An equation to represent all vector solutions to a system of equations with infinite solutions

Question

If both $x$ and $y$ are solutions to a system of linear equations with infinite solutions then

$$z = αx + (1 −α)y$$ is also a solution for any real α.

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I'm having some trouble understanding this. Can you explain how this is derived and also its intuition?

I understand that multiplying for example $x$ by a scalar usually necessitates a change in $y$ for $z$ to be a solution, but how do we know that scalar is $1-α$?

Answer 1

Let $x$ be a solution to $Ax=b$ and let $y$ be a solution too, i.e. $Ay=b$. Testing $z=\alpha x+(1-\alpha)y$ $$ Az=A(\alpha x+(1-\alpha)y)=\alpha Ax+(1-\alpha)Ay=\alpha b+(1-\alpha)b= (\alpha+1-\alpha)b=b. $$ Thus $z$ is a solution too.

Intuition: all solutions to $Ax=b$ span a linear manifold, think a plane in $\mathbb{R}^3$. The condition $z=\alpha x+(1-\alpha)y$ is a parameter equation for a line through points $x$ and $y$. Clearly if $x$ and $y$ are on the plane, then the whole line through them (any $z$ on the line) is also on the plane.

Answer 2

Assume you have a system of equations

$$ \mathbf{Ax} = \mathbf{b}$$

and assume furthermore that $\mathbf{x_1}, \mathbf{x_2}$ solve this system.

Then clearly $$\mathbf{A} (\alpha \mathbf{x_1} + (1- \alpha) \mathbf{x_2}) = \alpha \mathbf{b} + (1- \alpha) \mathbf{b} = \mathbf{b}.$$

Answer 3

What is the definition you have of a linear equation or a system of linear equations? That may sound a daft question, but we normally mean by this that the linear operator/system $L$ satisfies $L(\alpha x+\beta y)=\alpha L(x)+\beta L(y)$

Equivalently this can be written as $L(x+y)=L(x)+L(y)$ and $L(\lambda x)=\lambda L(x)$

Either way the $\alpha, \beta, \lambda$ are simply scalars, and the $x, y$ are variables or vectors (or elements of some other kind). These properties of $L$ are what we mean by a system being linear.

If you have $L(x)=L(y)$ then $L(\alpha x+(1-\alpha y))=\alpha L(x)+(1-\alpha)L(y)$ by the linearity property. Then put $L(y)=L(x)$ to get $L(x)$ on the right-hand side.

Answer 4

If the system of equations is written as

$$ \mathbf{A} x_k = b $$

then the system can only have infinite solutions, IF $\mathbf{A}^{-1}$ does not exist, as otherwise we would have

$$ x_k = \mathbf{A}^{-1} b, $$

having only one solution as $x_k = x_\ell$.

Considering $\mathbf{A}^{-1}$ does not exists, we get

$$ \mathbf{A} \Big[ \alpha x + (1-\alpha) y \Big] = \alpha \mathbf{A} x + \mathbf{A} y - \alpha \mathbf{A} y = \alpha b + b - \alpha b = b, $$

whence

$$ \alpha x + (1-\alpha) y $$

is also a solution.