Question

Find the orthogonal projection of the given vector into the given spaces (described as a span of a basis).

( 1 ) ( 0 ) ( 0 ) ( 1 ) ( 1 ) ( 1 )
a. ( -2 ) into ( 2 ), ( 1 ) b. ( 2 ), into ( 0 ), ( 1 )
( 1 ) ( 1 ) (-1 ) (-1) (-1) ( 1 )
(-1) (-1) ( 1 )

Answer 1

The projection of (1, 0, 0, 1, 1, 1) onto the subspace spanned by (-2, 2, 1) is (0, 0, 0), and the projection of (2, 0) onto the subspace spanned by (0, 1) is (0, 0).

To find the orthogonal projection of a vector onto a subspace spanned by a basis, you can use the formula:

`\[ \text{Projection of } \mathbf{v} \text{ onto the subspace} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \cdot \mathbf{u} \]`

Where:

`\(\mathbf{v}\)` is the given vector.

`\(\mathbf{u}\)` is the basis vector of the subspace.

Projection of (1, 0, 0, 1, 1, 1) onto the subspace spanned by (-2, 2, 1):

Let's compute the projection:

Given vector
`\(\mathbf{v}` = (1, 0, 0, 1, 1, 1)

Basis vector
`\(\mathbf{u}` = (-2, 2, 1)

The projection formula:

`\[ \text{Projection} = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \cdot \mathbf{u} \]`

Calculate the dot products:

`\[ \mathbf{v} \cdot \mathbf{u} = (1 \cdot (-2)) + (0 \cdot 2) + (0 \cdot 1) + (1 \cdot 0) + (1 \cdot 1) + (1 \cdot 1)`=
`-2 + 0 + 0 + 0 + 1 + 1 = 0 \]`

`\[ \mathbf{u} \cdot \mathbf{u} = (-2)^2 + 2^2 + 1^2 = 4 + 4 + 1 = 9 \]`

Compute the projection:

`\[ \text{Projection} = (0)/(9) \cdot (-2, 2, 1) = (0, 0, 0) \]`

Projection of (2, 0) onto the subspace spanned by (0, 1):

Given vector
`\(\mathbf{v}` = (2, 0)

Basis vector
`\(\mathbf{u}` = (0, 1)

Using the same projection formula:

`\[ \mathbf{v} \cdot \mathbf{u} = (2 \cdot 0) + (0 \cdot 1) = 0 \]`

`\[ \mathbf{u} \cdot \mathbf{u} = 0^2 + 1^2 = 1 \]`

Compute the projection:

`\[ \text{Projection} = (0)/(1) \cdot (0, 1) = (0, 0) \]`