Mixing
How to add vector with itself tranposed?
$begingroup$
So I'm solving basic linear algebra questions as part of review.
$$v=begin{bmatrix} 1 & 2 & 3 \ end{bmatrix}$$
When I do the operation $v+v^T$ according to matlab, numpy and wolfram alpha
spits out
$$v+v^T=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Originally I thought it would be like
$$ v + v^T = begin{bmatrix} 1 & 2 & 3 \ 1 & 2 & 3 \ 1 & 2 & 3\ end{bmatrix} + begin{bmatrix} 1 & 1 & 1 \ 2 & 2 & 2 \ 3 & 3 & 3\ end{bmatrix}=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Can someone explain to be how this makes any sense?
linear-algebra
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
$endgroup$
|
show 4 more comments
$begingroup$
So I'm solving basic linear algebra questions as part of review.
$$v=begin{bmatrix} 1 & 2 & 3 \ end{bmatrix}$$
When I do the operation $v+v^T$ according to matlab, numpy and wolfram alpha
spits out
$$v+v^T=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Originally I thought it would be like
$$ v + v^T = begin{bmatrix} 1 & 2 & 3 \ 1 & 2 & 3 \ 1 & 2 & 3\ end{bmatrix} + begin{bmatrix} 1 & 1 & 1 \ 2 & 2 & 2 \ 3 & 3 & 3\ end{bmatrix}=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Can someone explain to be how this makes any sense?
linear-algebra
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
$endgroup$
$begingroup$
How it makes sense? It does not. Your guess is probably correct as to how matlab interprets the mathematically meaningless expression.
$endgroup$
– darij grinberg
Jan 11 at 2:36
$begingroup$
The expression $v + v^t$ doesn't make sense. Maybe some systems will promote $v$ to a $3times 3$ matrix, but there's no mathematical justification for doing so.
$endgroup$
– anomaly
Jan 11 at 2:38
$begingroup$
Matrices can only be added if both dimensions agree. This is an illegal operation.
$endgroup$
– ncmathsadist
Jan 11 at 2:38
$begingroup$
Show us your alpha input.
$endgroup$
– Randall
Jan 11 at 2:38
$begingroup$
wolframalpha.com/input/… I'm looking at the alternative form
$endgroup$
– GrandFleet
Jan 11 at 2:40
|
show 4 more comments
$begingroup$
So I'm solving basic linear algebra questions as part of review.
$$v=begin{bmatrix} 1 & 2 & 3 \ end{bmatrix}$$
When I do the operation $v+v^T$ according to matlab, numpy and wolfram alpha
spits out
$$v+v^T=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Originally I thought it would be like
$$ v + v^T = begin{bmatrix} 1 & 2 & 3 \ 1 & 2 & 3 \ 1 & 2 & 3\ end{bmatrix} + begin{bmatrix} 1 & 1 & 1 \ 2 & 2 & 2 \ 3 & 3 & 3\ end{bmatrix}=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Can someone explain to be how this makes any sense?
linear-algebra
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
$endgroup$
So I'm solving basic linear algebra questions as part of review.
$$v=begin{bmatrix} 1 & 2 & 3 \ end{bmatrix}$$
When I do the operation $v+v^T$ according to matlab, numpy and wolfram alpha
spits out
$$v+v^T=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Originally I thought it would be like
$$ v + v^T = begin{bmatrix} 1 & 2 & 3 \ 1 & 2 & 3 \ 1 & 2 & 3\ end{bmatrix} + begin{bmatrix} 1 & 1 & 1 \ 2 & 2 & 2 \ 3 & 3 & 3\ end{bmatrix}=begin{bmatrix} 2 & 3 & 4\ 3 & 4 & 5\ 4 & 5 & 6\ end{bmatrix}$$
Can someone explain to be how this makes any sense?
linear-algebra
linear-algebra
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
asked Jan 11 at 2:34
GrandFleetGrandFleet
1083
1083
$begingroup$
How it makes sense? It does not. Your guess is probably correct as to how matlab interprets the mathematically meaningless expression.
$endgroup$
– darij grinberg
Jan 11 at 2:36
$begingroup$
The expression $v + v^t$ doesn't make sense. Maybe some systems will promote $v$ to a $3times 3$ matrix, but there's no mathematical justification for doing so.
$endgroup$
– anomaly
Jan 11 at 2:38
$begingroup$
Matrices can only be added if both dimensions agree. This is an illegal operation.
