Mixing
How can lists/vectors in $mathbb{R}^2$ be thought of as functions?
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It's written in my linear algebra textbook that $mathbb{R}^S$ is the set of all possible functions $f: S to mathbb{R}$, I fail to see how this applies when $S=2=lbrace {0,1} rbrace$ as it's not clear to me how a vector like $(72,15)$ is a function
linear-algebra
asked Jan 31 at 18:28
user168651user168651
62
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add a comment |
$begingroup$
It's written in my linear algebra textbook that $mathbb{R}^S$ is the set of all possible functions $f: S to mathbb{R}$, I fail to see how this applies when $S=2=lbrace {0,1} rbrace$ as it's not clear to me how a vector like $(72,15)$ is a function
linear-algebra
asked Jan 31 at 18:28
user168651user168651
62
$endgroup$
4
$begingroup$
I think $2={0,1}$, and then $(72,15)$ represents the function where $f(0)=72$ and $f(1)=15$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:31
1
$begingroup$
My comment should answer your question. Is there anything unclear about it?
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– Cheerful Parsnip
Jan 31 at 18:39
1
$begingroup$
@CheerfulParsnip If your comment answers the question, you should make it an answer! I’d gladly +1
$endgroup$
– Santana Afton
Jan 31 at 22:50
add a comment |
$begingroup$
It's written in my linear algebra textbook that $mathbb{R}^S$ is the set of all possible functions $f: S to mathbb{R}$, I fail to see how this applies when $S=2=lbrace {0,1} rbrace$ as it's not clear to me how a vector like $(72,15)$ is a function
linear-algebra
asked Jan 31 at 18:28
user168651user168651
62
$endgroup$
It's written in my linear algebra textbook that $mathbb{R}^S$ is the set of all possible functions $f: S to mathbb{R}$, I fail to see how this applies when $S=2=lbrace {0,1} rbrace$ as it's not clear to me how a vector like $(72,15)$ is a function
linear-algebra
linear-algebra
asked Jan 31 at 18:28
user168651user168651
62
asked Jan 31 at 18:28
user168651user168651
62
edited Feb 1 at 5:49
user168651
asked Jan 31 at 18:28
user168651user168651
62
asked Jan 31 at 18:28
user168651user168651
62
asked Jan 31 at 18:28
user168651user168651
62
62
4
$begingroup$
I think $2={0,1}$, and then $(72,15)$ represents the function where $f(0)=72$ and $f(1)=15$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:31
1
$begingroup$
My comment should answer your question. Is there anything unclear about it?
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:39
1
$begingroup$
@CheerfulParsnip If your comment answers the question, you should make it an answer! I’d gladly +1
$endgroup$
– Santana Afton
Jan 31 at 22:50
add a comment |
4
$begingroup$
I think $2={0,1}$, and then $(72,15)$ represents the function where $f(0)=72$ and $f(1)=15$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:31
1
$begingroup$
My comment should answer your question. Is there anything unclear about it?
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:39
1
$begingroup$
@CheerfulParsnip If your comment answers the question, you should make it an answer! I’d gladly +1
$endgroup$
– Santana Afton
Jan 31 at 22:50
4
4
$begingroup$
I think $2={0,1}$, and then $(72,15)$ represents the function where $f(0)=72$ and $f(1)=15$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:31
$begingroup$
I think $2={0,1}$, and then $(72,15)$ represents the function where $f(0)=72$ and $f(1)=15$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:31
1
1
$begingroup$
My comment should answer your question. Is there anything unclear about it?
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:39
$begingroup$
My comment should answer your question. Is there anything unclear about it?
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:39
1
1
$begingroup$
@CheerfulParsnip If your comment answers the question, you should make it an answer! I’d gladly +1
$endgroup$
– Santana Afton
Jan 31 at 22:50
$begingroup$
@CheerfulParsnip If your comment answers the question, you should make it an answer! I’d gladly +1
$endgroup$
– Santana Afton
Jan 31 at 22:50
add a comment |
1 Answer
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active
oldest
votes
$begingroup$
You can think of $mathbb R^n$ as lists of numbers but you can also think of it as functions $fcolon {1,ldots,n}to mathbb R$ where the $j$th coordinate can be regarded as $f(j)$. Then, if $|S|=s$ we have that $mathbb R^Scong mathbb R^s$.
edited Jan 31 at 23:09
Sambo
2,3012532
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
$endgroup$
1
$begingroup$
Small edit - changed $mathbb{R}^2$ to $mathbb{R}^n$ since you used $n$ afterwards.
