Mixing
Determine which of the following are Vector Spaces
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Going off the axioms, I'm trying to determine how to verify each of these. In the first set, do I take $u = (x_1, x_2, x_3)$ and $v = (y_1, y_2, y_3)$, or would it rather be $u = (x_1 + y_1, x_2 + y_2, x_3 + y_3), v = (x_4 + y_4, x_5 + y_5, x_6 + y_6)$?
Based on what I know about scalar multiplication, I would say the first set is not a vector space because $0(x_1, x_2, x_3)$ = $(0, 0, 0)$ which is not equivalent to $(0x_1, x_2, x_3)$, for non-zero values of $x_2$ and $x_3$. The second and third sets seem to define vector spaces, as I can't seem to figure out any axiom that directly fails (i.e. $u+v$ is in $V$, a unique additive identity exists, and the set is closed under scalar multiplication). Any help would be appreciated.
Edit: The third set is not a vector space.
linear-algebra vector-spaces vectors
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
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add a comment |
$begingroup$
Going off the axioms, I'm trying to determine how to verify each of these. In the first set, do I take $u = (x_1, x_2, x_3)$ and $v = (y_1, y_2, y_3)$, or would it rather be $u = (x_1 + y_1, x_2 + y_2, x_3 + y_3), v = (x_4 + y_4, x_5 + y_5, x_6 + y_6)$?
Based on what I know about scalar multiplication, I would say the first set is not a vector space because $0(x_1, x_2, x_3)$ = $(0, 0, 0)$ which is not equivalent to $(0x_1, x_2, x_3)$, for non-zero values of $x_2$ and $x_3$. The second and third sets seem to define vector spaces, as I can't seem to figure out any axiom that directly fails (i.e. $u+v$ is in $V$, a unique additive identity exists, and the set is closed under scalar multiplication). Any help would be appreciated.
Edit: The third set is not a vector space.
linear-algebra vector-spaces vectors
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
$endgroup$
add a comment |
$begingroup$
Going off the axioms, I'm trying to determine how to verify each of these. In the first set, do I take $u = (x_1, x_2, x_3)$ and $v = (y_1, y_2, y_3)$, or would it rather be $u = (x_1 + y_1, x_2 + y_2, x_3 + y_3), v = (x_4 + y_4, x_5 + y_5, x_6 + y_6)$?
Based on what I know about scalar multiplication, I would say the first set is not a vector space because $0(x_1, x_2, x_3)$ = $(0, 0, 0)$ which is not equivalent to $(0x_1, x_2, x_3)$, for non-zero values of $x_2$ and $x_3$. The second and third sets seem to define vector spaces, as I can't seem to figure out any axiom that directly fails (i.e. $u+v$ is in $V$, a unique additive identity exists, and the set is closed under scalar multiplication). Any help would be appreciated.
Edit: The third set is not a vector space.
linear-algebra vector-spaces vectors
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
$endgroup$
Going off the axioms, I'm trying to determine how to verify each of these. In the first set, do I take $u = (x_1, x_2, x_3)$ and $v = (y_1, y_2, y_3)$, or would it rather be $u = (x_1 + y_1, x_2 + y_2, x_3 + y_3), v = (x_4 + y_4, x_5 + y_5, x_6 + y_6)$?
Based on what I know about scalar multiplication, I would say the first set is not a vector space because $0(x_1, x_2, x_3)$ = $(0, 0, 0)$ which is not equivalent to $(0x_1, x_2, x_3)$, for non-zero values of $x_2$ and $x_3$. The second and third sets seem to define vector spaces, as I can't seem to figure out any axiom that directly fails (i.e. $u+v$ is in $V$, a unique additive identity exists, and the set is closed under scalar multiplication). Any help would be appreciated.
Edit: The third set is not a vector space.
linear-algebra vector-spaces vectors
linear-algebra vector-spaces vectors
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
edited Jan 22 at 5:46
Sanjoy The Manjoy
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
asked Jan 22 at 4:58
Sanjoy The ManjoySanjoy The Manjoy
493315
493315
add a comment |
add a comment |
1 Answer
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The third one is not a vector space. The zero vector in this space is $(-1,-1)$ but if you multiply this by $2$ you don't get back the zero vector.
