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Jul 18

Geometry of linear equations: structure of image and null space 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 $Ax=y.$ The special case $Ax=0$ is called a homogeneous equation. Obviously, $x=0$ satisfies it but there may be other solutions. The equation $Ax=y$ is called an inhomogeneous equation. We address the questions of existence and uniqueness of its solutions.

Structure of the image of $A$ $A$

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

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

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

Structure of the null space of $A$ $A$

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

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

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

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

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

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

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

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