# Why the $\lambda$-generalized eigenspace is invariant?

The following comes from a text book, I am very confused about the last sentence,

For a matrix $A$, a subspace $V$ is invariant w.r.t $A$ if $AV\subseteq V$. From my understanding we need to show $\forall x\in V_{\lambda_i},Ax\in V_{\lambda_i}$, i.e. $(A-\lambda_iI)^n(Ax)=0$.

• If $p$ is a polynomial, then $p(A)\cdot A=A\cdot p(A)$.
– user228113
Commented Oct 22, 2016 at 13:32

Indeed, you need to show $(A-\lambda_iI)^n(Ax)=0$. But this is easy since if $f(A)$ is a (analytic) function of operator $A$, then $[f(A),A]=0$, i.e. you have an operator identity $f(A) \cdot A = A \cdot f(A)$ and thus

$$(A-\lambda_iI)^n(Ax)=\left[(A-\lambda_iI)^n\cdot A\right]x=\left[A\cdot (A-\lambda_iI)^n\right]x=A\left((A-\lambda_iI)^nx\right)=0$$

In this problem, just by using the property that every operator commutes with $I$ and itself, you can show \begin{align*} (A-\lambda_iI)^n\cdot A &= \sum _{k=0}^n\binom{n}{k} A^k (-\lambda _i I)^{n-k} \cdot A=\sum _{k=0}^n \binom{n}{k} A^k \left(A\cdot (-\lambda _i I)^{n-k} \right) \\ &= \sum _{k=0}^n \binom{n}{k}\left(A^k \cdot A\right) (-\lambda _i I)^{n-k}=\sum _{k=0}^n \binom{n}{k}\left(A\cdot A^k \right) (-\lambda _i I)^{n-k}\\ &=A\cdot \sum _{k=0}^n \binom{n}{k}\ A^k (-\lambda _i I)^{n-k} =A\cdot (A-\lambda_iI)^n \end{align*}

Consider $x \in V_{\lambda_i}$ and the following computation, noting that we've subtracted zero from the first expression and factored the second.

\begin{align*} (A - \lambda_{i}I)^n(Ax) &= (A - \lambda_{i}I)^n(Ax) - \lambda_{i}(A - \lambda_{i}I)^{n}x \\ &= (A - \lambda_{i}I)^n(A - \lambda_{i}I)x \\ &= (A - \lambda_{i}I)(A - \lambda_{i}I)^{n}x \\ &= 0 \end{align*}

So $A$ maps elements of $V_{\lambda_i}$ to itself.

• Your proof is so elegant and refined, sir Commented Dec 2, 2023 at 13:09