Let VC4 Be The Vector Given By V1 I1 I Find The Mat

1 Let V C 4 Be The Vector Given By V 1 I 1 I Find The Mat

Let V ∈ ℂ⁴ be the vector given by v = (1, i, -1, -i). Find the matrix (with respect to the canonical basis on ℂ⁴) of the orthogonal projection P ∈ L(ℂ⁴) such that null(P) = {v}⊥.

Let U be the subspace of ℝ³ that coincides with the plane through the origin that is perpendicular to the vector n = (1, 1, 1) ∈ ℝ³. (a) Find an orthonormal basis for U. (b) Find the matrix (with respect to the canonical basis on ℝ³) of the orthogonal projection P ∈ L(ℝ³) onto U, i.e., such that range(P) = U.

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The problem posed encompasses two significant tasks in linear algebra: firstly, computing the orthogonal projection matrix onto the subspace in ℂ⁴ orthogonal to a given vector, and secondly, determining the orthogonal projection matrix onto a plane in ℝ³ perpendicular to a specific vector. Both problems require a solid understanding of orthogonal projections, orthonormal bases, and properties of linear transformations.

Orthogonal Projection in Complex Vector Space ℂ⁴

Given a vector v = (1, i, -1, -i) in ℂ⁴, we need to find the orthogonal projection matrix P such that the null space of P is exactly the orthogonal complement of v. Essentially, this means that P projects onto the subspace spanned by v. Since the null space of P is {v}⊥, the image of P must be the span of v. To construct P, we use the formula for the orthogonal projection onto a subspace spanned by a non-zero vector u:

P = (u u) / (u u)

where u* denotes the conjugate transpose of u. Here, u = v, so:

u* u = (1)(1) + (i)(-i) + (-1)(-1) + (-i)(i) = 1 + (i)(-i) + 1 + (-i)(i)

Note that (i)(-i) = -i² = -(-1) = 1, and similarly, (-i)(i) = -i² = 1. Therefore,:

u* u = 1 + 1 + 1 + 1 = 4

The conjugate transpose u* is:

u* = (1, -i, -1, i)

The matrix P is:

P = (1/4) u u = (1/4) [(1, i, -1, -i)ᵗ (1, -i, -1, i)]

Explicitly, the matrix components are computed as:

Calculating each entry yields:

Combining all entries and dividing by 4 yields the projection matrix P.

Orthogonal Projection onto a Plane in ℝ³

For the second task, the subspace U in ℝ³ is the plane perpendicular to n = (1, 1, 1). To find an orthonormal basis for U, we need vectors orthogonal to n and orthogonal among themselves. Since n is perpendicular to U, any vector u satisfying n ⋅ u = 0 lies within U. Choosing two linearly independent vectors orthogonal to n:

u₁ = (1, -1, 0)

u₂ = (1, 0, -1)

These vectors are orthogonal to n because:

n ⋅ u₁ = 11 + 1(-1) + 1*0 = 1 -1 + 0 = 0

n ⋅ u₂ = 11 + 10 + 1*(-1) = 1 + 0 - 1 = 0

Normalizing these vectors, we get:

e₁ = (1/√2, -1/√2, 0)

e₂ = (1/√3, 1/√3, -1/√3)

forming an orthonormal basis for U.

The projection matrix P projecting onto U is given by:

P = e₁ e₁ᵗ + e₂ e₂ᵗ

which can be computed explicitly by summing the rank-1 matrices formed by the outer products of these basis vectors.

Thus, the matrix P is:

P = (1/2) [[1, -1, 0], [-1, 1, 0], [0, 0, 0]] + (1/3) [[1, 1, -1], [1, 1, -1], [-1, -1, 1]]

Summing these two matrices gives the explicit projection matrix onto the subspace U in ℝ³.

In conclusion, these computations reveal the nature of orthogonal projections in both complex and real vector spaces, emphasizing the importance of orthonormal bases and the properties of projection matrices in linear algebra.

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