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 2×2 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?
Best Answer
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 $$ Ax=ax-by,\qquad Ay=bx+ay $$