Let $T: \mathbb{R}^3→\mathbb{R}^3$ be a linear transformation. Show that there is a line $L$ such that $T(L) = L$.
I am totally stuck on it.how can I able to solve this problem?please somebody help.thanks for your kind help.
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3The characteristic polynomial of T has degree 3, so it has a real root. – egreg Mar 04 '13 at 17:05
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5It can be zero. I think this is a problem. Example: T=0. – Aloizio Macedo Mar 04 '13 at 17:06
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still it is not clear to me.can you explain please. – priti Mar 04 '13 at 17:13
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1@AloizioMacedo Yeah, there needs to be some invertibility clause or else you can have a zero eigenvalue with eigenspace $L$, and $T(L) = {0}$. Otherwise this is true with $T(L) \subset L$. – A Blumenthal Mar 04 '13 at 17:14
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1@priti Try to solve this problem when you know that $T$ has an eigenvector with eigenvalue $\lambda \neq 0$. – A Blumenthal Mar 04 '13 at 17:15
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If you haven't studied the general theory of eigenvectors and eigenvalues, take into account these facts:
Every linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ can be described as $T(X)=AX$, where $A$ is an $m\times n$ matrix. In your case, it is a $3\times 3$ matrix.
For a linear transformation $T:\mathbb{R}^3\to\mathbb{R}^3$, with corresponding matrix $A$, the following are equivalent conditions:
(a) There is a line $L$ (that passes through the origin) such that $T(L)=L$
(b) For some nonzero vector $X$ in $L$, $T(X)$ is a nonzero multiple of $X$, that is, there is a nonzero real number $\lambda$ such that $T(X)=\lambda X$
(c) For some nonzero real number $\lambda$, the homogeneous system $(A-\lambda I)X=0$ has a nontrivial solution.
(d) For some nonzero real number $\lambda,\ \det (A-\lambda I)=0$
Now, in the $3\times 3$ case, the polynomial in $\lambda$ (called the characteristic polynomial of $A$) defined as $\det(A-\lambda I)$ has necessarily a real root, but, what if it has only a real root and it is zero? In this case it can't exist a line $L$ such that $T(L)=L$.
Take for example the matrix: $$A= \begin {pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix} $$
Then $\det(A-\lambda I)=-\lambda^3-\lambda$, that has roots $0,i,-i$, so in this case the corresponding linear transformation: $$T(x,y,z)=(0,-z,y) $$ cannot send a line $L$ onto itself.
Loronegro
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