Question about mapping of linear transformation The matrix of linear transformation with standard basis is given as $\begin{pmatrix} 0&a&b\\-a&0&c\\-b&-c&0\end{pmatrix}$,where $a,b,c$ are real numbers not all zero.Question is to check whether this maps any line through the origin onto itself.I was thinking to multiply the matrix by $(xt,yt,zt)$ and to check whether output is same$(xt,yt,zt)$.Is it correct? 
 A: Note that a linear transformation $T$ sends a line through the origin to itself if and only if that line is contained in an eigenspace (for a nonzero eigenvalue) of $T$. I.e., the line is of the form $\newcommand{\set}[1]{\left\{{#1}\right\}}\set{tv:t\in\newcommand{\RR}{\mathbb{R}}\RR}$ for some eigenvector $v$ with eigenvalue $\lambda\ne 0$ of $T$. One can see this from the fact that if $v$ is an eigenvector, then $T(tv)=t\lambda v$, which is still in that line (and this map is bijective from the line to itself if $\lambda\ne 0$), and if $T$ sends the line to itself, $Tv = tv$ for some $t\ne 0\in\RR$. That $t$ is the nonzero eigenvalue for which this $v$ is an eigenvector.
Thus to show that this matrix doesn't send any line through the origin to itself, it suffices to show that it has no nonzero real eigenvalues. Or to find a line through the origin that it sends to itself, we would need to find an eigenvector with nonzero eigenvalue. Either way, we need to compute the eigenvalues.
To compute the eigenvalues, we compute the characteristic polynomial of this matrix,
$$p_T(t) = \det (tI-T) = \det\begin{pmatrix} t&-a&-b\\a&t&-c\\b&c&t\end{pmatrix}= t^3+abc-abc +b^2t+c^2t+a^2t = t(t^2+a^2+b^2+c^2).$$
This has a single real root at $t=0$, and if one of $a,b,$ or $c$ is nonzero the other eigenvalues are complex. Even if $a=b=c=0$, the other eigenvalues are also 0, since then $T=0$. Thus in no case does $T$ send a line through the origin to itself.
A: The idea is correct. Let $M$ be your matrix. What you should be trying to determine is whether or not there is a vector $(x,y,z)\in\mathbb{R}^3$ such that $M.(x,y,z)$ is of the form $\lambda(x,y,z)$ for some $\lambda\neq0$. And there is no such vector because, as you can easily check, $(x,y,z)$ and $M.(x,y,z)$ are always orthogonal.
