### Inner Product Spaces: Orthogonal projections

### Orthogonal complement

An important concept in the theory of orthogonality is the orthogonal complement. We will discuss what it is and how we can construct the orthogonal complement of a linear subspace in a vector space.

Orthogonal complement Let #V# be an inner product space and let #W# be a linear subspace of #V#. The **orthogonal complement** of #W# is the set

\[

W^\perp =\left\{\vec{x}\in V\mid \dotprod{\vec{x}}{\vec{w}}=0\ \text{ for all }\ \vec{w}\in W\right\}

\]

Here are some key properties of the orthogonal complement.

Properties of the orthogonal complement Let #W# be a linear subspace of the vector space #V#.

- #W^\perp# is a linear subspace of #V#.
- #W\cap W^\perp =\{\vec{0}\}#; that is, a linear subspace #W# and its orthogonal complement #W^\perp# only have the zero vector in common.
- If #W=\linspan{\vec{a}_1,\ldots ,\vec{a}_n} #, then

\[

W^\perp=\left\{\vec{x}\in V\mid \dotprod{\vec{a}_i}{\vec{x}}=0\ \text{ for }\ i=1,\ldots ,n\right\}

\]

If #V# is a finite-dimensional vector space, then we can use the *orthogonal projection* and the *Gram-Schmidt procedure* to calculate the orthogonal complement of a subspace #W#.

Dimension Formula for the orthogonal complement Let #W# be an #m#-dimensional subspace of an #n#-dimensional vector space #V#. Suppose that #\basis{\vec{a}_1,\ldots,\vec{a}_m}# is a basis of #W# and that this basis extended by #\basis{\vec{a}_{m+1},\ldots,\vec{a}_n}# is a basis for the entire space #V#.

Then, the *Gram-Schmidt procedure* applied to the basis #\basis{\vec{a}_{1},\ldots,\vec{a}_n}# of #V# gives an *orthonormal basis* #\basis{\vec{e}_1,\ldots ,\vec{e}_n}# for #V# such that #W=\linspan{\vec{e}_1,\ldots ,\vec{e}_m}# and #W^\perp=\linspan{\vec{e}_{m+1},\ldots ,\vec{e}_n}#.

In particular, \[ \dim{V}=\dim{W}+\dim{W^{\perp}}\]

\[-9 x+2 y+5 z=0\]

Give your answer in the form of a list of basis vectors.

The subspace #W# consists of all vectors # \rv{x,y,z}# of #\mathbb{R}^3# with the property \(-9 x+2 y+5 z=0\); that is, #\dotprod{\rv{-9,2,5}}{\rv{x,y,z}} = 0#. This means #W = { \linspan{\rv{-9,2,5}}}^{\perp}#. As a consequence

\[ W ^\perp= \left(\linspan{\rv{-9,2,5}}^{\perp}\right)^\perp =\linspan{\rv{-9,2,5}}\] Thus, a basis is given by the vector #\rv{-9,2,5}#. It remains for us to normalize this basis vector to achieve an orthonormal basis.

\[\frac{1}{\norm{\rv{-9,2,5}}} \cdot \rv{-9,2,5} ={{1}\over{\sqrt{110}}}\cdot\rv{ -9 , 2 , 5 }\] This way we find the answer #\left\{{{1}\over{\sqrt{110}}}\cdot\rv{ -9 , 2 , 5 }\right\}#.

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