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Let $T : \mathbb{R}^n → \mathbb{R}^n$ be a linear transformation of $\mathbb{R}^n$ , where $n ≥ 3$, and let $λ_1, . . . , λ_n ∈ \mathbb{C}$ be the eigenvalues of T. Which of the following statements are true?
(a) If $λ_i = 0$, for some $i = 1, . . . , n$, then $T$ is not surjective.
(b) If $T$ is injective, then $λ_i = 1$ for some $i, 1 ≤ i ≤ n$.
(c) If there is a $3$-dimensional subspace $U$ of $V$ such that $T(U) = U$, then $λ_i ∈ \mathbb{R}$ for some $i$, $1 ≤ i ≤ n$.


(a) the product of the eigen values= the determinant of a matrix. If one eigen value is zero then determinant will be zero and hence the rank of matrix will be the less then the dimension of the space and hence the map is not surjective. so it is true.
(b)it is false obviously.
(c) But I am not sure here.

how can I able to tackle it?

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  • $\begingroup$ a) is too complicated. By the assumption, $Tx=0x=0$ has a non-trivial solution. So there is a non-trivial kernel vector. $\endgroup$ Jan 1, 2014 at 8:58
  • $\begingroup$ What is explanation for b) ? $\endgroup$
    – Error 404
    Jan 20, 2017 at 7:08

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The restriction of $T$ on $U$ is an endomorphism since $U$ is invariant by $T$ and its characteristic polynomial with degree $3$ has a real root (eigenvalue) by the intermediate value theorem hence the answer is YES.

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This is a clarification of what Sami wrote. The credit goes to him.

Look at what $T$ does to elements in $U$. It puts them back in $U$. So we can put blinders on our eyes and only consider the elements in $U$. Now $T$ will behave like a linear operator. Call this $T$ restricted to $U$. The important property you need is that the restricted $T$ will "inherit" 3 eigenvalues from $T$ (and also the associated eigenvectors, a result you do not need here).

So the restricted $T$ will have three eigenvalues, and since you are working with real numbers, complex eigenvalues will come in conjugate pair. So you will always have even number of complex eigenvalues. So, restricted $T$ must have a real eigenvalue.

Since the restricted $T$ inherits its eigenvalues from $T$, $T$ must have a real eigenvalue.

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