Question

Suppose \(A \in L\left(R^{n}, R^{*}\right)\), let \(r\) be the rank of \(A\). (a) Define \(S\) as in the proof of Theorem 9.32. Show that \(S A\) is a projection in \(R^{\circ}\) whose null space is \(\mathcal{N}(A)\) and whose range is \(\mathscr{R}(S) .\) Hint: By (68), SASA \(=S A\). (b) Use (a) to show that $$ \operatorname{dim} \mathcal{N}(A)+\operatorname{dim} \mathscr{A}(A)=n $$

Short answer

Based on the given solution, here's a possible question: Question: Let \(A\) be a linear transformation from \(R^n\) to \(R^\circ\), with a rank of \(r\). If we define a matrix \(S\) such that \(AS = I_r\), where \(I_r\) is the \(r\times r\) identity matrix, then show that the dimension of the null space of \(A\) plus the dimension of the annihilator of \(A\) is equal to \(n\). Answer: We can show this by first defining the matrix \(S\) and proving that \(SA\) is a projection matrix. Then, we find the null space and range of \(SA\) and use these results to prove the required equality: $$\operatorname{dim}\mathcal{N}(A) + \operatorname{dim}\mathscr{A}(A) = n$$

Step-by-step solution

Define the matrix S

Define the matrix \(S\) as in the proof of Theorem 9.32, such that \(AS = I_r\), where \(I_r\) is the \(r\times r\) identity matrix. This means that \(S\) acts as a right inverse of \(A\).

Show that SA is a projection matrix

We want to show that \(SA\) is a projection matrix. According to (68), we have \(SA(SA) = SA\). As the hint suggests, it means that \(SASA = SA\). A matrix is a projection matrix if it's idempotent, which means \(P^2=P\). In our case, \(P=SA\), and so we have shown \((SA)^2 = SA\).

Determine null space of SA

We now need to find the null space of SA. That is, we need to find all vectors \(x\) such that \((SA)x = 0\). Notice that \((SA)x = S(Ax)\). Since A is a linear transformation, every vector in \(\mathcal{N}(A)\) is mapped to the zero vector by A. So for any \(x \in \mathcal{N}(A)\), we have \((SA)x = S(0) = 0\). Thus, \(\mathcal{N}(A) \subseteq \mathcal{N}(SA)\). Now take any vector \(x \in \mathcal{N}(SA)\). Then \((SA)x = 0\), so we have \(S(Ax) = 0\). Since \(S\) is a right inverse of \(A\), it means that \(Ax\) must be in the null space of \(S\) for this equation to hold. But by the definition of \(S\), the null space of \(S\) is empty except for the zero vector. Therefore, \(Ax = 0\) and \(x \in \mathcal{N}(A)\). This means that \(\mathcal{N}(SA) \subseteq \mathcal{N}(A)\). Combining the two results, we conclude that \(\mathcal{N}(SA) = \mathcal{N}(A)\), which means the null space of the projection matrix \(SA\) is the same as the null space of \(A\).

Determine the range of SA

Now we want to find the range of the projection matrix \(SA\). Notice that for any vector in \(\mathscr{R}(S)\), we have \(y=Sx\) for some \(x\). So if we consider \((SA)y = (SA)(Sx) = S(AS)x = S(I_r x) = Sx\) and thus, \(y \in \mathscr{R}(SA)\). This means that \(\mathscr{R}(S) \subseteq \mathscr{R}(SA)\). It is also true that \(\mathscr{R}(SA) \subseteq \mathscr{R}(S)\) since columns of transformation matrix \(SA\) are linear combinations of columns of transformation matrix \(S\). Combining the results, we get that \(\mathscr{R}(SA) = \mathscr{R}(S)\).

Use results from (a) to prove the equality

Now, we use the results from part (a) to prove the equality: $$\operatorname{dim}\mathcal{N}(A) + \operatorname{dim}\mathscr{A}(A) = n$$ Recall that \(SA\) is a projection matrix whose null space is \(\mathcal{N}(A)\) and range space is \(\mathscr{R}(S)\). As a projection matrix, it maps each vector in \(R^\circ\) to a vector in its range. Moreover, \(S : R^\circ \to R^\circ\). Thus, we have \(\operatorname{dim}\mathcal{N}(A) + \operatorname{dim}\mathscr{R}(S) = n\). Now we only need to show that \(\operatorname{dim}\mathscr{R}(S)\) is equal to \(\operatorname{dim}\mathscr{A}(A)\). Since \(S\) is a matrix that satisfies \(AS = I_r\), we conclude that \(S\) acts as an isomorphism between \(\mathscr{A}(A)\) and \(R^\circ\). Hence, \(\operatorname{dim}\mathscr{A}(A) = \operatorname{dim}\mathscr{R}(S)\). Therefore, we have: $$\operatorname{dim}\mathcal{N}(A) + \operatorname{dim}\mathscr{A}(A) = \operatorname{dim}\mathcal{N}(A) + \operatorname{dim}\mathscr{R}(S) = n$$

Key concepts

Projection matrix

Understanding the concept of a Projection Matrix is vital for students of linear algebra. In essence, a projection matrix operates on a vector space in a way that projects a vector onto a subspace. A matrix is classified as a projection matrix if it satisfies the idempotency condition, which translates to the equation \(P^2 = P\). In practical terms, this means that applying the projection twice yields the same result as applying it once.

To visualize this, imagine shining a light on an object; the shadow it casts serves as a 'projection' onto the surface. Analogously, a projection matrix 'casts' vectors onto a subspace. The benefits of grasping this analogy can simplify the complex nature of vector transformations into more relatable concepts. Specifically, for a given matrix \(A\), the projection matrix \(SA\) created from \(S\) projects vectors onto the subspace spanned by \(A\)'s columns.

Rank of a matrix

The Rank of a Matrix is a fundamental characteristic that reveals the dimension of the column space (also known as the range) of the matrix, indicating the maximum number of independent column vectors. Essentially, rank identifies the number of dimensions in the image of the matrix transformation.

In our example, considering a matrix \(A\) and its rank \(r\), this measurement communicates the number of independent vectors after \(A\) has been applied to the original vector space. For students to truly appreciate the concept of rank, it's helpful to associate it with the concept of information in the system: the rank tells us how much unique information is present in a matrix, and therefore, how much the matrix can 'tell' us about the space it acts on.

Null space

The Null Space of a Matrix, denoted as \(\mathcal{N}(A)\), consists of all vectors that, when transformed by the matrix \(A\), result in the zero vector. In less technical terms, it's a collection of vectors that essentially 'disappear' under the transformation applied by the matrix.

Understanding this concept can be likened to solving a mystery: identifying the null space is akin to finding all the scenarios that lead to the same outcome - in this case, the zero vector. This is significant because it gives insights into the solutions of homogeneous systems of linear equations, which are represented by the equation \(Ax = 0\). Determining the null space is a critical step in understanding linear transformations and their properties.

Dimension theorem

The Dimension Theorem is a linchpin in linear algebra, presenting a fundamental relationship between the dimensions of a vector space. It states that for any finite-dimensional vector space, the dimension of the null space plus the dimension of the column space (or rank) equals the number of columns of the matrix, which can be written as \(\operatorname{dim}\mathcal{N}(A) + \operatorname{dim}\mathscr{R}(A) = n\).

Consider it a balancing act - it guarantees that as the dimension of the null space increases, the rank must correspondingly decrease to maintain equilibrium. Sending students down the path of conceptualizing dimensions as resources can provide a more accessible approach to this theorem: as resources are allocated to the null space, they must be drawn from the column space, and the total 'budget' remains fixed.