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I am quite confused with this problem.

"Let F be the field of complex numbers. Are the following two systems of linear equations equivalent ? If so, express each equation in each systems as a linear combination of the equations in the other system."

$$x_1-x_2=0, \qquad 3x_1+x_2=0;$$

on one hand, and on the other:

$$2x_1+x_2=0, \qquad x_1+x_2=0.$$

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Sure they are equivalent. But what does it mean to “express each equation in each systems as a linear combination of the equations in the other system.” We may express a vector as a linear combination of other vectors, but what about equations? Surely they are no vectors.

Have a look of the lhs of the first equation, namely $x_1-x_2$. This may be viewed as a linear form sending a vector $(a,b)$ to the complex number $a-b$ and the solution of the equation $x_1-x_2=0$ is the kernel of that linear form.

Now the solution set of a homogenous linear system is the intersection of the kernels of the linear forms involved. So the task now is to express the linear forms of the first system in terms of the linear form of the second system. Since a finite dimensional vector space is isomorphic to its dual space, you may view the linear form $x_1-x_2$ as the vector $(1,-1)^T$. To express the first linear form of the second system you have to solve $$r\begin{pmatrix}1\\ -1\end{pmatrix}+s\begin{pmatrix}3\\ 1\end{pmatrix}= \begin{pmatrix}2\\ 1\end{pmatrix}$$ to find $$\frac{-1}{4}\begin{pmatrix}1\\ -1\end{pmatrix}+\frac{3}{4}\begin{pmatrix}3\\ 1\end{pmatrix}= \begin{pmatrix}2\\ 1\end{pmatrix}.$$

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  • $\begingroup$ Thanks a lot @Michael Hoppe $\endgroup$ – mahavir Mar 11 '14 at 16:26
  • $\begingroup$ @mahavir You're welcome; the question was slightly ill-posed. $\endgroup$ – Michael Hoppe Mar 11 '14 at 16:44

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