$endgroup$
– ncmathsadist
Jan 11 at 2:38
$begingroup$
Show us your alpha input.
$endgroup$
– Randall
Jan 11 at 2:38
$begingroup$
wolframalpha.com/input/… I'm looking at the alternative form
$endgroup$
– GrandFleet
Jan 11 at 2:40
|
show 4 more comments
$begingroup$
How it makes sense? It does not. Your guess is probably correct as to how matlab interprets the mathematically meaningless expression.
$endgroup$
– darij grinberg
Jan 11 at 2:36
$begingroup$
The expression $v + v^t$ doesn't make sense. Maybe some systems will promote $v$ to a $3times 3$ matrix, but there's no mathematical justification for doing so.
$endgroup$
– anomaly
Jan 11 at 2:38
$begingroup$
Matrices can only be added if both dimensions agree. This is an illegal operation.
$endgroup$
– ncmathsadist
Jan 11 at 2:38
$begingroup$
Show us your alpha input.
$endgroup$
– Randall
Jan 11 at 2:38
$begingroup$
wolframalpha.com/input/… I'm looking at the alternative form
$endgroup$
– GrandFleet
Jan 11 at 2:40
$begingroup$
How it makes sense? It does not. Your guess is probably correct as to how matlab interprets the mathematically meaningless expression.
$endgroup$
– darij grinberg
Jan 11 at 2:36
$begingroup$
How it makes sense? It does not. Your guess is probably correct as to how matlab interprets the mathematically meaningless expression.
$endgroup$
– darij grinberg
Jan 11 at 2:36
$begingroup$
The expression $v + v^t$ doesn't make sense. Maybe some systems will promote $v$ to a $3times 3$ matrix, but there's no mathematical justification for doing so.
$endgroup$
– anomaly
Jan 11 at 2:38
$begingroup$
The expression $v + v^t$ doesn't make sense. Maybe some systems will promote $v$ to a $3times 3$ matrix, but there's no mathematical justification for doing so.
$endgroup$
– anomaly
Jan 11 at 2:38
$begingroup$
Matrices can only be added if both dimensions agree. This is an illegal operation.
$endgroup$
– ncmathsadist
Jan 11 at 2:38
$begingroup$
Matrices can only be added if both dimensions agree. This is an illegal operation.
$endgroup$
– ncmathsadist
Jan 11 at 2:38
$begingroup$
Show us your alpha input.
$endgroup$
– Randall
Jan 11 at 2:38
$begingroup$
Show us your alpha input.
$endgroup$
– Randall
Jan 11 at 2:38
$begingroup$
wolframalpha.com/input/… I'm looking at the alternative form
$endgroup$
– GrandFleet
Jan 11 at 2:40
$begingroup$
wolframalpha.com/input/… I'm looking at the alternative form
$endgroup$
– GrandFleet
Jan 11 at 2:40
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Here is what Wolfram alpha is doing when you ask it for $v+v^T$ when $v$ is, say, a $3$-component vector. If $v= (a, b, c)$ it will interpret $v+v^T$ as a sequence of three separate "shifts" of $v^T$: first by $a$, then by $b$, then by $c$. It puts these three shifts as columns of a matrix (in order), giving a $3 times 3$ matrix as output. So, if you shift $v^T$ by $a$ you get $(2a,a+b, a+c)^T$, and this will be the first column of the output matrix. The second column is the shift by $b$ to get $(a+b, 2b, b+c)^T$, etc. If you do this for your example, you get your claimed output above. (I deduced this by following the step-by-step link in alpha where they show you what the computation $v+v^T$ is doing, so this is not speculation on my part.)
However, TO BE CLEAR, this is NOT traditional vector addition in any way whatsoever. This is certainly undefined in your case. (And all cases, except when your vectors have only one component, which is hardly a vector.) It would be defined for square matrices of the same size.
Moral: never trust output unless you REALLY know what the machine is doing.
answered Jan 11 at 2:47
RandallRandall
9,71111230
$endgroup$
$begingroup$
This is explained rather tersely in the Wolfram Language documentation forPlus: “Plusalso threads element-wise over lists.“ There’s one other thing that W|A does that contributes to this behavior: a one-level deep list can’t be transposed, so an input of $(a,b,c)^T$ gets interpreted asTranspose[{{a,b,c}}], with an extra layer of nesting. Moral: RTFD.
