0
$\begingroup$

I was studying about system of linear systems and had come across homogeneous systems . What I wanted to ask was , given a homogeneous system of n variables , like this having 4 variables : $$a_1x + a_2y + a_3z + a_4w = 0$$ $$a_2x + a_3y + a_4z + a_1w = 0$$ $$a_3x + a_4y + a_1z + a_2w = 0$$ $$a_4x + a_1y + a_2z + a_3w = 0$$ Here , as we know that this system has zero solution and we can see that the coefficients are rotating in each of the linear equation .

So , is there any general form for solution of this system other than the zero solution ?

Also , what if we are given the non-zero solution then can we find the values of $$(a_1,a_2,a_3,a_4) $$

$\endgroup$
7
  • $\begingroup$ There is no general form, in a sense that the system must be solved, meaning there are different outcomes to the solution space. IF they (the rows) are all linearly independent, then the solution is just the 0 vector. If there is one linearly dependent row, the null space is a line, 2 linearly dependent row gives a planar subspace, etc. $\endgroup$ Commented Dec 9, 2013 at 12:58
  • $\begingroup$ Can we not somehow use the property that the coefficients are rotating ? $\endgroup$
    – Mod
    Commented Dec 9, 2013 at 13:00
  • $\begingroup$ they are linearly independent. No other solution $\endgroup$ Commented Dec 9, 2013 at 13:05
  • $\begingroup$ So you mean to say it will only have zero solution ? $\endgroup$
    – Mod
    Commented Dec 9, 2013 at 13:08
  • $\begingroup$ Yes, now what do you mean by the inverse? The inverse of the coefficient matrix? If a matrix is invertible, then the system only has the 0 vector as solution. If your solution vector is nonzero, your null space (solution to the homogeneous system) depends on the rows (or columns) and their linear dependence. $\endgroup$ Commented Dec 9, 2013 at 13:12

1 Answer 1

1
$\begingroup$

General sollution to $AX=0$ is the kernel $Ker(A)$.

To say $AX=0$ only have zero solution (trivial kernel $Ker(A)=\left\{ 0 \right\}$) is equivalent to verify the row reduced echelon form of matrix A:

$$rref(A)=I$$

In your case:

$$A = \begin{pmatrix}a1&a2&a3&a4\\a2&a3&a4&a1\\a3&a4&a1&a2\\a4&a1&a2&a3\end{pmatrix}$$

Just need to verify:

$$rref(A) = \begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$

For second question, treat $a_i$ as unknowns: $$xa_1 + ya_2 + za_3 + wa_4 = 0$$

$$wa_1 + xa_2 + ya_3 + za_4 = 0$$

$$za_1 + wa_2 + xa_3 + wa_4 = 0$$

$$ya_1 + za_2 + wa_3 + xa_4 = 0$$

Then it is another linear system you can solve.

$\endgroup$
1
  • $\begingroup$ Thank you for the answer , and could please help on the second part which is if we are given a solution , can we deduce all $a_i$ . $\endgroup$
    – Mod
    Commented Dec 9, 2013 at 13:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .