Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Orthogonal Projections

  • Last Updated : 16 Oct, 2021

Orthogonal Sets:

A set of vectors \left \{ u_1, u_2, ... u_p \right \}    in \mathbb{R^n}     is called orthogonal set, if u_i \cdot u_j =0    . if i \neq j

Orthogonal Basis

An orthogonal basis for a subspace W of \mathbb{R^n}     is a basis for W that is also an orthogonal set.

Let S = \left \{ u_1, u_2, ... u_p \right \}     be the orthogonal basis for a W of \mathbb{R^n}     is a basis for W that is also a orthogonal set. We need to calculate c_1, c_2, ... c_p     such that :

y = c_1 u_1 + c_2 u_2 + ... c_p u_p

Let’s take the dot product of u_1 both side.

y \cdot u_1 = (c_1 u_1 + c_2 u_2 + ... c_p u_p) \cdot u_1

y \cdot u_1 = c_1 (u_1 \cdot u_1) + c_2 (u_2 \cdot u_1) + ... c_p (u_p \cdot u_1)

Since, this is orthogonal basis u_2 \cdot u_1 = u_3 \cdot u_1 = ... =  u_p \cdot u_1 =0    . This gives c_1    :

c_1 = \frac{y \cdot u_1}{ u_1 \cdot u_1}

We can generalize the above equation

c_j = \frac{y \cdot u_j}{ u_j \cdot u_j}

Orthogonal Projections

Suppose {u_1, u_2,… u_n} is an orthogonal basis for W in \mathbb{R^n}    . For each y in W:

y =\left ( \frac{y \cdot u_1}{u_1 \cdot u_1} \right ) u_1 + ... +  \left ( \frac{y \cdot u_p}{u_p \cdot u_p } \right )u_p

Let’s take \left \{ u_1, u_2, u_3  \right \}     is an orthogonal basis for \mathbb{R^3}     and W = span \left \{ u_1, u_2 \right \}    . Let’s try to write a write y in the form \hat{y}     belongs to W space, and z that is orthogonal to W.

y =\left ( \frac{y \cdot u_1}{u_1 \cdot u_1} \right ) u_1  +  \left ( \frac{y \cdot u_2}{u_2\cdot u_2 } \right )u_2 + \left ( \frac{y \cdot u_3}{u_3 \cdot u_3} \right ) u_3  \\


\hat{y} = \left ( \frac{y \cdot u_1}{u_1 \cdot u_1} \right ) u_1  +  \left ( \frac{y \cdot u_2}{u_2\cdot u_2 } \right )u_2


z = \left ( \frac{y \cdot u_3}{u_3 \cdot u_3} \right ) u_3     [Tex]y= \hat{y} + z[/Tex]

Now, we can see that z is orthogonal to both u_1     and u_2     such that:

z \cdot u_1 =0 \\ z \cdot u_2 =0

Orthogonal Decomposition Theorem:

Let W be the subspace of \mathbb{R^n}    . Then each y in \mathbb{R^n}     can be uniquely represented in the form:

y = \hat{y} + z

where \hat{y}    is in W and z in W^{\perp}. If \left \{ u_1, u_2, ... u_p \right \}      is an orthogonal basis of W. then,

\hat{y} =\left ( \frac{y \cdot u_1}{u_1 \cdot u_1} \right ) u_1 + ... +  \left ( \frac{y \cdot u_p}{u_p \cdot u_p } \right )u_p


z = y - \hat{y}

Then, \hat{y}    is the orthogonal projection of y in W.

Best Approximation Theorem 

Let W is the subspace of \mathbb{R^n}    , y any vector in \mathbb{R^n}    . Let v in W and different from \hat{y}     . Then \left \| v-\hat{y} \right \|     also in W.

z = y - \hat{y}     is orthogonal to W, and also orthogonal to v=\hat{y}    .  Then y-v  can be written as:

y-v =  (y- \hat{y}) + (\hat{y} -v)


\left \| y-v \right \|^{2} = \left \| y- \hat{y} \right \|^{2} + \left \| \hat{y}-v \right \|^{2}

Thus, this can be written as:

\left \| y-v \right \|^{2} > \left \| y- \hat{y} \right \|^{2}


\left \| y-v \right \| > \left \| y- \hat{y} \right \|


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!