$endgroup$
– amd
Jan 11 at 8:40
add a comment |
$begingroup$
Can someone explain to be how this makes any sense?
It doesn't make any sense.
If you're talking about tansposition, then you're implicitly viewing your vectors as matrices with one of the dimensions being $1$.
Then your $v$ is a $1times 3$ matrix and $v^T$ is a $3times 1$ matrix.
Addition of matrices that don't have the same dimension is not defined.
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Hmm, then why does matlab, numpy and/or wolfram alpha return an answer?
$endgroup$
– GrandFleet
Jan 11 at 2:39
$begingroup$
@GrandFleet: Either brecause they're written to try their darned best to come up with some meaningful interpretation when they're given nonsense, even if that means they need to invent operations that are not there in the input -- or possibly because you're misunderstanding their input syntax and are actually asking them something different from what you think you're asking.
$endgroup$
– Henning Makholm
Jan 11 at 2:41
add a comment |
$begingroup$
A thought here as to why those systems would do that: there's an occasionally useful operation on a vector these systems support, namely adding a constant to each term. So then if $a$ is a constant and $v$ is a vector, the systems abbreviate this operator as $a+v$ or $v+a$. In effect, we promote $a$ to a vector with each element equal.
Now, if we're working with a row vector $u^T$ and a column vector $v$? In order to add the constants in each column of $u^T$ to $v$, we promote each of those length-1 columns to full length. In order to add the constants in each row of $v$ to $u^T$, we promote those length-1 rows to full length. This amounts to the same sum of matrices hypothesized in the question.
This is, of course, not standard vector/matrix addition.
answered Jan 11 at 2:50
jmerryjmerry
7,535921
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is what Wolfram alpha is doing when you ask it for $v+v^T$ when $v$ is, say, a $3$-component vector. If $v= (a, b, c)$ it will interpret $v+v^T$ as a sequence of three separate "shifts" of $v^T$: first by $a$, then by $b$, then by $c$. It puts these three shifts as columns of a matrix (in order), giving a $3 times 3$ matrix as output. So, if you shift $v^T$ by $a$ you get $(2a,a+b, a+c)^T$, and this will be the first column of the output matrix. The second column is the shift by $b$ to get $(a+b, 2b, b+c)^T$, etc. If you do this for your example, you get your claimed output above. (I deduced this by following the step-by-step link in alpha where they show you what the computation $v+v^T$ is doing, so this is not speculation on my part.)
However, TO BE CLEAR, this is NOT traditional vector addition in any way whatsoever. This is certainly undefined in your case. (And all cases, except when your vectors have only one component, which is hardly a vector.) It would be defined for square matrices of the same size.
Moral: never trust output unless you REALLY know what the machine is doing.
answered Jan 11 at 2:47
RandallRandall
9,71111230
$endgroup$
$begingroup$
This is explained rather tersely in the Wolfram Language documentation forPlus: “Plusalso threads element-wise over lists.“ There’s one other thing that W|A does that contributes to this behavior: a one-level deep list can’t be transposed, so an input of $(a,b,c)^T$ gets interpreted asTranspose[{{a,b,c}}], with an extra layer of nesting. Moral: RTFD.
$endgroup$
– amd
Jan 11 at 8:40
add a comment |
$begingroup$
Here is what Wolfram alpha is doing when you ask it for $v+v^T$ when $v$ is, say, a $3$-component vector. If $v= (a, b, c)$ it will interpret $v+v^T$ as a sequence of three separate "shifts" of $v^T$: first by $a$, then by $b$, then by $c$. It puts these three shifts as columns of a matrix (in order), giving a $3 times 3$ matrix as output. So, if you shift $v^T$ by $a$ you get $(2a,a+b, a+c)^T$, and this will be the first column of the output matrix. The second column is the shift by $b$ to get $(a+b, 2b, b+c)^T$, etc. If you do this for your example, you get your claimed output above. (I deduced this by following the step-by-step link in alpha where they show you what the computation $v+v^T$ is doing, so this is not speculation on my part.)
However, TO BE CLEAR, this is NOT traditional vector addition in any way whatsoever. This is certainly undefined in your case. (And all cases, except when your vectors have only one component, which is hardly a vector.) It would be defined for square matrices of the same size.
