4
$\begingroup$

I have a problem and a proposed solution. Please tell me if I'm correct.

Problem: For A,B real matrices, prove that if there is a solution in the complex numbers then there is also a real solution.

Solution: A and B are real matrices. Therefore, they are not defined over the complex plane, and a solution in C is not possible for the system AX=B. Hence, the "if" part is false by default and the "then" part is always true, making the implication always true and rendering the problem statement as proven.

Thanks!

$\endgroup$
  • 2
    $\begingroup$ The reals are a subset of the complex numbers, so your proof doesn't hold. $\endgroup$ – coffeemath Jul 19 '13 at 12:20
  • 1
    $\begingroup$ But a real number is, in particular, a complex number (i.e., the real number $a$ corresponds to the complex number $a + i0$), so a real matrix can always be viewed as a complex matrix. $\endgroup$ – Branimir Ćaćić Jul 19 '13 at 12:20
  • 3
    $\begingroup$ The equation $x+y=17$ can be written as $Ax=b$ with $A$ and $b$ real, but it has the nonreal solution $x=3+i$, $y=14-i$, so your logic is faulty. $\endgroup$ – Gerry Myerson Jul 19 '13 at 12:20
  • $\begingroup$ Hmm. What should I do differently then? $\endgroup$ – user85362 Jul 19 '13 at 12:22
  • 1
    $\begingroup$ If $A$ is a square matrix with nonzero determinant, then Cramer's rule shows the solutions are in the same field as are the coefficients of the matrices. $\endgroup$ – coffeemath Jul 19 '13 at 12:24
2
$\begingroup$

The matrix equation boils down to a system of linear equations of the form $$a_1z_1+ a_2z_2 + \cdots a_nz_n=c,$$ where you're assuming the $a_i$ and $c$ are real numbers. There will typically be systems of such equations which must hold all at once and mention the same $z_i$.

Now write each $z_k=x_k+iy_k$ and note that each equation above implies (by taking real parts) that if all the $y_k$ are replaced by $0$ the equation still holds, giving a real solution also.

$\endgroup$
4
$\begingroup$

Let $X$ be a (possibly) complex solution to $AX=B$. Since $A$ and $B$ are real, then $\overline{X}$ (component-wise conjugate) is also a solution, so that $\frac{X+\overline{X}}{2}$ is solution but this a real matrix. In this way we have constructed a real solution starting from a complex one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy