1
$\begingroup$

Question :

Suppose $T : \mathbb R^n \rightarrow \mathbb R^n$ is a linear transformation and $T$ has non zeroes distinct eigen values. Then which of the following is not necessarily true ?

(1) There exist $\lambda \in \mathbb R$ such that $T + \lambda I_n$ is invertible .

(2) $T$ is invertible.

(3) There exist $\lambda \in \mathbb R$ such that $T + \lambda I_n$ is diagonalizable .

(4) Exept for finite many values of $\lambda \in \mathbb R$, $T + \lambda I_n$ is invertible,

I have tried :

Since Eigen values of $T$ are distinct and they are nonzeroes and n eigen values says $\lambda_1 \lambda_2 \cdots ,\lambda_n$ , for all $\lambda \in \mathbb R$ except $\lambda_i$, $T + \lambda I_n$ are invertible, Since all the eigen value of $T$ are non zeroes. So $T$ is invertible. Thus (1), (2) and (4) option are true.

Please tell me abut (3).

Any help would be appreciated. Thank you

$\endgroup$
1
$\begingroup$

Because the eigenvalues are distinct, $T$ itself is diagonalizable, so that solves the problem. One way to see this is that any eigenvalue must always have a geometric multiplicity of at least $1$, and so if the (algebraic) eigenvalues are distinct then the geometric multiplicities add up to $n$.

As an aside, actually $T+\lambda I_n$ is diagonalizable for every $\lambda$, because its eigenvalues are those of $T$ shifted by $\lambda$, so they are also distinct.

$\endgroup$
  • $\begingroup$ @ Ian :Which option is false $\endgroup$ – user120386 May 17 '15 at 19:18
  • $\begingroup$ @user120386 All four are true. $\endgroup$ – Ian May 17 '15 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.