- #1

Robin04

- 260

- 16

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- I
- Thread starter Robin04
- Start date

- #1

Robin04

- 260

- 16

- #2

- 17,607

- 18,183

Only in case the ##e_i=(\delta_{ij})_j##. If we project onto a basis vector, we will get its component ##v_k##, so? And what is ##\vec{e}##, i.e. which coordinates does it have?

- #3

Robin04

- 260

- 16

I realized what I misunderstood: I thought that ##e_ke_l## is the scalar product of the basis vectors, but they're are components of a single basis vector into which we project. But how can ##P_{kl}## be ##\delta_{kl}##? I can't imagine it. You said it is only in the case when ##e_i=(\delta{ij})_j##, but what does the last ##j## index mean?Only in case the ##e_i=(\delta_{ij})_j##. If we project onto a basis vector, we will get its component ##v_k##, so? And what is ##\vec{e}##, i.e. which coordinates does it have?

- #4

- 17,607

- 18,183

... which I think it exactly is ...I realized what I misunderstood: I thought that ##e_ke_l## is the scalar product of the basis vectors,...

Not sure what you mean, as you haven't said what ##e_k## are, or what ##\vec{e}## should be. You have a general formula plus a calculation with coordinates without saying what your vectors according to this basis are. So there is necessarily guesswork going on.... but they're are components of a single basis vector into which we project.

Say we have an ##n-##dimensional vector space with basis vectors ##(1,0,\ldots,0)\, , \,(0,1,0,\ldots, 0) \, , \,\ldots## Then ##(\delta_{ij})_j = (\delta_{ij})_{1\leq j\leq n}= (\delta_{i1},\ldots,\delta_{ij},\ldots,\delta_{in})## is another way to write them. I suspect that the ##e_k## are exactly those vectors: ##e_k=(\delta_{kl})_l## where ##k## is fixed and ##l## runs from ##1## to ##n##. Of course this still doesn't explain what ##\vec{e}## is - one of them or ##\vec{e}=\sum_j c_je_j\,?##But how can ##P_{kl}## be ##\delta_{kl}##? I can't imagine it. You said it is only in the case when ##e_i=(\delta{ij})_j##, but what does the last ##j## index mean?

- #5

eys_physics

- 268

- 72

$$\vec{e}=\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}, \quad \vec{e}^T=\begin{pmatrix}1 & 0 & 0 \end{pmatrix} \rightarrow P=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$.

- #6

WWGD

Science Advisor

Gold Member

- 6,207

- 7,746

Are you projecting onto the axes? You may project onto lines, planes, etc. and not necessarily orthogonally .

- #7

Robin04

- 260

- 16

Yes, that's what I missed, thank you! We also called this dyadic product.

$$\vec{e}=\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}, \quad \vec{e}^T=\begin{pmatrix}1 & 0 & 0 \end{pmatrix} \rightarrow P=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$.

... which I think it exactly is ...

Not sure what you mean, as you haven't said what ##e_k## are, or what ##\vec{e}## should be. You have a general formula plus a calculation with coordinates without saying what your vectors according to this basis are. So there is necessarily guesswork going on.

We project onto a line given by the vector ##\vec{e}## and ##e_k## are its components.

Are you projecting onto the axes? You may project onto lines, planes, etc. and not necessarily orthogonally .

Yes, we learned those too, I just had some trouble understanding this particular case.

Last edited:

- #8

- 17,607

- 18,183

A dyadic (outer) product is the tensor product of two vectors: ##u\otimes v##. In coordinates you can achieve the result as the matrix multiplication column times row: ##u \cdot v^\tau.## It is necessarily a matrix of rank one.We also called this dyadic product.

Share:

- Last Post

- Replies
- 4

- Views
- 601

- Replies
- 5

- Views
- 220

- Last Post

- Replies
- 9

- Views
- 609

- Last Post

- Replies
- 1

- Views
- 518

- Last Post

- Replies
- 3

- Views
- 634

- Replies
- 1

- Views
- 98

- Replies
- 1

- Views
- 339

- Last Post

- Replies
- 12

- Views
- 468

- Last Post

- Replies
- 3

- Views
- 709

- Last Post

- Replies
- 10

- Views
- 525