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Jul 18

Geometry of linear equations: structure of image and null space

category: Matrix algebra, MT3042 Optimisation theory, Optimisation theory No Comments

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

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

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

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

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 a 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.

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