Is the answer wrong: linear algebra done right Ex. 2B Q4
0
$begingroup$
Here is the answer. The underlined vector is the place where I am confused.
Beware the little ~ before the equation, It is not a minus sign.
In part (a):
$$U={(z_1, 6z_1, -2z_4-3z_5, z_4, z_5) in mathbb{R}^3}$$
Then the solution to part (a) is $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1)$ why the answer is $(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1)$? (just multiply $-1$?)
In part (b)
Should the answer be $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1), (0,1,0,0,0), (0,0,1,0,0)$ since we need to add $z_2$ and $z_3$? (This is how I think. Some dimensions are missing, and then add them back. Please correct me if I am wrong.)
linear-algebra
asked Jan 2 at 23:08
JOHN JOHN
4119
$endgroup$
|
show 2 more comments
0
$begingroup$
Here is the answer. The underlined vector is the place where I am confused.
Beware the little ~ before the equation, It is not a minus sign.
In part (a):
$$U={(z_1, 6z_1, -2z_4-3z_5, z_4, z_5) in mathbb{R}^3}$$
Then the solution to part (a) is $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1)$ why the answer is $(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1)$? (just multiply $-1$?)
In part (b)
Should the answer be $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1), (0,1,0,0,0), (0,0,1,0,0)$ since we need to add $z_2$ and $z_3$? (This is how I think. Some dimensions are missing, and then add them back. Please correct me if I am wrong.)
linear-algebra
asked Jan 2 at 23:08
JOHN JOHN
4119
$endgroup$
1
$begingroup$
There is no difference between $(0,0,2,-1,0)$ and $(0,0,-2,1,0)$ as a basis vector. If $vec{a}$ is part of a basis set, so is $lambda vec{a}$ for any $lambda neq 0$. And why do you think some dimensions are missing?
$endgroup$
– stressed out
Jan 2 at 23:12
3
$begingroup$
OK. I get it now. Well, you have found $3$ vectors so far but you need $5$. So, you should add two linearly independent vectors to your current set of $3$ vectors. There are many ways to do this. Your book chooses an easy way: check the two most obvious ones: $(1,0,0,0,0)$ and $(0,0,1,0,0)$. If they work (i.e. they're linearly independent from the other three), you're done. Now the only thing that's left to check is whether they're linearly independent or not. Can you proceed from there? To know whether it's wrong or not you should check whether the proposed set is linearly independent or not.
$endgroup$
– stressed out
Jan 2 at 23:46
1
$begingroup$
There are many ways to obtain a basis for a vector space. A vector space doesn't have a unique basis, so it's possible to find many different bases for the same vector space. If you think your answer works, you should prove that the set you have found is linearly independent and it spans the space. If this condition is satisfied, your answer is correct. Otherwise, it's not. And just because your answer is correct, it won't disprove other answers because as I said, there always exist different bases for the same vector spaces.
$endgroup$
– stressed out
Jan 3 at 1:26
1
$begingroup$
@stressedout got it!
$endgroup$
– JOHN
Jan 3 at 1:30
1
$begingroup$
$U$ is most definitely not $mathbb C^3$. It is a three-dimensional subspace of $mathbb C^5$. Try to get out of the bad habit of calling any old three-dimensional complex vector space “$mathbb C^3$.”
$endgroup$
– amd
Jan 3 at 1:38
|
show 2 more comments
0
0
0
$begingroup$
Here is the answer. The underlined vector is the place where I am confused.
Beware the little ~ before the equation, It is not a minus sign.
In part (a):
$$U={(z_1, 6z_1, -2z_4-3z_5, z_4, z_5) in mathbb{R}^3}$$
Then the solution to part (a) is $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1)$ why the answer is $(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1)$? (just multiply $-1$?)
In part (b)
Should the answer be $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1), (0,1,0,0,0), (0,0,1,0,0)$ since we need to add $z_2$ and $z_3$? (This is how I think. Some dimensions are missing, and then add them back. Please correct me if I am wrong.)
linear-algebra
asked Jan 2 at 23:08
JOHN JOHN
4119
$endgroup$
Here is the answer. The underlined vector is the place where I am confused.
Beware the little ~ before the equation, It is not a minus sign.
In part (a):
$$U={(z_1, 6z_1, -2z_4-3z_5, z_4, z_5) in mathbb{R}^3}$$
Then the solution to part (a) is $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1)$ why the answer is $(1,6,0,0,0), (0,0,2,-1,0), (0,0,3,0,-1)$? (just multiply $-1$?)
