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I have this problem :

This is a linear system with two equations, and four unknown.

$$a_1x_1+a_2x_2+a_3x_3+a_4x_4=a$$ $$b_1x_1+b_2x_2+b_3x_3+b_4x_4=b$$

These vectors $(a_1,a_2,a_3,a_4),(b_1,b_2,b_3,b_4)$ are independent.

Proof that the system has a solution and proof that there are three vectors: $$u_0,u_1,u_2$$

so any solution $x$ of the system is :

$$x=u_0+\alpha u_1+\lambda u_2$$

while $\lambda,\alpha \in R$.

I don't know how to approach to this problem, since this is underdetermined linear system, in case of non-homogeneous it could have no solution.

Any help will be appreciated.

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  • $\begingroup$ The matrix from the equations has full row rank. The system is consistent always. $\endgroup$ – Karthik Upadhya Dec 2 '14 at 22:39
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First, since the vectors are independent, the matrix of the system is full rank. let's call it $A = \begin{bmatrix}a_1&a_2&a_3&a_4\\b_1&b_2&b_3&b_4\end{bmatrix}$ and the vector $t :=\begin{bmatrix}a\\b\end{bmatrix}$. Your system has then the form $Ax = t$.

Now because $A$ is full rank, there exists a $u_0$ such that $Au_0 = t$. Now assume there is another solution $u_0 + u_s$ then $A(u_0+u_s)=t$ has to hold, which infers though: \begin{align} \Leftrightarrow && A(u_0+u_s)&=t\\ \Leftrightarrow && Au_s+t&=t\\ \Leftrightarrow && Au_s&=0\\ \end{align} Hence, it is necessary that $u_s$ lies in the nullspace of $A$. By the rank-nullity theorem Rank-Nullity Theorem, we know that the dimension of our matrix $A$ is $2$. Now taking any two basis vectors of this nullspace $u_1$, $u_2$, we get, that $u_s$ has to look like $\alpha u_1 + \beta u_2$ for any $\alpha$ and $\beta$ and therefore every solution of the $Ax=t$ looks like $x = u_0 + \alpha u_1 + \beta u_2$.

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  • $\begingroup$ Do you conclude that there are two basis of the nullspaces because its $R^4$? since we conclude that the rank is $2$ so rank+nullity=$2+2=4$? $\endgroup$ – JaVaPG Dec 3 '14 at 0:20
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Since $(a_1,a_2,a_3,a_4)$ and $(b_1,b_2,b_3,b_4)$ are linearly independent we have that

$$\mathrm{ran}\left(\begin{array}{cccc}a_1 & a_2 & a_3 & a_4 \\b_1 & b_2 & b_3 & b_4 \end{array}\right)=2.$$ Thus,

$$\mathrm{ran}\left(\begin{array}{cccc}a_1 & a_2 & a_3 & a_4 \\b_1 & b_2 & b_3 & b_4 \end{array}\right)=\mathrm{ran}\left(\begin{array}{ccccc}a_1 & a_2 & a_3 & a_4& a \\b_1 & b_2 & b_3 & b_4 & b \end{array}\right)=2.$$ Using the Rouche-Capelli theorem we know that the system has solution $u_0.$

Now, consider the homogeneous system

$$\left\{\begin{array}{c}a_1x_1+a_2x_2+a_3x_3+a_4x_4=0\\ b_1x_1+b_2x_2+b_3x_3+b_4x_4=0\end{array}\right.$$ Since $(a_1,a_2,a_3,a_4)$ and $(b_1,b_2,b_3,b_4)$ are linearly independent we have that the matrix

$$\left(\begin{array}{cccc}a_1 & a_2 & a_3 & a_4 \\b_1 & b_2 & b_3 & b_4 \end{array}\right)$$ has a nonzero minor of order two. Say

$$\left|\begin{array}{cccc}a_1 & a_2 \\b_1 & b_2 \end{array}\right|\ne 0.$$ Write the system as

$$\left\{\begin{array}{c}a_1x_1+a_2x_2=-a_3x_3-a_4x_4\\ b_1x_1+b_2x_2=-b_3x_3-b_4x_4\end{array}\right.$$ Then writing $x_3=\lambda,x_4=\mu,$ we get the solutions of the system

$$\left\{\begin{array}{l} x_1= \frac{\left|\begin{array}{cccc}-a_3\lambda-a_4\mu & a_2 \\-b_3\lambda-b_4\mu & b_2 \end{array}\right|}{\left|\begin{array}{cccc}a_1 & a_2 \\b_1 & b_2 \end{array}\right|} \\ x_2 =\frac{\left|\begin{array}{cccc}a_1 & -a_3\lambda-a_4\mu \\ b_1& -b_3\lambda-b_4\mu \end{array}\right|}{\left|\begin{array}{cccc}a_1 & a_2 \\b_1 & b_2 \end{array}\right|} \\ x_3=\lambda \\x_4=\mu \end{array}\right.$$

This system has two linearly independent solutions $u_1,u_2$ (one for $\lambda=0,\mu=1$ and other for $\lambda=1,\mu=0$). Since it is homogenous $\alpha_1 u_1+\alpha_2 u_2$ is a solution for any $\alpha_1,\alpha_2.$ Thus,

$$u_0+\alpha_1 u_1+\alpha_2 u_2$$ is a solution of the initial system for any $\alpha_1,\alpha_2.$

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  • $\begingroup$ we conclude that there is non-trival solution for the homogeneous system since it underdetermined so exists $u_1$ it's clear that $a_1u_1$ is also a solution for the homogeneous system. Using Rouché–Capelli theorem there is solution to the non-homogeneous systems, therefore $u_0$ is a solution for non-homogeneous, so $u_0+a_1u_1$ is a solution. I don't understand how did you get $a_2$ can you explain please? I think that with $u_0+a_1u_1$ we can show any solution. Is it correct? Thanks! $\endgroup$ – JaVaPG Dec 2 '14 at 23:51
  • $\begingroup$ I have edited the answer to clarify such a point. $\endgroup$ – mfl Dec 2 '14 at 23:59

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