Compute a scalar such that the set given is linearly independent

Question

Let vector set $\{u_1, u_2, u_3, u_4\}$ is linearly independent. The vector set $\{u_1+au_2, u_2+au_3, u_3+au_4, u_4+au_1\}$ is also linearly independent if the value of $a = \cdots$

I have applied the definition of linearly independent vectors in the set, but found a hard way when dealing with the second vector set, assuming that we have $k_1 = k_2 = k_3 = k_4 = 0$ from the first one. Do you have any idea?

Answer 1

make the ansatz $$\alpha(\vec{u_1}+a\vec{u_2})+\beta(\vec{u_2}+a\vec{u_3})+\gamma(\vec{u_3}+a\vec{u_4})+\delta(\vec{u_4}+a\vec{u_1})=\vec{0}$$ where $$\alpha,\beta,\gamma,\delta$$ are real numbers then you will get $$\alpha+a\delta=0$$ $$\alpha a+\beta=0$$ $$\beta a+\gamma=0$$ $$\gamma a+\delta=0$$

Answer 2

Since the first set of vectors is linearly independent, it forms a basis for the set’s span. The coordinates of the second set of vectors relative to this basis are $[1,a,0,0]^T$ etc. The vectors are linearly independent iff the determinant of the matrix that has these coordinate vectors as its columns (or rows) is nonzero. The latter matrix is, not coincidentally, also the coefficient matrix of the system of linear equations in Dr. Graubner’s answer.