In part (b)
Should the answer be $(1,6,0,0,0), (0,0,-2,1,0), (0,0,-3,0,1), (0,1,0,0,0), (0,0,1,0,0)$ since we need to add $z_2$ and $z_3$? (This is how I think. Some dimensions are missing, and then add them back. Please correct me if I am wrong.)
linear-algebra
linear-algebra
asked Jan 2 at 23:08
JOHN JOHN
4119
asked Jan 2 at 23:08
JOHN JOHN
4119
asked Jan 2 at 23:08
JOHN JOHN
4119
asked Jan 2 at 23:08
JOHN JOHN
4119
asked Jan 2 at 23:08
JOHN JOHN
4119
4119
1
$begingroup$
There is no difference between $(0,0,2,-1,0)$ and $(0,0,-2,1,0)$ as a basis vector. If $vec{a}$ is part of a basis set, so is $lambda vec{a}$ for any $lambda neq 0$. And why do you think some dimensions are missing?
$endgroup$
– stressed out
Jan 2 at 23:12
3
$begingroup$
OK. I get it now. Well, you have found $3$ vectors so far but you need $5$. So, you should add two linearly independent vectors to your current set of $3$ vectors. There are many ways to do this. Your book chooses an easy way: check the two most obvious ones: $(1,0,0,0,0)$ and $(0,0,1,0,0)$. If they work (i.e. they're linearly independent from the other three), you're done. Now the only thing that's left to check is whether they're linearly independent or not. Can you proceed from there? To know whether it's wrong or not you should check whether the proposed set is linearly independent or not.
$endgroup$
– stressed out
Jan 2 at 23:46
1
$begingroup$
There are many ways to obtain a basis for a vector space. A vector space doesn't have a unique basis, so it's possible to find many different bases for the same vector space. If you think your answer works, you should prove that the set you have found is linearly independent and it spans the space. If this condition is satisfied, your answer is correct. Otherwise, it's not. And just because your answer is correct, it won't disprove other answers because as I said, there always exist different bases for the same vector spaces.
$endgroup$
– stressed out
Jan 3 at 1:26
1
$begingroup$
@stressedout got it!
$endgroup$
– JOHN
Jan 3 at 1:30
1
$begingroup$
$U$ is most definitely not $mathbb C^3$. It is a three-dimensional subspace of $mathbb C^5$. Try to get out of the bad habit of calling any old three-dimensional complex vector space “$mathbb C^3$.”
$endgroup$
– amd
Jan 3 at 1:38
|
show 2 more comments
1
$begingroup$
There is no difference between $(0,0,2,-1,0)$ and $(0,0,-2,1,0)$ as a basis vector. If $vec{a}$ is part of a basis set, so is $lambda vec{a}$ for any $lambda neq 0$. And why do you think some dimensions are missing?
$endgroup$
– stressed out
Jan 2 at 23:12
3
$begingroup$
OK. I get it now. Well, you have found $3$ vectors so far but you need $5$. So, you should add two linearly independent vectors to your current set of $3$ vectors. There are many ways to do this. Your book chooses an easy way: check the two most obvious ones: $(1,0,0,0,0)$ and $(0,0,1,0,0)$. If they work (i.e. they're linearly independent from the other three), you're done. Now the only thing that's left to check is whether they're linearly independent or not. Can you proceed from there? To know whether it's wrong or not you should check whether the proposed set is linearly independent or not.
$endgroup$
– stressed out
Jan 2 at 23:46
1
$begingroup$
There are many ways to obtain a basis for a vector space. A vector space doesn't have a unique basis, so it's possible to find many different bases for the same vector space. If you think your answer works, you should prove that the set you have found is linearly independent and it spans the space. If this condition is satisfied, your answer is correct. Otherwise, it's not. And just because your answer is correct, it won't disprove other answers because as I said, there always exist different bases for the same vector spaces.
$endgroup$
– stressed out
Jan 3 at 1:26
1
$begingroup$
@stressedout got it!
$endgroup$
– JOHN
Jan 3 at 1:30
1
$begingroup$
$U$ is most definitely not $mathbb C^3$. It is a three-dimensional subspace of $mathbb C^5$. Try to get out of the bad habit of calling any old three-dimensional complex vector space “$mathbb C^3$.”
