Ytukyg

Show that if $(A+2I)^2=0$, then $A+lambda I$ is invertible for $lambda ne 2$.

I tried to solve this by treating $(A+lambda I)v=0$ as linear equation system, and proving that $v$ must be $0$ (trivial solution) therefore $A+lambda I$ echelon form is I and it's invertible..

Would love to hear another solutions, and tips to my own proof:

$$A+lambda I=A+2I+(lambda-2) I.$$

$$(A+lambda I)v=0 Rightarrow (A+2I)v+(lambda-2)Iv=0.$$

$$(A+2I)(A+2I)v+(A+2I)(lambda-2)Iv=0 Rightarrow (lambda-2)(A+2I)Iv=0 Rightarrow (A+2I)Iv=0$$ (multiply by $A+2I$ and we know $lambda ne 2$).

$$(A+2I)v+(lambda-2)Iv=0 wedge (A+2I)Iv=0Rightarrow (λ−2)Iv=0 Rightarrow v=0$$

Therefore if v is solution for $(A+lambda I)v=0$ it must be 0.

(We have also shown that for $lambda = 2$, $v$ can be $(A+2I)$ which shows $A+2I$ isn't invertible)

BTW If A is such that $(A+2I)^2=0$, prove that $A+I$ is invertible. solves the basic case for $lambda = 1$

Below is a more general statement. I have provided three solutions to your question. The first two use the proposition, and the last one is as you requested.

Proposition. Let $n$ be positive integer and $A$ an $n$-by-$n$ matrix over a field $mathbb{K}$. Write $I$ for the $n$-by-$n$ identity matrix. For $muin K$, the matrix $A-mu, I$ is not invertible if and only if $mu$ is an eigenvalue of $A$ (or equivalently, $det(A-mu, I)=0$, or $ker(A-mu,I)neq {0}$).

Proof. For each $muin mathbb{K}$, let $V_muin mathbb{K}^n$ denote the kernel (i.e., the nullspace) of $A-mu,I$. If $A-mu,I$ is invertible, then for any $vin V_mu$, we have $(A-mu,I),v=0$ and so $$v=(A-mu,I)^{-1}(0)=0,,$$ implying that $V_mu={0}$. Conversely, if $V_mu={0}$, then $A-mu,I$ is an injective linear map from $mathbb{K}^n$ to itself. It is well known that any injective linear map on a finite-dimensional vector space is also surjective, whence bijective and so invertible. (This well known result is not true for infinite-dimensional vector spaces, by the way.) Therefore, $A-mu,I$ is invertible.

First Solution.

In your case, the only eigenvalue of $A$ is $-2$. Thus, $A-mu,I$ is invertible if and only if $muneq -2$, which is equivalent to saying that $A+lambda, I$ is invertible if and only if $lambdaneq 2$.

Second Solution.

We shall prove that $ker(A+lambda,I)={0}$ when $lambdaneq 2$. Suppose $vin ker(A+lambda,I)$. Then, $(A+lambda,I),v=0$, so $Av=-lambda,v$ and $A^2v=A(Av)=A(-lambda,v)=-lambda,(Av)=-lambda,(-lambda v)=lambda^2,v$. Since $(A+2,I)^2=0$, we also have $(A+2,I)^2,v=0$. Therefore, $$lambda^2,v-4,lambda,v+4,v=A^2v+4,Av,+4,v=0,.$$
Ergo, $(lambda-2)^2,v=0$. Since $lambdaneq 2$, $v=0$, which implies $ker(A+lambda ,I)={0}$.

Third Solution.

Since $lambda neq 2$, we have
$$frac{(x+2)^2-(x+4-lambda)(x+lambda)}{(lambda-2)^2}=1,,$$
where $x$ is a dummy variable. Therefore, $$frac{(A+2,I)^2-big(A+(4-lambda),Ibig),(A+lambda , I)}{(lambda-2)^2}=I,.$$
As $(A+2,I)^2=0$, we get
$$left(-frac{1}{(lambda-2)^2},big(A+(4-lambda),Ibig)right),(A+lambda,I)=I,.$$ Thence, $A+lambda,I$ is invertible and
$$(A+lambda,I)^{-1}=-frac{1}{(lambda-2)^2},big(A+(4-lambda),Ibig),.$$

Have you heard about eigenvalues? YOur equation for $A$ shows that $-2$ is the only eigenvalue of A(because any eigenvalue would satisfy $(x+2)^2=0$, the equation satisfied by $A$), so no other $lambda$ could be an eigenvalue for $A$, qed.