$A$ is nilpotent, then $I+\lambda A$ is invertible for any $\lambda \in \mathbb{R}$

Question

I need help in this problem
Let $A$ is a square real matrix such that $A^{n}=0$ for some positive integer $n .$ Such a matrix is called nilpotent. Show that if $A$ is nilpotent, then $I+\lambda A$ is invertible for any $\lambda \in \mathbb{R}$

I know that for $\lambda = 1$ it is true by
$$\left(A+I\right)\left(I-A+A^2-...+(-1)^n A^{n-1}\right) = I +(-1)^{n-1} A^n = I$$ Thus proving that $A+I$ is invertible for any nilpotent $A$.

But I don't know if this is true for any $\lambda$.

Answer 1

We have $(I+\lambda A)(I-\lambda A+\lambda^2A^2-...)=I$.

Answer 2

Hint: Since you already know if $A$ is nilpotent, then $I+A$ is a unit, you may observe that $\lambda A$ is nilpotent and conclude.

Answer 3

You can also show it by contradiction as follows:

For $\lambda = 0$ there is nothing to show.

So, let's $\lambda \neq 0$ and assume that $I+\lambda A$ is not invertible. $$\Rightarrow \det (I+\lambda A)=0 \Rightarrow \det \left(A - \mu I\right)=0 \text{ with } \mu = -\frac 1{\lambda} \neq 0$$

So, $\mu\neq 0$ would be an eigenvalue of $A$. Since $A$ is nilpotent, we would have $\mu = 0$, which is a contradiction.

Answer 4

I'm assuming $A$ is over $\mathbb{R}$ , with nilpotent index $k$

in case $\lambda = 0$ : $(I - \lambda A) = I$ and is invertible.
in case $\lambda \not= 0$ : we will use this next lemma

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lemma: if $B$ is a real nilpotent matrix, $B$ has no real non-zero eigenvalues.
proof:

let us assume $B$ is nilpotent of index $r$.
also we assume it does have a real eigenvalue - $\xi$

by definition, it means there exists a non-zero vector $v$ such that-

$Bv = \xi v$

so, $B^2v = B(B(v)) = B(\xi v) = \xi B(v) = \xi^2 v$
in general, it isn't hard to show by induction that
$\forall n \in \mathbb{N} : B^n (v) = \xi ^n v$

with that, we get : $B ^r (v) = \xi ^r v$
but also $B ^r (v) = O (v) = \underline0$

so - $\xi ^r v = \underline0$.
but $v \not= \underline0$ , so we must get $\xi ^r = 0$ - which overall means that :
$\xi = 0$
$Q.E.D$

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Now, apply this lemma on $A$, and get that $A$ has no non-zero eigenvalues.

let us examine the matrix $I - \lambda A$ , remembering that here $\lambda \not= 0$.

This matrix is invertible, if and only if it has a non-zero determinant.
Let us calculate its determinant:

$det(I - \lambda A) = \lambda det(\lambda ^{-1}I - A)$

assume this equals $0$ and get:
$\lambda det(\lambda ^{-1}I - A) = 0$
$det(\lambda ^{-1}I - A) = 0$

and so - $\lambda ^{-1}$ is an eigenvalue of $A$, which is non-zero. Contrediction.

therefore, $\lambda det(\lambda ^{-1}I - A) \not= 0$ and so -
$det(I - \lambda A) \not = 0$ and the matrix is invertible.