Eigenvector Question Let $A$ be a fixed $3 \times 3$ matrix and define a linear map $T : M_{33} \to M_{33}$ by $T(X) = AX$. If $\lambda$ is a real eigenvalue of $T$ corresponding to an invertible eigenvector $X$ find $\lambda$ in terms of $\det(A)$.
I have trouble just seeing how the map works and the solution. Help appreciated, thanks!
 A: HINT :
given map is $T(B)=AB$ ,
as $\lambda$ is an eigenvalue for an "INVERTIBLE MATRIX" B, so you have   $T(B)=\lambda B$ 
i.e., $AB=\lambda B$ 
i.e., $\det(AB)=\det(\lambda B)$ 
i.e., $\det(A)\det(B)=(\lambda^{???})\det(B)$
as $B$ is invertible, $\det(B)$ can be cancelled out...
Then, you have $\det(A)=(\lambda^{???})$
A: In general, such invertible $X$ cannot exist.  Why?  If $X$ satisfies the equation $AX =\lambda X$, then it is easy to see that the columns of $X$ are in fact the ordinary eigenvectors of $A$ associated with the eigenvalue $\lambda$.  An $N \times N$ square matrix typically has $N$ distinct eigenvalues, and since eigenvectors corresponding to distinct eigenvalues are linearly independent, each such eigenvalue will have a one-dimensional eigenspace, in which the columns of $X$ must all lie.  This being the case, the columns of $X$ are linearly dependent, whence $\det X = 0$ and $X$ is singular; noninvertible.  Essentially the same situation occurs in the event that has $X$ has any more that one (distinct) eigenvalue; again the dimensions of the corresponding eigenspaces must be less than $N$, so any selection of $N$ vectors for the columns of $X$ from any eigenspace must exhibit a linear dependence, forcing $\det X = 0$ and $X$ to be singular.  Evidently the only possible way $X$ can be invertible under these circumstances is for $A$ to have precisely one eigenvalue $\lambda$ with an $N$-dimensional eigenspace.  But then $A = \lambda I$ is a multiple of the identity matrix;  $\det A = \lambda^N$ in this situation.
Be apprised that in the case of $X$ having more than one eigenvalue, the determinant cannot be computed from the equation $AX = \lambda X$, since $\det X = 0$.  Furthermore, there will be other eigevalues which bear on $\det A$ which are not addressed by the single equation $AX = \lambda X$.
Nota Bene:  The above argument is stated for matrices of size $N$, to which it apparently applies; taking $N = 3$ is in no-wise essential to this result.
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!
