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Chapter 8: Problem 11

For the matrix \(P=I-A^{\mathrm{T}}\left(A A^{\mathrm{T}}\right)^{-1} A\), show that if \(x\) is in the nullspace of \(A\), then \(P x=x\). The nullspace stays unchanged under this projection.

Short Answer

Expert verified
If \(x\) is in the nullspace of \(A\), then \(Px = x\), showing nullspace is unchanged under \(P\).

Step by step solution

01

Understand the Nullspace Condition

The nullspace of a matrix \(A\) consists of all vectors \(x\) such that \(Ax = 0\). We need to check what happens to these vectors when matrix \(P\) is applied to them.
02

Analyze the Matrix Expression

The matrix \(P\) is given as \(P = I - A^{\mathrm{T}}(A A^{\mathrm{T}})^{-1}A\). We need to examine what this matrix does to vectors in the nullspace of \(A\).
03

Substitute the Nullspace Condition

Starting from \(Px = x\), substitute \(x\) such that \(Ax = 0\). The expression becomes \(Px = (I - A^{\mathrm{T}}(A A^{\mathrm{T}})^{-1}A)x = x\).
04

Simplify the Matrix Application

Since \(Ax = 0\), then \(A^{\mathrm{T}}(A A^{\mathrm{T}})^{-1}Ax = A^{\mathrm{T}}(A A^{\mathrm{T}})^{-1} \, 0 = 0\). So the expression \(P x = x - 0 = x\).
05

Conclude with the Projection Property

As shown in Step 4, when a vector \(x\) is in the nullspace of \(A\), \(Px = x\), meaning \(P\) leaves \(x\) unchanged. Thus, \(P\) retains nullspace vectors, which confirms that the nullspace remains unchanged under this projection.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nullspace
The nullspace of a matrix is a fundamental concept in linear algebra. It is the set of all vectors that, when multiplied by a given matrix, result in a zero vector. Specifically, for a matrix \( A \), the nullspace consists of vectors \( x \) such that \( Ax = 0 \). This can be written as:
  • The solution to \( Ax = 0 \) provides all the members of the nullspace.
  • The nullspace itself is a vector space and can be visualized as a line, plane, or higher-dimensional space, depending on the dimensions of \( A \).
Understanding the nullspace is essential for analyzing linear transformations, solving systems of linear equations, and many other applications. It tells us which vectors are "invisible" to the matrix transformation and can reveal redundancies in systems.
Matrix multiplication
Matrix multiplication is a fundamental operation that combines two matrices to produce a new matrix. The rules for multiplying matrices are straightforward but require attention to detail. Here is how it works:
  • Each element in the product matrix is the dot product of a row from the first matrix and a column from the second matrix.
  • The number of columns in the first matrix must equal the number of rows in the second matrix for multiplication to be valid.
  • The result is a matrix with the same number of rows as the first matrix and columns as the second matrix.
Matrix multiplication is not commutative, meaning \( AB \) is not necessarily equal to \( BA \). It plays a crucial role in various applications, including transforming geometric data and solving linear systems.
Projection matrix
A projection matrix is a special type of matrix used to "project" vectors onto a certain subspace. In linear algebra, if we have a matrix \( A \), the associated projection matrix \( P \) can be written as:\[ P = I - A^{\mathrm{T}}(A A^{\mathrm{T}})^{-1}A \]
  • The identity matrix \( I \) ensures that the projection matrix is properly defined.
  • Projection matrices are idempotent, meaning applying them multiple times has the same effect as applying them once.
  • They can "leave" parts of vectors unchanged while altering others to fit within a particular subspace.
In the context of the exercise, \( P \) is used to project onto the nullspace of \( A \), proving that vectors in this nullspace remain unchanged. This highlights the precise action of projection matrices in maintaining certain vector characteristics.
Orthogonal projection
Orthogonal projection refers to the process of projecting a vector onto another vector or subspace such that the resulting projection is perpendicular (orthogonal) to the orthogonal complement of that subspace. Mathematically, this often involves a projection matrix.
  • Orthogonal projections minimize the distance from the original vector to the projection.
  • They are especially valuable in solving least squares problems, where minimizing projection error is essential.
  • Implementing orthogonal projections using matrices involves using the transpose and inverse operations, as seen in the formula \( P = I - A^{\mathrm{T}}(A A^{\mathrm{T}})^{-1}A \).
Orthogonal projections can be applied in data analysis, computer graphics, and many fields requiring dimensionality reduction and simplification of data while retaining significant features.

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Most popular questions from this chapter

In three dimensions, can you find a set of six vectors whose cone of nonnegative combinations fills the whole space? What about four vectors?

Add a single inequality constraint to \(x \geq 0, y \geq 0\) such that the feasible set contains only one point.

(a) Suppose every row and every column contains exactly two \(1 \mathrm{~s}\). Prove that a complete matching is possible. (Show that the 1s cannot be covered by less than \(n\) lines.) (b) Find an example with two or more \(1 \mathrm{~s}\) in each row and column, for which a complete matching is impossible.

(a) Why does the greedy algorithm work for the spanning tree problem? (b) Show by example that the greedy algorithm could fail to find the shortest path from \(s\) to \(t\), by starting with the shortest edge.

Sketch the feasible set with constraints \(x+2 y \geq 6,2 x+y \geq 6, x \geq 0, y \geq 0\). What points lie at the three "corners" of this set?
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