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


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.

  • $\begingroup$ @ 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

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