}{a&c\\ b&d}- \matp{cc}{\lambda&0\\ 0&\lambda}= \matp{cc}{(a- \lambda)&c\\ b& (d - \lambda)}$, so $\det(\lambda I - A) = (a - \lambda)(d - \lambda) - bc = \lambda ^ 2 - (a+d) \lambda + (ad-bc)$.

Notice that the constant term is $\det A$. The coefficient of $\lambda$ is $-(a+d)$. The trace of a matrix means the sum of the diagonal entries, so the coefficient of $\lambda$ is minus the trace.

Now you can see this fact:

Observation. If $A$ is $2 \times 2$, then $p _ A(\lambda) = \lambda ^ 2 -$   trace$(A) \lambda + \det A$.

Now it becomes easy to write down the characteristic polynomials in the examples from the early sections of this handout:

For $A = \matp{rr}{4&1\\ 1&4}$ we get $p _ A(\lambda) = \lambda ^ 2 -$   trace$(A) \lambda + \det A = \lambda ^ 2 - 8 \lambda + 15$.

For $A = \matp{rr}{5&0\\ 0&3}$ we get $p _ A(\lambda) = \lambda ^ 2 -$   trace$(A) \lambda + \det A = \lambda ^ 2 - 8 \lambda + 15$.

For $A = \matp{rr}{7&-4\\ 2&1}$ we get $p _ A(\lambda) = \lambda ^ 2 -$   trace$(A) \lambda + \det A = \lambda ^ 2 - 8 \lambda + 15$.

This explains why all three examples had the same eigenvalues: All three had the same characteristic polynomial.

Problem U-7. Invent two more $2 \times 2$ matrices with characteristic polynomial $\lambda ^ 2 - 8 \lambda + 15$.