$endgroup$
– Sambo
Jan 31 at 23:10
$begingroup$
@Sambo, oops. Thanks.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:14
add a comment |
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1 Answer
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$begingroup$
You can think of $mathbb R^n$ as lists of numbers but you can also think of it as functions $fcolon {1,ldots,n}to mathbb R$ where the $j$th coordinate can be regarded as $f(j)$. Then, if $|S|=s$ we have that $mathbb R^Scong mathbb R^s$.
edited Jan 31 at 23:09
Sambo
2,3012532
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
$endgroup$
1
$begingroup$
Small edit - changed $mathbb{R}^2$ to $mathbb{R}^n$ since you used $n$ afterwards.
$endgroup$
– Sambo
Jan 31 at 23:10
$begingroup$
@Sambo, oops. Thanks.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:14
add a comment |
$begingroup$
You can think of $mathbb R^n$ as lists of numbers but you can also think of it as functions $fcolon {1,ldots,n}to mathbb R$ where the $j$th coordinate can be regarded as $f(j)$. Then, if $|S|=s$ we have that $mathbb R^Scong mathbb R^s$.
edited Jan 31 at 23:09
Sambo
2,3012532
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
$endgroup$
1
$begingroup$
Small edit - changed $mathbb{R}^2$ to $mathbb{R}^n$ since you used $n$ afterwards.
$endgroup$
– Sambo
Jan 31 at 23:10
$begingroup$
@Sambo, oops. Thanks.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:14
add a comment |
$begingroup$
You can think of $mathbb R^n$ as lists of numbers but you can also think of it as functions $fcolon {1,ldots,n}to mathbb R$ where the $j$th coordinate can be regarded as $f(j)$. Then, if $|S|=s$ we have that $mathbb R^Scong mathbb R^s$.
edited Jan 31 at 23:09
Sambo
2,3012532
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
$endgroup$
You can think of $mathbb R^n$ as lists of numbers but you can also think of it as functions $fcolon {1,ldots,n}to mathbb R$ where the $j$th coordinate can be regarded as $f(j)$. Then, if $|S|=s$ we have that $mathbb R^Scong mathbb R^s$.
edited Jan 31 at 23:09
Sambo
2,3012532
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
edited Jan 31 at 23:09
Sambo
2,3012532
edited Jan 31 at 23:09
Sambo
2,3012532
edited Jan 31 at 23:09
Sambo
2,3012532
2,3012532
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
answered Jan 31 at 23:03
Cheerful ParsnipCheerful Parsnip
21.2k23598
21.2k23598
1
$begingroup$
Small edit - changed $mathbb{R}^2$ to $mathbb{R}^n$ since you used $n$ afterwards.
$endgroup$
– Sambo
Jan 31 at 23:10
$begingroup$
@Sambo, oops. Thanks.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:14
add a comment |
1
$begingroup$
Small edit - changed $mathbb{R}^2$ to $mathbb{R}^n$ since you used $n$ afterwards.
$endgroup$
– Sambo
Jan 31 at 23:10
$begingroup$
@Sambo, oops. Thanks.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:14
1
1
$begingroup$
Small edit - changed $mathbb{R}^2$ to $mathbb{R}^n$ since you used $n$ afterwards.
$endgroup$
– Sambo
Jan 31 at 23:10
$begingroup$
Small edit - changed $mathbb{R}^2$ to $mathbb{R}^n$ since you used $n$ afterwards.
$endgroup$
– Sambo
Jan 31 at 23:10
$begingroup$
@Sambo, oops. Thanks.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:14
$begingroup$
@Sambo, oops. Thanks.
$endgroup$
– Cheerful Parsnip
Jan 31 at 23:14
add a comment |
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$begingroup$
I think $2={0,1}$, and then $(72,15)$ represents the function where $f(0)=72$ and $f(1)=15$.
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:31
1
$begingroup$
My comment should answer your question. Is there anything unclear about it?
$endgroup$
– Cheerful Parsnip
Jan 31 at 18:39
1
$begingroup$
@CheerfulParsnip If your comment answers the question, you should make it an answer! I’d gladly +1
$endgroup$
– Santana Afton
Jan 31 at 22:50