16.-Linear-Algebra-by-Kenneth-Hoffman · Problem 8.2.6

Inner Product Spaces

Mathematics — Linear Algebra · orthogonal-projection

Problem

Let WW be the subspace of R2R^2 spanned by the vector (3,4)(3, 4). Using the standard inner product, let EE be the orthogonal projection of R2R^2 onto WW. Find (a) a formula for E(x1,x2)E(x_1, x_2); (b) the matrix of EE in the standard ordered basis; (c) W⊥W^\perp; (d) an orthonormal basis in which EE is represented by the matrix [1000].\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}.
Intuitive explanation

Orthogonal projection onto a line through the origin is fundamentally about decomposing any vector into its "shadow" along that line and the perpendicular remainder. The key insight is that projecting (x1,x2)(x_1, x_2) onto the direction (3,4)(3, 4) uses the formula projv(x)=x⋅vv⋅vv\text{proj}_{\mathbf{v}}(\mathbf{x}) = \frac{\mathbf{x} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}\mathbf{v}—this scales the direction vector by how much the input "aligns" with it. Once you have the projection formula, the matrix follows immediately, the orthogonal complement is perpendicular to (3,4)(3,4), and choosing an orthonormal basis aligned with the line and its perpendicular gives you the diagonal form.

Intuitive explanation

Orthogonal projection onto a line through the origin is fundamentally about decomposing any vector into its "shadow" along that line and the perpendicular remainder. The key insight is that projecting (x1,x2)(x_1, x_2) onto the direction (3,4)(3, 4) uses the formula projv(x)=x⋅vv⋅vv\text{proj}_{\mathbf{v}}(\mathbf{x}) = \frac{\mathbf{x} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}\mathbf{v}—this scales the direction vector by how much the input "aligns" with it. Once you have the projection formula, the matrix follows immediately, the orthogonal complement is perpendicular to (3,4)(3,4), and choosing an orthonormal basis aligned with the line and its perpendicular gives you the diagonal form.

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