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
$endgroup$
$begingroup$
Ahh thank you! My Linear Algebra is very rusty right now, so these problems are helping me refresh. For the first set, what would you set the $u$ and $v$ vectors as when checking for $u+v$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:46
1
$begingroup$
@SanjoyTheManjoy In the first example all properties of addition are satisfied. It is teh scalar multiplication (and its relation to addition) that is a problem.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 5:51
$begingroup$
Right, I figured as much, but to verify that axiom of addition holds, do I let u = $(x_1, x_2, x_3)$ or rather u = $(x_1 + y_1,...)$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:52
add a comment |
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1 Answer
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$begingroup$
The third one is not a vector space. The zero vector in this space is $(-1,-1)$ but if you multiply this by $2$ you don't get back the zero vector.
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
$endgroup$
$begingroup$
Ahh thank you! My Linear Algebra is very rusty right now, so these problems are helping me refresh. For the first set, what would you set the $u$ and $v$ vectors as when checking for $u+v$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:46
1
$begingroup$
@SanjoyTheManjoy In the first example all properties of addition are satisfied. It is teh scalar multiplication (and its relation to addition) that is a problem.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 5:51
$begingroup$
Right, I figured as much, but to verify that axiom of addition holds, do I let u = $(x_1, x_2, x_3)$ or rather u = $(x_1 + y_1,...)$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:52
add a comment |
$begingroup$
The third one is not a vector space. The zero vector in this space is $(-1,-1)$ but if you multiply this by $2$ you don't get back the zero vector.
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
$endgroup$
$begingroup$
Ahh thank you! My Linear Algebra is very rusty right now, so these problems are helping me refresh. For the first set, what would you set the $u$ and $v$ vectors as when checking for $u+v$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:46
1
$begingroup$
@SanjoyTheManjoy In the first example all properties of addition are satisfied. It is teh scalar multiplication (and its relation to addition) that is a problem.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 5:51
$begingroup$
Right, I figured as much, but to verify that axiom of addition holds, do I let u = $(x_1, x_2, x_3)$ or rather u = $(x_1 + y_1,...)$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:52
add a comment |
$begingroup$
The third one is not a vector space. The zero vector in this space is $(-1,-1)$ but if you multiply this by $2$ you don't get back the zero vector.
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
$endgroup$
The third one is not a vector space. The zero vector in this space is $(-1,-1)$ but if you multiply this by $2$ you don't get back the zero vector.
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
answered Jan 22 at 5:34
Kavi Rama MurthyKavi Rama Murthy
65.7k42767
65.7k42767
$begingroup$
Ahh thank you! My Linear Algebra is very rusty right now, so these problems are helping me refresh. For the first set, what would you set the $u$ and $v$ vectors as when checking for $u+v$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:46
1
$begingroup$
@SanjoyTheManjoy In the first example all properties of addition are satisfied. It is teh scalar multiplication (and its relation to addition) that is a problem.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 5:51
$begingroup$
Right, I figured as much, but to verify that axiom of addition holds, do I let u = $(x_1, x_2, x_3)$ or rather u = $(x_1 + y_1,...)$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:52
add a comment |
$begingroup$
Ahh thank you! My Linear Algebra is very rusty right now, so these problems are helping me refresh. For the first set, what would you set the $u$ and $v$ vectors as when checking for $u+v$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:46
1
$begingroup$
@SanjoyTheManjoy In the first example all properties of addition are satisfied. It is teh scalar multiplication (and its relation to addition) that is a problem.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 5:51
$begingroup$
Right, I figured as much, but to verify that axiom of addition holds, do I let u = $(x_1, x_2, x_3)$ or rather u = $(x_1 + y_1,...)$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:52
$begingroup$
Ahh thank you! My Linear Algebra is very rusty right now, so these problems are helping me refresh. For the first set, what would you set the $u$ and $v$ vectors as when checking for $u+v$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:46
$begingroup$
Ahh thank you! My Linear Algebra is very rusty right now, so these problems are helping me refresh. For the first set, what would you set the $u$ and $v$ vectors as when checking for $u+v$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:46
1
1
$begingroup$
@SanjoyTheManjoy In the first example all properties of addition are satisfied. It is teh scalar multiplication (and its relation to addition) that is a problem.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 5:51
$begingroup$
@SanjoyTheManjoy In the first example all properties of addition are satisfied. It is teh scalar multiplication (and its relation to addition) that is a problem.
$endgroup$
– Kavi Rama Murthy
Jan 22 at 5:51
$begingroup$
Right, I figured as much, but to verify that axiom of addition holds, do I let u = $(x_1, x_2, x_3)$ or rather u = $(x_1 + y_1,...)$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:52
$begingroup$
Right, I figured as much, but to verify that axiom of addition holds, do I let u = $(x_1, x_2, x_3)$ or rather u = $(x_1 + y_1,...)$
$endgroup$
– Sanjoy The Manjoy
Jan 22 at 5:52
add a comment |
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