Mixing
Using diagonality in Einstein notation
Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives
$$
D_{ij} = delta_{ijk} d_k
$$
where
$$
delta_{ijk} =
begin{cases}
1 & text{if } i = j = k \
0 & text{ otherwise}
end{cases}
$$
If I now apply this in a matrix multiplication, e.g.
$$
(AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
$$
or
$$
(ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
$$
The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.
This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:
- Where do I go wrong in my reasoning?
- Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?
linear-algebra index-notation
asked Nov 21 '18 at 11:24
user495268
183
add a comment |
Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives
$$
D_{ij} = delta_{ijk} d_k
$$
where
$$
delta_{ijk} =
begin{cases}
1 & text{if } i = j = k \
0 & text{ otherwise}
end{cases}
$$
If I now apply this in a matrix multiplication, e.g.
$$
(AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
$$
or
$$
(ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
$$
The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.
This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:
- Where do I go wrong in my reasoning?
- Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?
linear-algebra index-notation
asked Nov 21 '18 at 11:24
user495268
183
add a comment |
Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives
$$
D_{ij} = delta_{ijk} d_k
$$
where
$$
delta_{ijk} =
begin{cases}
1 & text{if } i = j = k \
0 & text{ otherwise}
end{cases}
$$
If I now apply this in a matrix multiplication, e.g.
$$
(AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
$$
or
$$
(ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
$$
The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.
This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:
- Where do I go wrong in my reasoning?
- Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?
linear-algebra index-notation
asked Nov 21 '18 at 11:24
user495268
183
Given a diagonal matrix $D$, with diagonal elements given by vector $mathbf{d}$. Representing this in Einstein notation gives
$$
D_{ij} = delta_{ijk} d_k
$$
where
$$
delta_{ijk} =
begin{cases}
1 & text{if } i = j = k \
0 & text{ otherwise}
end{cases}
$$
If I now apply this in a matrix multiplication, e.g.
$$
(AD)_{lj} = A_{li} D_{ij} = A_{li} delta_{ijk} d_k = A_{li} d_i
$$
or
$$
(ADA)_{ml} = A_{mi} D_{ij} A_{jl} = A_{mi} delta_{ijk} d_k A_{jl} = A_{mi} d_i A_{il}
$$
The first example makes sense entry-wise, but only if there is no summation over $i$. Also, the indices on the LHS and RHS do not match anymore. The same is true for the second example, but only if there is summation over all three $i$'s.
This obviously violates Einstein notation, but I don't see in which step a false assumption is made. My questions are therefore:
- Where do I go wrong in my reasoning?
- Is there some other way to exploit the fact that $D$ is diagonal (in index notation)?
linear-algebra index-notation
linear-algebra index-notation
asked Nov 21 '18 at 11:24
user495268
183
asked Nov 21 '18 at 11:24
user495268
183
asked Nov 21 '18 at 11:24
user495268
183
asked Nov 21 '18 at 11:24
user495268
183
asked Nov 21 '18 at 11:24
user495268
183
183
add a comment |
add a comment |
1 Answer
1
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oldest
votes
What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this
$$
D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
$$
which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option
begin{eqnarray}
(A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
end{eqnarray}
answered Nov 21 '18 at 20:40
caverac
13.8k21130
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this
$$
D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
$$
which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option
begin{eqnarray}
(A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
end{eqnarray}
answered Nov 21 '18 at 20:40
caverac
13.8k21130
add a comment |
What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this
$$
D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
$$
which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option
begin{eqnarray}
(A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
end{eqnarray}
answered Nov 21 '18 at 20:40
caverac
13.8k21130
add a comment |
What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this
$$
D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
$$
which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option
begin{eqnarray}
(A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
end{eqnarray}
answered Nov 21 '18 at 20:40
caverac
13.8k21130
What you're doing when calculating the value of $(AD)_{lj}$ is the equivalent of doing this
$$
D_{ij} = delta_{ijk}d_k color{red}{stackrel{!!}{=}} d_i
$$
which clearly shows the problem much earlier than you noticed: expanding the symbol $delta_{ijk}$ is the issue here. Einstein's notation is useful, but it doesn't mean you need to use it everywhere, here's an option
begin{eqnarray}
(A D)_{lj} &=& sum_{i}A_{li}color{blue}{D_{ij}} = sum_iA_{li}color{blue}{delta_{ij}d_j} = A_{lj}d_j ~~~mbox{(sum not implied)}
end{eqnarray}
answered Nov 21 '18 at 20:40
caverac
13.8k21130
answered Nov 21 '18 at 20:40
caverac
13.8k21130
answered Nov 21 '18 at 20:40
caverac
13.8k21130
answered Nov 21 '18 at 20:40
caverac
13.8k21130
13.8k21130
add a comment |
add a comment |
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