# Linear transformations and orthogonal basis

1. Mar 22, 2006

### stunner5000pt

Let {E1,E2,...En} be an orthogonal basis of Rn. Given k, 1<=k<=n, define Pk: Rn -> Rn by $P_{k} (r_{1} E_{1} + ... + r_{n} E_{n}) = r_{k} E_{k}.$ Show that $P_{k} = proj_{U} ()$ where U = span {Ek}

well $$\mbox{proj}_{U} \vec{m}= \sum_{i} \frac{ m \bullet u_{i}}{||u_{i}||^2} \vec{u}$$
right?
here we have Pk transforming linear combination of the orthogonal basis into rk Ek same index as the subscript of P

would it turn into
$$\mbox{proj}_{U} \vec{m}= \frac{ m \bullet E_{1}}{||E_{1}||^2}\vec{E_{1}} + ... + \frac{ m \bullet E_{n}}{||E_{n}||^2}\vec{E_{n}}$$
and the whole stuff in front of each Ei can be interpreted as the Ri, a scalar multiple yes?

2. Mar 22, 2006

### 0rthodontist

You can approach it with fewer symbols by simply noting that the orthogonal projection of a vector is the unique vector that is orthogonal to (the vector minus the projection fo the vector). That is, iff v is a vector and u is its orthogonal projection onto some subspace U, then (v - u) . u = 0, and u is in U.

3. Apr 6, 2006

### stunner5000pt

why is u = 0???

4. Apr 6, 2006

### Euclid

u is not 0
(v-u).u is 0 (i.e., v-u is orthogonal to u)

5. Apr 6, 2006

### stunner5000pt

ok so (v-u).u = 0
but how does this equal to the Pk?

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