# Finding a base field for diagonalized linear transformation and justifying

So I encountered this long question that asks you to find bases and such, I searched up

Find the eigenvalues for the linear transformation and base associated to each eigenvalue. If possible find the base field so that the matrix representation of the Linear Transformation is diagonal AND if there is no such base, justify why.

I tried to solve for $F$ and ended up with 3 equations where the only eigenvalue that worked was $\lambda=-1$. $$F(x, y, z)=(-x-2y+2z,-y,-x-3y-4z) \tag{0}$$ $$\lambda x=-x-3y+2z \tag{1}$$ $$\lambda z=-x-3y-4z\tag{2}$$ $$\lambda y=-y\tag{3}$$

The only other case you could find other eigenvalues would be for $y=0$, which yielded no eigenvalues in my search. Is there no base for the diagonal matrix? How do you justify this to an examiner?

• You found one of the eigenvalues, $\lambda = -1$, but there are two more. You should find an eigenvector for each of the three eigenvalues, and these three vectors together will form a basis for which the representation of the linear transformation is a diagonal matrix (with diagonal entries holding the eigenvalues). – hardmath Dec 14 '14 at 13:50
• @hardmath using $\lambda =-1$ leaves me with $(-6,1,1)$ as a possible eigenvector. Using equations $(1)$ and $(2)$ don't give a numerical value for $\lambda$... – FemtoComm Dec 14 '14 at 14:37
• The possible values for $\lambda$ are those which make $A - \lambda I$ singular where $A$ is a matrix representing the linear transformation. That is, if $(A-\lambda I)v = 0$, then $Av = \lambda v$. If $v$ is a nonzero vector, then it would be an eigenvector corresponding to eigenvalue $\lambda$. The polynomial $p(\lambda) = \det (A - \lambda I)$ is the characteristic polynomial of $A$ (and also of the linear transformation $F$), and eigenvalues are roots of this polynomial. – hardmath Dec 14 '14 at 15:09

The matrix representation of $\;F\;$ wrt the canonical basis is
$$[F]=\begin{pmatrix}-1&-2&2\\ 0&-1&0\\-1&-3&-4\end{pmatrix}$$
$$\det(xI-[F])=\begin{vmatrix}x+1&2&-2\\ 0&x+1&0\\1&3&x+4\end{vmatrix}=(x+1)\begin{vmatrix}x+1&-2\\1&x+4\end{vmatrix}=$$
$$=(x+1)(x^2+5x+6)=(x+1)(x+2)(x+3)$$
Thus no matter on what field is your vector space, and thus your transformation $\;F\;$, defined, it contains all three eigenvalues of the matrix (which are two if the field's characteristic is$\;2\;$).