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Let S and T be subspaces in Rⁿ such that S⊂T. Let Pₛ and Pₜ represent the orthogonal projectors onto S and T, respectively. (a) Prove that Pₛ Pₜ=Pₛ. (b) Show tha Pₜ Pₛ is not necessarily equal to Pₛ and provide a counterexample to illustrate this. (c) Explain the concept of an orthogonal projector in the context of subspaces and how it is used in linear algebra.

User Gerard Yin
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Final answer:

An orthogonal projector onto a subspace S is a linear transformation that projects any vector onto S in such a way that the projected vector is orthogonal to the complement of S. (a) We can prove that Pₛ Pₜ=Pₛ, (b) but Pₜ Pₛ is not necessarily equal to Pₛ. (c) Orthogonal projectors are used to decompose vectors into components that lie within a subspace and components that lie outside the subspace.

Step-by-step explanation:

(a) Prove that Pₛ Pₜ=Pₛ:

To prove this, we need to show that the composition of the orthogonal projectors onto S and T, denoted as Pₛ Pₜ, is equal to the orthogonal projector onto S, denoted as Pₛ.

Since S⊂T, any vector projected onto S will also be projected onto T. Therefore, Pₛ Pₜ will project any vector onto T and then onto S, which is equivalent to just projecting it onto S. Thus, Pₛ Pₜ=Pₛ.

(b) Show that Pₜ Pₛ is not necessarily equal to Pₛ:

To show this, we need a counterexample. Let's consider R² with S=span{(1,0)} and T=span{(1,1)}. In this case, Pₛ projects any vector onto the x-axis, while Pₜ projects any vector onto the line y=x. Therefore, Pₜ Pₛ will project any vector onto the line y=x and then onto the x-axis, which is equivalent to just projecting it onto the x-axis. Thus, Pₜ Pₛ=Pₛ. Hence, Pₛ Pₜ is not necessarily equal to Pₛ.

(c) Explain the concept of an orthogonal projector in the context of subspaces and how it is used in linear algebra:

An orthogonal projector onto a subspace S is a linear transformation that projects any vector onto S in such a way that the projected vector is orthogonal to the complement of S. In linear algebra, orthogonal projectors are used to decompose vectors into components that lie within a subspace and components that lie outside the subspace. This decomposition is useful for solving systems of linear equations, finding least squares solutions, and studying properties of subspaces.

User Lajos Veres
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