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- Thread starter Robin04
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fresh_42

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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?

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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?

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fresh_42

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... 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?

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

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

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

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

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

Yes, we learnt those too, I just had some trouble understanding this particular case.Are you projecting onto the axes? You may project onto lines, planes, etc. and not necessarily orthogonally .

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fresh_42

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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.

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