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If a \(3 \times 3\) matrix \(A\) represents the projection onto a plane in \(\mathbb{R}^{3},\) what is rank \((A) ?\)

Short Answer

Expert verified
The rank of the projection matrix \(A\) is 2.

Step by step solution

01

Understanding Projection Matrices

A projection matrix is a square matrix that gives a vector space projection from \(R^n\) to a subspace. The rank of a projection matrix is equal to the dimension of the subspace onto which it projects.
02

Interpreting the Given Problem

Since matrix \(A\) is a projection onto a plane in \(\mathbb{R}^{3},\) it reduces the dimensionality of any vector in \(\mathbb{R}^{3}\) onto a two-dimensional subspace (the plane).
03

Determining the Rank of the Projection Matrix

The rank of matrix \(A\) is the dimension of the plane onto which vectors are projected, which in \(\mathbb{R}^{3}\) is 2. Thus, rank \(A = 2\).

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Key Concepts

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

Projection Matrix
Understanding the concept of a projection matrix is crucial when dealing with linear transformations in vector spaces. Imagine you have a vector in a high dimensional space, like \(\mathbb{R}^{3}\), and you want to 'project' it onto a plane or line within that space. That's where a projection matrix comes to play.

A projection matrix is a type of square matrix that when multiplied by a vector in the space, \(\mathbb{R}^{n}\), results in a new vector that lies on a certain lower-dimensional subspace, such as a line or a plane. This is analogous to casting a shadow of the original vector onto the subspace. It's an idempotent matrix, meaning when applied twice to any vector, the result will be the same as a single application \(P^{2}=P\). The rank of this matrix tells us the dimensionality of the 'shadow', or more formally, the dimension of the image subspace onto which it projects.
Vector Space Projection
Vector space projection is the process of transforming a vector in \(\mathbb{R}^{n}\) onto a subspace within that same space. This concept is akin to beaming a light onto an object and observing the shadow it casts onto a surface, with the shadow representing the projection of the object.

Projection onto a subspace can be visualized by considering a line or plane within a higher-dimensional space. Any vector in the space can be 'broken down' or decomposed into two components - one that lies on the subspace and another that is perpendicular to it. The part on the subspace is the 'projection' of the vector. The key here is that projection maintains all the properties of vector addition and scalar multiplication, which is why it's such a fundamental concept in linear algebra. Understanding projections helps us to solve various problems, including least squares and optimizing functions subject to constraints.
Dimension of a Subspace
The dimension of a subspace is a measure of its 'size' in terms of the minimum number of vectors needed to span the space. In more colloquial terms, think of it as the number of independent directions you can move within that space.

For instance, a line is a one-dimensional subspace since you need only one vector to define any point on the line. Similarly, a plane is a two-dimensional subspace because it requires two vectors to describe any point within the plane. Continuing this analogy, in \(\mathbb{R}^{3}\), if a third, non-coplanar vector is added, we now have a three-dimensional subspace which encompasses the entire space. In the context of the projection matrix, the rank of the matrix directly correlates to the dimension of the subspace. Furthermore, for a projection onto a plane in \(\mathbb{R}^{3}\), this planar subspace is necessarily two-dimensional, because that's the minimum number of vectors needed to define a plane. Therefore, the rank of the projection matrix that projects vectors onto this plane is also two.

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Most popular questions from this chapter

In Exercises 19 through \(24,\) find the matrix \(B\) of the linear transformation \(T(\vec{x})=A \vec{x}\) with respect to the basis \(\mathfrak{B}=\left(\vec{v}_{1}, \vec{v}_{2}\right) .\) For practice, solve each problem in three ways: (a) Use the formula \(B=S^{-1} A S,(b)\) use a commutative diagram (as in Examples 3 and 4), and (c) construct \(\boldsymbol{B}^{*}\)column by column." $$A=\left[\begin{array}{rr} 13 & -20 \\ 6 & -9 \end{array}\right] ; \vec{v}_{1}=\left[\begin{array}{l} 2 \\ 1 \end{array}\right], \vec{v}_{2}=\left[\begin{array}{l} 5 \\ 3 \end{array}\right]$$

Consider the matrices $$A=\left[\begin{array}{llllll} 1 & 0 & 2 & 0 & 4 & 0 \\ 0 & 1 & 3 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 & 6 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right]$$ and $$B=\left[\begin{array}{llllll} 1 & 0 & 2 & 0 & 4 & 0 \\ 0 & 1 & 3 & 0 & 5 & 0 \\ 0 & 0 & 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right]$$ Show that the kernels of matrices \(A\) and \(B\) are different. Hint: Think about ways to write the fifth column as a linear combination of the preceding columns.

Consider an \(n \times n\) matrix \(A .\) Show that there exist scalars \(c_{0}, c_{1}, \ldots, c_{n}\) (not all zero) such that the ma\(\operatorname{trix} c_{0} I_{n}+c_{1} A+c_{2} A^{2}+\cdots+c_{n} A^{n}\) is noninvertible. Hint: Pick an arbitrary nonzero vector \(\vec{v}\) in \(\mathbb{R}^{n}\). Then the \(n+1\) vectors \(\vec{v}, A \vec{v}, A^{2} \vec{v}, \ldots, A^{n} \vec{v}\) will be linearly dependent. (Much more is true: There are scalars \(c_{0}, c_{1}, \ldots, c_{n},\) not all zero, such that \(c_{0} I_{n}+\) \(c_{1} A+c_{2} A^{2}+\cdots+c_{n} A^{n}=0 .\) You are not asked to demonstrate this fact here.)

Consider an \(n \times p\) matrix \(A\) and a \(p \times m\) matrix \(B\) a. What is the relationship between \(\operatorname{ker}(A B)\) and \(\operatorname{ker}(B) ?\) Are they always equal? Is one of them always contained in the other? b. What is the relationship between im(A) and \(\operatorname{im}(A B) ?\)

Show that if a \(3 \times 3\) matrix \(A\) represents the reflection about a plane, then \(A\) is similar to the matrix \(\left[\begin{array}{rrr}1 & 0 & 0 \\\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right].\)

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