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Linear Algebra 6.2.2 Orthogonal Projections

SawFin Mathematics · YouTube · 8:44

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Use orthogonal projection onto a line to express a vector as two perpendicular components.

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Chapters

0:00Projection onto a line3:00A worked projection7:30Verify orthogonality

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For the line spanned by a nonzero v, the projection of y is [(y⋅vy\cdot v)/(v⋅vv\cdot v)]v. The residual y minus this projection is perpendicular to the line.

With y=(7,6)y=(7,6) and u=(2,1)u=(2,1), the coefficient is 20/5=420/5=4. Thus the projected vector is (8,4), which lies in the span of u.

The residual is (−1,2), and (8,4)·(−1,2)=−8+8=08+8=0. Adding the projected vector and residual recovers (7,6).

Knowledge cards

01

Projection onto a line

For the line spanned by a nonzero v, the projection of y is [(y⋅vy\cdot v)/(v⋅vv\cdot v)]v. The residual y minus this projection is perpendicular to the line.

02

A worked projection

With y=(7,6)y=(7,6) and u=(2,1)u=(2,1), the coefficient is 20/5=420/5=4. Thus the projected vector is (8,4), which lies in the span of u.

03

Verify orthogonality

The residual is (−1,2), and (8,4)·(−1,2)=−8+8=08+8=0. Adding the projected vector and residual recovers (7,6).

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  • Orthogonal projections ExplanationAt 0:00
    Why this connection?

    For the line spanned by a nonzero v, the projection of y is [(y⋅vy\cdot v)/(v⋅vv\cdot v)]v. The residual y minus this projection is perpendicular to the line. With y=(7,6)y=(7,6) and u=(2,1)u=(2,1), the coefficient is 20/5=420/5=4. Thus the projected vector is (8,4), which lies in the span of u. The residual is (−1,2), and (8,4)·(−1,2)=−8+8=08+8=0. Adding the projected vector and residual recovers (7,6).

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