Raising the bar

Geometry of linear equations: structure of image and null space

Definition 1. The subspace notion allows us to describe the algebraic structure of the set of solutions of $latex Ax=y.$ The special case $latex Ax=0$ is called a homogeneous equation. Obviously, $latex x=0$ satisfies it but there may be other solutions. The equation $latex Ax=y$ is called an inhomogeneous equation. We address the questions of existence and uniqueness of its solutions.

Structure of the image of $latex A$

Recall Basic observation 1: The image $latex \text{Img}(A)=\{ Ax:x\in R^k\}$ is the set of $latex y=Ax$ for which the inhomogeneous equation has solutions.

Exercise 1. $latex \text{Img}(A)$ is a linear subspace in $latex R^n.$

Proof. This follows from the first characterization of the matrix image. Here is a direct proof. Suppose $latex y_1,y_2\in \text{Img}(A).$ Then there exist $latex x_1,x_2\in R^k$ such that $latex Ax_1=y_1,$ $latex Ax_2=y_2.$ By linearity this gives $latex ay_1+by_2=aAx_1+bAx_2=A(ax_1+bx_2).$ Thus for $latex y=ay_1+by_2$ we have found $latex x=ax_1+bx_2$ such that $latex y=Ax,$ which means $latex ay_1+by_2\in \text{Img}(A)$ for any $latex a,b$ and $latex \text{Img}(A)$ is a subspace.

Structure of the null space of $latex A$

Definition 2. The set of solutions $latex x$ of the homogeneous equation $latex Ax=0$ is called the null space of $latex A.$ It is denoted $latex N(A)=\{ x:Ax=0\}.$

Exercise 2. The null space of $latex A$ is a linear subspace of $latex D(A).$

Proof. Suppose $latex x_1,x_2\in N(A)$ so that $latex Ax_1=0,$ $latex Ax_2=0.$ Then by linearity $latex A(ax_1+bx_2)=aAx_1+bAx_2=0,$ so $latex ax_1+bx_2\in N(A)$ for any $latex a,b$ and $latex N(A)$ is a linear subspace.

Description of the set of solutions of $latex Ax=y$

Intuition. In $latex R^{3},$ straight lines and planes that don't contain the origin can be obtained by shifting straight lines and planes that do (geometry should dominate the algebra at this point, see Figure 1). This is generalized in the next definition.

Figure 1. Shifting a subspace gives a hyperplane

Definition 3. Let $latex x\in R^n$ be any vector and let $latex L$ be a subspace. $latex x+L=\{x+l:l\in L\}$ denotes a shift of $latex L$ by $latex x$ and it is obtained by adding to $latex x$ all elements of $latex L.$ Some people call $latex x+L$ a hyperplane.

Exercise 3. As we know, the equation $latex Ax=y$ has solutions if and only if $latex y\in \text{Img}(A).$ Let us fix $latex y\in \text{Img}(A)$ and let $latex x_{0}$ be some solution of $latex Ax=y$. Then the set of all solutions of this equation is $latex
x_0+N(A)$ (in detail: any other solution $latex x$ of that equation can be obtained by adding an element $latex z$ of the null space to $latex x_0$: $latex x=x_0+z$). This is written as $latex \{ x:Ax=y\}=x_0+N(A).$

Proof. Let $latex Ax_0=y$ and let $latex x$ be any solution of $latex Ax=y.$ Then $latex A(x-x_0)=0,$ $latex x-x_0\in N(A)$ and $latex x-x_{0}=z$ for some $latex z\in N(A).$ We obtain $latex x=x_0+z$ which proves the inclusion $latex \{ x:Ax=y\}\subset x_0+N(A).$ Conversely, if $latex x=x_0+z$ with $latex z\in N(A)$, then $latex Ax=Ax_0+Az=y$ and we obtain $latex x_0+N(A)\subset \{ x:Ax=y\}.$

Conclusion. If $latex N(A)=\{0\}$, then $latex A$ is one-to-one and we have uniqueness of solutions of the inhomogeneous equation; otherwise, $latex N(A)$ can serve as a measure of non-uniqueness.

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