# Why can we substitute matrix and eigenvalues into other than the characteristic polynomial (Cayley Hamilton)

Generally speaking, this is a question about when we can swap $$A \rightarrow\lambda$$.

In problem 2-5(c) of Applied Optimal Estimation, we are asked to consider the matrix $$A = \left[\begin{matrix} 1 & 2 \\ 3 & 4\end{matrix}\right],$$ whose characteristic polynomial is $$p(\lambda) = \lambda^2-5\lambda-2.$$ We then Taylor expand $$e^{At}$$ and use Cayley-Hamilton to simplify, yielding: $$e^{At} = a_1(t)I + a_2(t)A.$$

Part (c) of the question is getting a closed form expression for $$a_1(t)$$ and $$a_2(t)$$. To do that, this solution pdf swaps out $$\lambda_1$$ and then $$\lambda_2$$ for $$A$$, which gives two equations for the two unknowns.

Why can one swap out $$\lambda_i$$ for $$A$$ in that expression, if it is not the characteristic polynomial? Does this hold in general, that while we can't necessarily swap $$A$$ for $$\lambda$$, we can always do the reverse?

You obtained that expression using only the fact that $$A$$ is a "zero" of the characteristic polynomial. Similarly $$\lambda_1$$ and $$\lambda_2$$ are zeros of the characteristic polynomial.

If you are still not convinced, just expand $$e^{\lambda_1 t}$$ and use what you know about $$\lambda_1$$ to simplify in exactly the same manner as you did with $$A$$.

• Ah, I see. They aren't using some property I don't understand about Cayley-Hamilton or polynomial division, it's just that (obviously) you can use the CP to substitute expressions for $\lambda$, just as we can use it for $A$. Sep 28, 2021 at 21:57