Showing $Y(t)$ is a solution to $X'=AX$

Question

I am struggling with the problem:

Let $A$ be a $n\times n$ matrix with eigenvalue $\lambda$ with multiplicity 3 and suppose the associated eigen vector $X_0\in N((A-\lambda I)^3)$ (Nullspace). Show that $$Y(t)=e^{\lambda t}\bigg[I+t(A-\lambda I)+\frac{t^2(A-\lambda I)^2}{2!}\bigg]X_0$$ is a solution to $X'=AX$ (without the use of a power series).

My attempt: I know that for $Y(t)$ to be a solution, it must satisfy $$Y'(t)=AY(t).$$ Finding $Y'(t)$ we have $$Y'(t)=\lambda e^{\lambda t}\bigg[I+t(A-\lambda I)+\frac{t^2(A-\lambda I)^2}{2!}\bigg]X_0+e^{\lambda t}\bigg[I+A-\lambda I+t(A-\lambda I)^2\bigg]X_0$$ and expanding the right hand side we have $$AY(t)=e^{\lambda t}A\bigg[I+t(A-\lambda I)+\frac{t^2(A-\lambda I)^2}{2!}\bigg]X_0$$

I tried many things to get the LHS and RHS to equal. I know we have to use the fact that $(A-\lambda I)^3X_0=0$ somewhere, but i just cannot figure it out. Please if anyone can help out, it would be much appreciated.

Answer 1

Note that $Y(t) = e^{At} X_0 = e^{\lambda t} e^{(A-\lambda I)t} X_0$.

Now expand $e^{(A-\lambda I)t} = \sum_{k=0}^\infty {t^k \over k!} (A-\lambda I )^k X_0$.

Without power series is basically the same idea.

Let $Z(t) = (\sum_{k=0}^2 {t^k \over k!} (A -\lambda I)^k) X_0 $ and note that $Z'(t) = (A- \lambda I) Z(t)$.

Note that $Y(t) = e^{\lambda t} Z(t)$ and so $Y'(t) = e^{\lambda t} ( (A- \lambda I) + \lambda I ) Z(t) = A Y(t)$.