# Show that a 2x2 matrix A with complex eigenvalues has a special similarity

My question is: if $$A$$ is a 2×2 matrix, with complex eigenvalues $$(a±bi)$$, show that there is a real matrix $$P$$ for which $$P^{−1}AP = \left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)$$.

UPDATE: working with the discriminant of the characteristic polynomial of a 2x2 matrix, I can see that $$\left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)$$ constructs a real matrix with complex eigenvalues, since its discriminant $$= -4b^2 < 0$$.

I also have a theorem that if two matrices $$A$$ and $$B$$ represent the same linear transformation, then $$A$$ and $$B$$ are similar, i.e. $$A = P^{-1}BP$$. But how can we guarantee that $$\left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)$$ can represent the same transformation as any $$A$$?

Is it something like: does this matrix necessarily have the coordinates for $$A$$'s transformation, but instead of $$A$$ being similar to a diagonal matrix (with eigenvalues on the diagonal), the complex eigenvalues need two entries?

• Take a (complex) eigenvector, and consider the matrix built from its real and imaginary parts as columns. – Lord Shark the Unknown Apr 20 at 6:29

The matrix $$A$$ should be real in order that a real $$P$$ exists.
Suppose the eigenvalues are $$a+bi$$ and $$a-bi$$, with $$b\ne0$$. The relation can also be written as $$AP=P\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$$ If $$v$$ is a complex eigenvector relative to $$a+bi$$, then, due to $$A$$ being real, $$\bar{v}$$ is an eigenvector relative to $$a-bi$$. These two vectors are linearly independent, being relative to distinct eigenvalues. Thus also $$x=\frac{1}{2}(v+\bar{v}),\qquad y=\frac{1}{2i}(v-\bar{v})$$ are linearly independent. Note that $$v=x+iy$$ and $$Ax+iAy=Av=(a+ib)(x+iy)=(ax-by)+i(bx+ay)$$ Equating the real and imaginary parts, we get $$Ax=ax-by,\qquad Ay=bx+ay$$
Thus the matrix with respect to the basis $$\{x,y\}$$ is exactly $$\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$$
This is not necessarily true. If it were, then $$A= P\left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)P^{-1}$$ would have to be real, and in the case when $$A$$ is diagonal (with $$a\pm bi$$ on the diagonal) it is not.
• Ok, A has to be real, and it also has to have complex eigenvalues (this is true when the discriminant of the characteristic polynomial < 0). I still need to show that the real $P$ exists in this case. – user636164 Apr 20 at 21:28
• You should edit the question and remove the requirement that "$A$ is not necessarily real." – ancientmathematician Apr 21 at 6:18