Please help me understand the following notation The 2019 Stack Overflow Developer Survey Results Are InWhy not use the identity matrix instead of the Kronecker delta?Matrix Transformation Onto?Need help understanding matrix norm notationDeterminants and the following questionLinear Algebra matrix notationUnderstanding an eigen decomposition notationHelp Determinant Binary MatrixStrange math notation, vertical bar with parentheses.Confused about diagonal matix notationInterpretation of Einstein notation for matrix multiplication
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Please help me understand the following notation
The 2019 Stack Overflow Developer Survey Results Are InWhy not use the identity matrix instead of the Kronecker delta?Matrix Transformation Onto?Need help understanding matrix norm notationDeterminants and the following questionLinear Algebra matrix notationUnderstanding an eigen decomposition notationHelp Determinant Binary MatrixStrange math notation, vertical bar with parentheses.Confused about diagonal matix notationInterpretation of Einstein notation for matrix multiplication
$begingroup$
Can someone kindly tell me the meaning of the following notation:
A book defined the following matrix $(a_ij)_3times 3$ :
$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$
where $d_ij$ are elements defined for $1le i,jle 3$.
I dont understand the case for $i=j$
Does it mean $d_11=d_11+d_12+d_13$?
linear-algebra matrices proof-verification proof-explanation
edited Mar 23 at 5:25
David G. Stork
12.2k41836
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
$endgroup$
add a comment |
$begingroup$
Can someone kindly tell me the meaning of the following notation:
A book defined the following matrix $(a_ij)_3times 3$ :
$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$
where $d_ij$ are elements defined for $1le i,jle 3$.
I dont understand the case for $i=j$
Does it mean $d_11=d_11+d_12+d_13$?
linear-algebra matrices proof-verification proof-explanation
edited Mar 23 at 5:25
David G. Stork
12.2k41836
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
$endgroup$
1
$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57
2
$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58
$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22
add a comment |
$begingroup$
Can someone kindly tell me the meaning of the following notation:
A book defined the following matrix $(a_ij)_3times 3$ :
$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$
where $d_ij$ are elements defined for $1le i,jle 3$.
I dont understand the case for $i=j$
Does it mean $d_11=d_11+d_12+d_13$?
linear-algebra matrices proof-verification proof-explanation
edited Mar 23 at 5:25
David G. Stork
12.2k41836
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
$endgroup$
Can someone kindly tell me the meaning of the following notation:
A book defined the following matrix $(a_ij)_3times 3$ :
$a_ij=begincases d_ij& ineq j\d_ii+sum_j=1^3 d_ij&i=jendcases$
where $d_ij$ are elements defined for $1le i,jle 3$.
I dont understand the case for $i=j$
Does it mean $d_11=d_11+d_12+d_13$?
linear-algebra matrices proof-verification proof-explanation
linear-algebra matrices proof-verification proof-explanation
edited Mar 23 at 5:25
David G. Stork
12.2k41836
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
edited Mar 23 at 5:25
David G. Stork
12.2k41836
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
edited Mar 23 at 5:25
David G. Stork
12.2k41836
edited Mar 23 at 5:25
David G. Stork
12.2k41836
edited Mar 23 at 5:25
David G. Stork
12.2k41836
12.2k41836
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
asked Mar 23 at 3:55
Math_FreakMath_Freak
535
535
1
$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57
2
$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58
$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22
add a comment |
1
$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57
2
$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58
$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22
1
1
$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57
$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57
2
2
$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58
$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58
$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22
$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
$endgroup$
add a comment |
$begingroup$
beginaligned
a_11 & = d_11 + d _11 + d_12 + d _13 \
a_12 & = d_12 \
a_13 & = d_13 \
a_21 & = d_21 \
a_22 & = d_22 + d _21 + d_22 + d _23 \
a_23 & = d_23 \
a_31 & = d_31 \
a_32 & = d_32 \
a_33 & = d_33 + d _31 + d_32 + d _33
endaligned
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
$endgroup$
add a comment |
$begingroup$
We have for $1leq i,jleq 3$
beginalign*
a_ij&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
endcasestag1\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
endcasestag2\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_k=1^3 d_ikquadqquad i=j
endcases\
endalign*
These variants are all valid and represent the same.
In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.
The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
$endgroup$
add a comment |
$begingroup$
I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
$endgroup$
add a comment |
$begingroup$
I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
$endgroup$
I agree with Theo Bendit's comment re: the reuse of a variable in the summation, as it tends to confuse people. Thus, I will rephrase it to give that, for $i = j$, the notation means $a_ii = d_ii + sum_k=1^3 d_ik = d_ii + d_i1 + d_i2 + d_i3$. Thus, for example, $a_11 = d_11 + d_11 + d_12 + d_13 = 2d_11 + d_12 + d_13$.
