Projection Matrix: Expressing Operator with Vectors

[Robin04]

If we express the projection operator with vectors, we get $\hat{P}\vec{v} = \vec{e}(\vec{e}\vec{v})$ which means that we project $\vec{v}$ onto $\vec{e}$. We can write this as $\hat{P}\vec{v} = e_k \sum_{l} e_lv_l = \sum_l (e_ke_l )v_l$. In my class we said that the matrix for the projection operator is $P_ {kl}=e_ke_l$, so $\hat{P}\vec{v}=\sum_l P_{kl} v_l$. But isn't $e_ke_l$ equal to $\delta_{kl}$?

 

[fresh_42]

If we express the projection operator with vectors, we get $\hat{P}\vec{v} = \vec{e}(\vec{e}\vec{v})$ which means that we project $\vec{v}$ onto $\vec{e}$. We can write this as $\hat{P}\vec{v} = e_k \sum_{l} e_lv_l = \sum_l (e_ke_l )v_l$. In my class we said that the matrix for the projection operator is $P_ {kl}=e_ke_l$, so $\hat{P}\vec{v}=\sum_l P_{kl} v_l$. But isn't $e_ke_l$ equal to $\delta_{kl}$?

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?

 

[Robin04]

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?

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?

 

[fresh_42]

I realized what I misunderstood: I thought that $e_ke_l$ is the scalar product of the basis vectors,...

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

... but they're are components of a single basis vector into which we project.

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 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?

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 $\vec{e}=\sum_j c_je_j\,?$

 

[eys_physics]

I believe, that it should be $P=\vec{e}\vec{e}^T$, where $\vec{e}$ is a basis vector, e.g.
$$\vec{e}=\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}, \quad \vec{e}^T=\begin{pmatrix}1 & 0 & 0 \end{pmatrix} \rightarrow P=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$.

 

[WWGD]

If we express the projection operator with vectors, we get $\hat{P}\vec{v} = \vec{e}(\vec{e}\vec{v})$ which means that we project $\vec{v}$ onto $\vec{e}$. We can write this as $\hat{P}\vec{v} = e_k \sum_{l} e_lv_l = \sum_l (e_ke_l )v_l$. In my class we said that the matrix for the projection operator is $P_ {kl}=e_ke_l$, so $\hat{P}\vec{v}=\sum_l P_{kl} v_l$. But isn't $e_ke_l$ equal to $\delta_{kl}$?

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

 

[Robin04]

I believe, that it should be $P=\vec{e}\vec{e}^T$, where $\vec{e}$ is a basis vector, e.g.
$$\vec{e}=\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}, \quad \vec{e}^T=\begin{pmatrix}1 & 0 & 0 \end{pmatrix} \rightarrow P=\begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$.

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

... 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.

 

[fresh_42]

We also called this dyadic product.

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.