Moral: never trust output unless you REALLY know what the machine is doing.
answered Jan 11 at 2:47
RandallRandall
9,71111230
$endgroup$
$begingroup$
This is explained rather tersely in the Wolfram Language documentation forPlus: “Plusalso threads element-wise over lists.“ There’s one other thing that W|A does that contributes to this behavior: a one-level deep list can’t be transposed, so an input of $(a,b,c)^T$ gets interpreted asTranspose[{{a,b,c}}], with an extra layer of nesting. Moral: RTFD.
$endgroup$
– amd
Jan 11 at 8:40
add a comment |
$begingroup$
Here is what Wolfram alpha is doing when you ask it for $v+v^T$ when $v$ is, say, a $3$-component vector. If $v= (a, b, c)$ it will interpret $v+v^T$ as a sequence of three separate "shifts" of $v^T$: first by $a$, then by $b$, then by $c$. It puts these three shifts as columns of a matrix (in order), giving a $3 times 3$ matrix as output. So, if you shift $v^T$ by $a$ you get $(2a,a+b, a+c)^T$, and this will be the first column of the output matrix. The second column is the shift by $b$ to get $(a+b, 2b, b+c)^T$, etc. If you do this for your example, you get your claimed output above. (I deduced this by following the step-by-step link in alpha where they show you what the computation $v+v^T$ is doing, so this is not speculation on my part.)
However, TO BE CLEAR, this is NOT traditional vector addition in any way whatsoever. This is certainly undefined in your case. (And all cases, except when your vectors have only one component, which is hardly a vector.) It would be defined for square matrices of the same size.
Moral: never trust output unless you REALLY know what the machine is doing.
answered Jan 11 at 2:47
RandallRandall
9,71111230
$endgroup$
Here is what Wolfram alpha is doing when you ask it for $v+v^T$ when $v$ is, say, a $3$-component vector. If $v= (a, b, c)$ it will interpret $v+v^T$ as a sequence of three separate "shifts" of $v^T$: first by $a$, then by $b$, then by $c$. It puts these three shifts as columns of a matrix (in order), giving a $3 times 3$ matrix as output. So, if you shift $v^T$ by $a$ you get $(2a,a+b, a+c)^T$, and this will be the first column of the output matrix. The second column is the shift by $b$ to get $(a+b, 2b, b+c)^T$, etc. If you do this for your example, you get your claimed output above. (I deduced this by following the step-by-step link in alpha where they show you what the computation $v+v^T$ is doing, so this is not speculation on my part.)
However, TO BE CLEAR, this is NOT traditional vector addition in any way whatsoever. This is certainly undefined in your case. (And all cases, except when your vectors have only one component, which is hardly a vector.) It would be defined for square matrices of the same size.
Moral: never trust output unless you REALLY know what the machine is doing.
answered Jan 11 at 2:47
RandallRandall
9,71111230
edited Jan 11 at 2:54
answered Jan 11 at 2:47
RandallRandall
9,71111230
answered Jan 11 at 2:47
RandallRandall
9,71111230
answered Jan 11 at 2:47
RandallRandall
9,71111230
9,71111230
$begingroup$
This is explained rather tersely in the Wolfram Language documentation forPlus: “Plusalso threads element-wise over lists.“ There’s one other thing that W|A does that contributes to this behavior: a one-level deep list can’t be transposed, so an input of $(a,b,c)^T$ gets interpreted asTranspose[{{a,b,c}}], with an extra layer of nesting. Moral: RTFD.
$endgroup$
– amd
Jan 11 at 8:40
add a comment |
$begingroup$
This is explained rather tersely in the Wolfram Language documentation forPlus: “Plusalso threads element-wise over lists.“ There’s one other thing that W|A does that contributes to this behavior: a one-level deep list can’t be transposed, so an input of $(a,b,c)^T$ gets interpreted asTranspose[{{a,b,c}}], with an extra layer of nesting. Moral: RTFD.
$endgroup$
– amd
Jan 11 at 8:40
$begingroup$
This is explained rather tersely in the Wolfram Language documentation for
Plus: “Plus also threads element-wise over lists.“ There’s one other thing that W|A does that contributes to this behavior: a one-level deep list can’t be transposed, so an input of $(a,b,c)^T$ gets interpreted as Transpose[{{a,b,c}}], with an extra layer of nesting. Moral: RTFD.$endgroup$
– amd
Jan 11 at 8:40
$begingroup$
This is explained rather tersely in the Wolfram Language documentation for
Plus: “Plus also threads element-wise over lists.“ There’s one other thing that W|A does that contributes to this behavior: a one-level deep list can’t be transposed, so an input of $(a,b,c)^T$ gets interpreted as Transpose[{{a,b,c}}], with an extra layer of nesting. Moral: RTFD.$endgroup$
– amd
Jan 11 at 8:40
add a comment |
$begingroup$
Can someone explain to be how this makes any sense?
