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

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.

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

• Yes this does make sense, Thank you! However I should have included in my original post that the question requires that we do this without any use of a power series. I edited my post to include this. Is there another way to do this without the use of a power series? Oct 15, 2020 at 20:23
• It is the same basic idea. See my edits. Oct 15, 2020 at 20:44
• Awesome Thanks alot!! Oct 15, 2020 at 20:48
• Why is $Z'(t)= (A-\lambda I)Z(t)$. If I take the derivative, I get $$Z'(t)=(A-\lambda I)(X_0 +(A-\lambda I)X_0)$$ I do not have the $\frac{t^2}{2}(A-\lambda I)^2X_0$ term. What am I missing here? Oct 15, 2020 at 21:09
• What happens if you multiply that term by $(A- \lambda I)$??? Oct 15, 2020 at 21:34