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$T : \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation of $\mathbb{R}^n$, where $n \ge 3$ and $ \lambda_1, .. ,\lambda_n \in \mathbb{C}$ be eigen values of T. Then which of the following statements are true?

(A) If $\lambda_i = 0$ for some $i$, then T is not surjective.

(B) If T is injective, then $\lambda_i = 1$ for some $i$.

(C) If there is a 3-dimensional subspace $U$ of $V$ s.t. $T(U)=U$ then $\lambda_i \in \mathbb{R}$ for some $i$.

My attempt :

(A) $\lambda_i = 0$ then $det(T)=0$, which means $T$ is singular and $dim(kerT) \gt 0$. Therefore not surjective.

(B) Let, $\lambda_i$'s are distinct non-zero eigen values of $T$(all of them are real) such that none of them is equal to $1$.

Then, $det(T) \ne 0$. Thus T is injective. Therefore the statement is false.

Bit about C I couldn't proceed much further. I thought if the restriction of $T$ to $U$, namely $T_U$.

Since $T_U(U) = U$. Then $T_U$ is non-singular. But does that say that it must have at least one real eigen value? And also, is it true that eigen values of $T_U$ is also eigen values of $T$? I don't know surely.

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When $T(U)=U$ for a $3$-dimensional subspace $U\subseteq\mathbb R^n$, the restriction $T|_U\colon U\to U$ has $3$ of the eigenvalues of $T$ (counting algebraic multiplicity). The non-real complex eigenvalues of a real linear transform always come in pairs: When $\lambda\in\mathbb C$ is a complex eigenvalue, then so is the complex conjugate $\overline \lambda$. Hence, the number of non-real complex eigenvalues of any real linear transform is even. Since $3$ is odd, $T|_U$ has one or three real eigenvalues.

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  • $\begingroup$ How do you know that eigen value of $T$ restricted is also a eigenvalue of $T$? $\endgroup$
    – nonuser
    Dec 17, 2020 at 18:22
  • $\begingroup$ I just explained it in the other comment thread. $\endgroup$
    – Christoph
    Dec 17, 2020 at 18:22
  • $\begingroup$ No, $U$ is a subspace of $V$. This means every element of $U$ is an element of $V$. $\endgroup$
    – Christoph
    Dec 17, 2020 at 18:26

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