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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?

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  • $\begingroup$ Take a (complex) eigenvector, and consider the matrix built from its real and imaginary parts as columns. $\endgroup$ – Lord Shark the Unknown Apr 20 at 6:29
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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 $$\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$$

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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.

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  • $\begingroup$ Then how can we show that A has to be real? $\endgroup$ – user636164 Apr 20 at 13:57
  • $\begingroup$ I have edited it. $\endgroup$ – ancientmathematician Apr 20 at 16:33
  • $\begingroup$ 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. $\endgroup$ – user636164 Apr 20 at 21:28
  • $\begingroup$ You should edit the question and remove the requirement that "$A$ is not necessarily real." $\endgroup$ – ancientmathematician Apr 21 at 6:18

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