Let 𝐮1, …, 𝐮k be an orthonormal basis for the subspace W ⊂ℝ^m. Let A=(𝐮1 𝐮2 …𝐮k) be the m ×k mat

step 1 In this problem, we have an orthogonal set of vectors, and we are told that these vectors are al

Let 𝐮1, …, 𝐮k be an orthonormal basis for the subspace W...

Let $\mathbf{u}_1, \ldots, \mathbf{u}_k$ be an orthonormal basis for the subspace $W \subset \mathbb{R}^m$. Let $A=\left(\mathbf{u}_1 \mathbf{u}_2 \ldots \mathbf{u}_k\right)$ be the $m \times k$ matrix whose columns are the orthonormal basis vectors, and define $P=A A^T$ to be the corresponding projection matrix. (a) Given $\mathbf{v} \in \mathbb{R}^n$, prove that its orthogonal projection $\mathbf{w} \in W$ is given by matrix multiplication: $\mathbf{w}=P \mathbf{v}$. (b) Prove that $P=P^T$ is symmetric. (c) Prove that $P$ is idempotent: $P^2=P$. Give a geometrical explanation of this fact. (d) Prove that $\operatorname{rank} P=k$. (e) Write out the projection matrix corresponding to the subspaces spanned by (i) $\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$, (ii) $\left(\begin{array}{r}\frac{2}{3} \\ -\frac{2}{3} \\ \frac{1}{3}\end{array}\right)$ (iii) $$ \left(\begin{array}{c} \frac{1}{\sqrt{6}} \\ -\frac{2}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \end{array}\right),\left(\begin{array}{c} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array}\right) $$ (iv) $$ \left(\begin{array}{r} \frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \end{array}\right), \quad\left(\begin{array}{r} \frac{1}{2} \\ -\frac{1}{2} \\ \frac{1}{2} \\ \frac{1}{2} \end{array}\right), \quad\left(\begin{array}{r} \frac{1}{2} \\ \frac{1}{2} \\ -\frac{1}{2} \\ \frac{1}{2} \end{array}\right) . $$

Let $\mathbf{u}_1, \ldots, \mathbf{u}_k$ be an orthonormal basis for the subspace $W \subset \mathbb{R}^m$. Let $A=\left(\mathbf{u}_1 \mathbf{u}_2 \ldots \mathbf{u}_k\right)$ be the $m \times k$ matrix whose columns are the orthonormal basis vectors, and define $P=A A^T$ to be the corresponding projection matrix. (a) Given $\mathbf{v} \in \mathbb{R}^n$, prove that its orthogonal projection $\mathbf{w} \in W$ is given by matrix multiplication: $\mathbf{w}=P \mathbf{v}$.
(b) Prove that $P=P^T$ is symmetric. (c) Prove that $P$ is idempotent: $P^2=P$. Give a geometrical explanation of this fact. (d) Prove that $\operatorname{rank} P=k$. (e) Write out the projection matrix corresponding to the subspaces spanned by
(i) $\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$,
(ii)
$\left(\begin{array}{r}\frac{2}{3} \\ -\frac{2}{3} \\ \frac{1}{3}\end{array}\right)$
(iii)
$$
\left(\begin{array}{c}
\frac{1}{\sqrt{6}} \\
-\frac{2}{\sqrt{6}} \\
\frac{1}{\sqrt{6}}
\end{array}\right),\left(\begin{array}{c}
\frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{3}} \\
\frac{1}{\sqrt{3}}
\end{array}\right)
$$
(iv)
$$
\left(\begin{array}{r}
\frac{1}{2} \\
\frac{1}{2} \\
\frac{1}{2} \\
-\frac{1}{2}
\end{array}\right), \quad\left(\begin{array}{r}
\frac{1}{2} \\
-\frac{1}{2} \\
\frac{1}{2} \\
\frac{1}{2}
\end{array}\right), \quad\left(\begin{array}{r}
\frac{1}{2} \\
\frac{1}{2} \\
-\frac{1}{2} \\
\frac{1}{2}
\end{array}\right) .
$$

Key Concepts

Geometrical Interpretation of Projection

The geometric interpretation of projection, particularly in terms of projection matrices, is that the operation maps any given vector to the 'shadow' it casts onto the subspace. This 'shadow' is nothing else than the closest point in the subspace to the original vector. The idempotence property aligns with this interpretation: once the shadow is cast, further projecting it again does not alter its position.

Matrix Rank

The rank of a matrix is the dimension of its column space, representing the number of linearly independent columns. For a projection matrix constructed from an orthonormal basis of a k-dimensional subspace, the rank is k. This reflects the fact that the projection does not change the dimensionality of vectors in the subspace while reducing the dimensionality of vectors outside of it.

Idempotent Matrix

An idempotent matrix is one that satisfies the condition P^2 = P, meaning that once the projection operation is applied, applying it again does not change the result. This property characterizes projection matrices because once a vector is projected onto the subspace, further projections leave it unchanged. Idempotency is a hallmark of many operators that are designed to 'stabilize' a particular subspace.

Orthogonal Projection

Orthogonal projection is the process of finding the closest point within a subspace to a given point in the ambient space. This operation minimizes the distance between the given point and points in the subspace and is fundamental in least squares approximations, signal processing, and optimization. It involves decomposing a vector into a component in the subspace and another component orthogonal to the subspace.

Projection Matrix

A projection matrix, especially one of the form P = AA^T where the columns of A form an orthonormal basis for the subspace, is a linear operator that maps any vector to its projection onto that subspace. This matrix encapsulates the operation of orthogonally projecting vectors, and it plays a critical role in numerical methods, computer graphics, and various applications involving linear transformations.

Orthonormal Basis

An orthonormal basis is a set of vectors that are all mutually orthogonal and each of unit length. This property greatly simplifies many linear algebra computations, including those involving projections, because the inner products between different basis vectors vanish and the norm of each vector is one. They are especially useful in constructing projection matrices and simplifying matrix decompositions.

Symmetric Matrix

A symmetric matrix is one that is equal to its transpose, i.e., A = A^T. In the context of projection matrices, symmetry is a natural consequence of the projection being an orthogonal projection. Symmetric matrices have real eigenvalues and their eigenvectors can be chosen to be orthonormal, which makes them ideal for various decompositions in linear algebra.

Please give Ace some feedback