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
answered Mar 23 at 5:18
John OmielanJohn Omielan
4,8962216
4,8962216
add a comment |
add a comment |
$begingroup$
beginaligned
a_11 & = d_11 + d _11 + d_12 + d _13 \
a_12 & = d_12 \
a_13 & = d_13 \
a_21 & = d_21 \
a_22 & = d_22 + d _21 + d_22 + d _23 \
a_23 & = d_23 \
a_31 & = d_31 \
a_32 & = d_32 \
a_33 & = d_33 + d _31 + d_32 + d _33
endaligned
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
$endgroup$
add a comment |
$begingroup$
beginaligned
a_11 & = d_11 + d _11 + d_12 + d _13 \
a_12 & = d_12 \
a_13 & = d_13 \
a_21 & = d_21 \
a_22 & = d_22 + d _21 + d_22 + d _23 \
a_23 & = d_23 \
a_31 & = d_31 \
a_32 & = d_32 \
a_33 & = d_33 + d _31 + d_32 + d _33
endaligned
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
$endgroup$
add a comment |
$begingroup$
beginaligned
a_11 & = d_11 + d _11 + d_12 + d _13 \
a_12 & = d_12 \
a_13 & = d_13 \
a_21 & = d_21 \
a_22 & = d_22 + d _21 + d_22 + d _23 \
a_23 & = d_23 \
a_31 & = d_31 \
a_32 & = d_32 \
a_33 & = d_33 + d _31 + d_32 + d _33
endaligned
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
$endgroup$
beginaligned
a_11 & = d_11 + d _11 + d_12 + d _13 \
a_12 & = d_12 \
a_13 & = d_13 \
a_21 & = d_21 \
a_22 & = d_22 + d _21 + d_22 + d _23 \
a_23 & = d_23 \
a_31 & = d_31 \
a_32 & = d_32 \
a_33 & = d_33 + d _31 + d_32 + d _33
endaligned
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
answered Mar 23 at 5:44
MarianDMarianD
2,2611618
2,2611618
add a comment |
add a comment |
$begingroup$
We have for $1leq i,jleq 3$
beginalign*
a_ij&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
endcasestag1\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
endcasestag2\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_k=1^3 d_ikquadqquad i=j
endcases\
endalign*
These variants are all valid and represent the same.
In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.
The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
$begingroup$
We have for $1leq i,jleq 3$
beginalign*
a_ij&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
endcasestag1\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
endcasestag2\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_k=1^3 d_ikquadqquad i=j
endcases\
endalign*
These variants are all valid and represent the same.
In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.
The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
add a comment |
$begingroup$
We have for $1leq i,jleq 3$
beginalign*
a_ij&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
endcasestag1\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
endcasestag2\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_k=1^3 d_ikquadqquad i=j
endcases\
endalign*
These variants are all valid and represent the same.
In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.
The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
$endgroup$
We have for $1leq i,jleq 3$
beginalign*
a_ij&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_colorbluej=1^3 d_icolorbluejquadqquad i=j
endcasestag1\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+left(sum_colorbluej=1^3 d_icolorbluejright)quad,,,, i=j
endcasestag2\
&=
begincases
d_ijqquadqquadqquadqquad ineq j\
d_ii+sum_k=1^3 d_ikquadqquad i=j
endcases\
endalign*
These variants are all valid and represent the same.
In (1) and (2) readability is reduced somewhat due to the multiple usage of $j$, on the one hand as bound index variable and on the other hand as free variable.
The scope of the index variable $colorbluej$ is indicated in (2) by parenthesis.
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
answered Mar 23 at 18:01
Markus ScheuerMarkus Scheuer
64.1k460152
64.1k460152
add a comment |
add a comment |
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1
$begingroup$
What is k in this problem? You haven't defined it.
$endgroup$
– Don Thousand
Mar 23 at 3:57
2
$begingroup$
Also, not a fan of using $j$ as the dummy variable in the sum, given that $j = i$ is already fixed.
$endgroup$
– Theo Bendit
Mar 23 at 3:58
$begingroup$
@TheoBendit,i have edited the question,can you help now
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
@DonThousand,can you help now,i have edited it
$endgroup$
– Math_Freak
Mar 23 at 4:09
$begingroup$
Why the downvotes
$endgroup$
– Math_Freak
Mar 23 at 4:22