# $T : \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation of $\mathbb{R}^n$. Then which of the following statements are true?

$$T : \mathbb{R}^n \to \mathbb{R}^n$$ 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.

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.
• How do you know that eigen value of $T$ restricted is also a eigenvalue of $T$? Dec 17, 2020 at 18:22
• No, $U$ is a subspace of $V$. This means every element of $U$ is an element of $V$. Dec 17, 2020 at 18:26