It doesn't make any sense.
If you're talking about tansposition, then you're implicitly viewing your vectors as matrices with one of the dimensions being $1$.
Then your $v$ is a $1times 3$ matrix and $v^T$ is a $3times 1$ matrix.
Addition of matrices that don't have the same dimension is not defined.
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Hmm, then why does matlab, numpy and/or wolfram alpha return an answer?
$endgroup$
– GrandFleet
Jan 11 at 2:39
$begingroup$
@GrandFleet: Either brecause they're written to try their darned best to come up with some meaningful interpretation when they're given nonsense, even if that means they need to invent operations that are not there in the input -- or possibly because you're misunderstanding their input syntax and are actually asking them something different from what you think you're asking.
$endgroup$
– Henning Makholm
Jan 11 at 2:41
add a comment |
$begingroup$
Can someone explain to be how this makes any sense?
It doesn't make any sense.
If you're talking about tansposition, then you're implicitly viewing your vectors as matrices with one of the dimensions being $1$.
Then your $v$ is a $1times 3$ matrix and $v^T$ is a $3times 1$ matrix.
Addition of matrices that don't have the same dimension is not defined.
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Hmm, then why does matlab, numpy and/or wolfram alpha return an answer?
$endgroup$
– GrandFleet
Jan 11 at 2:39
$begingroup$
@GrandFleet: Either brecause they're written to try their darned best to come up with some meaningful interpretation when they're given nonsense, even if that means they need to invent operations that are not there in the input -- or possibly because you're misunderstanding their input syntax and are actually asking them something different from what you think you're asking.
$endgroup$
– Henning Makholm
Jan 11 at 2:41
add a comment |
$begingroup$
Can someone explain to be how this makes any sense?
It doesn't make any sense.
If you're talking about tansposition, then you're implicitly viewing your vectors as matrices with one of the dimensions being $1$.
Then your $v$ is a $1times 3$ matrix and $v^T$ is a $3times 1$ matrix.
Addition of matrices that don't have the same dimension is not defined.
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
$endgroup$
Can someone explain to be how this makes any sense?
It doesn't make any sense.
If you're talking about tansposition, then you're implicitly viewing your vectors as matrices with one of the dimensions being $1$.
Then your $v$ is a $1times 3$ matrix and $v^T$ is a $3times 1$ matrix.
Addition of matrices that don't have the same dimension is not defined.
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
answered Jan 11 at 2:37
Henning MakholmHenning Makholm
240k17305541
240k17305541
$begingroup$
Hmm, then why does matlab, numpy and/or wolfram alpha return an answer?
$endgroup$
– GrandFleet
Jan 11 at 2:39
$begingroup$
@GrandFleet: Either brecause they're written to try their darned best to come up with some meaningful interpretation when they're given nonsense, even if that means they need to invent operations that are not there in the input -- or possibly because you're misunderstanding their input syntax and are actually asking them something different from what you think you're asking.
$endgroup$
– Henning Makholm
Jan 11 at 2:41
add a comment |
$begingroup$
Hmm, then why does matlab, numpy and/or wolfram alpha return an answer?
$endgroup$
– GrandFleet
Jan 11 at 2:39
$begingroup$
@GrandFleet: Either brecause they're written to try their darned best to come up with some meaningful interpretation when they're given nonsense, even if that means they need to invent operations that are not there in the input -- or possibly because you're misunderstanding their input syntax and are actually asking them something different from what you think you're asking.
$endgroup$
– Henning Makholm
Jan 11 at 2:41
$begingroup$
Hmm, then why does matlab, numpy and/or wolfram alpha return an answer?
$endgroup$
– GrandFleet
Jan 11 at 2:39
$begingroup$
Hmm, then why does matlab, numpy and/or wolfram alpha return an answer?
$endgroup$
– GrandFleet
Jan 11 at 2:39
$begingroup$
@GrandFleet: Either brecause they're written to try their darned best to come up with some meaningful interpretation when they're given nonsense, even if that means they need to invent operations that are not there in the input -- or possibly because you're misunderstanding their input syntax and are actually asking them something different from what you think you're asking.
