# homogeneous linear system

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 $$(a_1,a_2,a_3,a_4)$$

• 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. Commented Dec 9, 2013 at 12:58
• Can we not somehow use the property that the coefficients are rotating ?
– Mod
Commented Dec 9, 2013 at 13:00
• they are linearly independent. No other solution Commented Dec 9, 2013 at 13:05
• So you mean to say it will only have zero solution ?
– Mod
Commented Dec 9, 2013 at 13:08
• 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. Commented Dec 9, 2013 at 13:12

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:

$$rref(A)=I$$

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

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.

• 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 $a_i$ .
– Mod
Commented Dec 9, 2013 at 13:37