$endgroup$
– Henning Makholm
Jan 11 at 2:41
$begingroup$
@GrandFleet: Either brecause they're written to try their darned best to come up with some meaningful interpretation when they're given nonsense, even if that means they need to invent operations that are not there in the input -- or possibly because you're misunderstanding their input syntax and are actually asking them something different from what you think you're asking.
$endgroup$
– Henning Makholm
Jan 11 at 2:41
add a comment |
$begingroup$
A thought here as to why those systems would do that: there's an occasionally useful operation on a vector these systems support, namely adding a constant to each term. So then if $a$ is a constant and $v$ is a vector, the systems abbreviate this operator as $a+v$ or $v+a$. In effect, we promote $a$ to a vector with each element equal.
Now, if we're working with a row vector $u^T$ and a column vector $v$? In order to add the constants in each column of $u^T$ to $v$, we promote each of those length-1 columns to full length. In order to add the constants in each row of $v$ to $u^T$, we promote those length-1 rows to full length. This amounts to the same sum of matrices hypothesized in the question.
This is, of course, not standard vector/matrix addition.
answered Jan 11 at 2:50
jmerryjmerry
7,535921
$endgroup$
add a comment |
$begingroup$
A thought here as to why those systems would do that: there's an occasionally useful operation on a vector these systems support, namely adding a constant to each term. So then if $a$ is a constant and $v$ is a vector, the systems abbreviate this operator as $a+v$ or $v+a$. In effect, we promote $a$ to a vector with each element equal.
Now, if we're working with a row vector $u^T$ and a column vector $v$? In order to add the constants in each column of $u^T$ to $v$, we promote each of those length-1 columns to full length. In order to add the constants in each row of $v$ to $u^T$, we promote those length-1 rows to full length. This amounts to the same sum of matrices hypothesized in the question.
This is, of course, not standard vector/matrix addition.
answered Jan 11 at 2:50
jmerryjmerry
7,535921
$endgroup$
add a comment |
$begingroup$
A thought here as to why those systems would do that: there's an occasionally useful operation on a vector these systems support, namely adding a constant to each term. So then if $a$ is a constant and $v$ is a vector, the systems abbreviate this operator as $a+v$ or $v+a$. In effect, we promote $a$ to a vector with each element equal.
Now, if we're working with a row vector $u^T$ and a column vector $v$? In order to add the constants in each column of $u^T$ to $v$, we promote each of those length-1 columns to full length. In order to add the constants in each row of $v$ to $u^T$, we promote those length-1 rows to full length. This amounts to the same sum of matrices hypothesized in the question.
This is, of course, not standard vector/matrix addition.
answered Jan 11 at 2:50
jmerryjmerry
7,535921
$endgroup$
A thought here as to why those systems would do that: there's an occasionally useful operation on a vector these systems support, namely adding a constant to each term. So then if $a$ is a constant and $v$ is a vector, the systems abbreviate this operator as $a+v$ or $v+a$. In effect, we promote $a$ to a vector with each element equal.
Now, if we're working with a row vector $u^T$ and a column vector $v$? In order to add the constants in each column of $u^T$ to $v$, we promote each of those length-1 columns to full length. In order to add the constants in each row of $v$ to $u^T$, we promote those length-1 rows to full length. This amounts to the same sum of matrices hypothesized in the question.
This is, of course, not standard vector/matrix addition.
answered Jan 11 at 2:50
jmerryjmerry
7,535921
answered Jan 11 at 2:50
jmerryjmerry
7,535921
answered Jan 11 at 2:50
jmerryjmerry
7,535921
answered Jan 11 at 2:50
jmerryjmerry
7,535921
7,535921
add a comment |
add a comment |
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$begingroup$
How it makes sense? It does not. Your guess is probably correct as to how matlab interprets the mathematically meaningless expression.
$endgroup$
– darij grinberg
Jan 11 at 2:36
$begingroup$
The expression $v + v^t$ doesn't make sense. Maybe some systems will promote $v$ to a $3times 3$ matrix, but there's no mathematical justification for doing so.
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– anomaly
Jan 11 at 2:38
$begingroup$
Matrices can only be added if both dimensions agree. This is an illegal operation.
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– ncmathsadist
Jan 11 at 2:38
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Show us your alpha input.
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– Randall
Jan 11 at 2:38
$begingroup$
wolframalpha.com/input/… I'm looking at the alternative form
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– GrandFleet
Jan 11 at 2:40