$endgroup$
– amd
Jan 3 at 1:38
1
1
$begingroup$
There is no difference between $(0,0,2,-1,0)$ and $(0,0,-2,1,0)$ as a basis vector. If $vec{a}$ is part of a basis set, so is $lambda vec{a}$ for any $lambda neq 0$. And why do you think some dimensions are missing?
$endgroup$
– stressed out
Jan 2 at 23:12
$begingroup$
There is no difference between $(0,0,2,-1,0)$ and $(0,0,-2,1,0)$ as a basis vector. If $vec{a}$ is part of a basis set, so is $lambda vec{a}$ for any $lambda neq 0$. And why do you think some dimensions are missing?
$endgroup$
– stressed out
Jan 2 at 23:12
3
3
$begingroup$
OK. I get it now. Well, you have found $3$ vectors so far but you need $5$. So, you should add two linearly independent vectors to your current set of $3$ vectors. There are many ways to do this. Your book chooses an easy way: check the two most obvious ones: $(1,0,0,0,0)$ and $(0,0,1,0,0)$. If they work (i.e. they're linearly independent from the other three), you're done. Now the only thing that's left to check is whether they're linearly independent or not. Can you proceed from there? To know whether it's wrong or not you should check whether the proposed set is linearly independent or not.
$endgroup$
– stressed out
Jan 2 at 23:46
$begingroup$
OK. I get it now. Well, you have found $3$ vectors so far but you need $5$. So, you should add two linearly independent vectors to your current set of $3$ vectors. There are many ways to do this. Your book chooses an easy way: check the two most obvious ones: $(1,0,0,0,0)$ and $(0,0,1,0,0)$. If they work (i.e. they're linearly independent from the other three), you're done. Now the only thing that's left to check is whether they're linearly independent or not. Can you proceed from there? To know whether it's wrong or not you should check whether the proposed set is linearly independent or not.
$endgroup$
– stressed out
Jan 2 at 23:46
1
1
$begingroup$
There are many ways to obtain a basis for a vector space. A vector space doesn't have a unique basis, so it's possible to find many different bases for the same vector space. If you think your answer works, you should prove that the set you have found is linearly independent and it spans the space. If this condition is satisfied, your answer is correct. Otherwise, it's not. And just because your answer is correct, it won't disprove other answers because as I said, there always exist different bases for the same vector spaces.
$endgroup$
– stressed out
Jan 3 at 1:26
$begingroup$
There are many ways to obtain a basis for a vector space. A vector space doesn't have a unique basis, so it's possible to find many different bases for the same vector space. If you think your answer works, you should prove that the set you have found is linearly independent and it spans the space. If this condition is satisfied, your answer is correct. Otherwise, it's not. And just because your answer is correct, it won't disprove other answers because as I said, there always exist different bases for the same vector spaces.
$endgroup$
– stressed out
Jan 3 at 1:26
1
1
$begingroup$
@stressedout got it!
$endgroup$
– JOHN
Jan 3 at 1:30
$begingroup$
@stressedout got it!
$endgroup$
– JOHN
Jan 3 at 1:30
1
1
$begingroup$
$U$ is most definitely not $mathbb C^3$. It is a three-dimensional subspace of $mathbb C^5$. Try to get out of the bad habit of calling any old three-dimensional complex vector space “$mathbb C^3$.”
$endgroup$
– amd
Jan 3 at 1:38
$begingroup$
$U$ is most definitely not $mathbb C^3$. It is a three-dimensional subspace of $mathbb C^5$. Try to get out of the bad habit of calling any old three-dimensional complex vector space “$mathbb C^3$.”
$endgroup$
– amd
Jan 3 at 1:38
|
show 2 more comments
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1
$begingroup$
There is no difference between $(0,0,2,-1,0)$ and $(0,0,-2,1,0)$ as a basis vector. If $vec{a}$ is part of a basis set, so is $lambda vec{a}$ for any $lambda neq 0$. And why do you think some dimensions are missing?
$endgroup$
– stressed out
Jan 2 at 23:12
3
$begingroup$
OK. I get it now. Well, you have found $3$ vectors so far but you need $5$. So, you should add two linearly independent vectors to your current set of $3$ vectors. There are many ways to do this. Your book chooses an easy way: check the two most obvious ones: $(1,0,0,0,0)$ and $(0,0,1,0,0)$. If they work (i.e. they're linearly independent from the other three), you're done. Now the only thing that's left to check is whether they're linearly independent or not. Can you proceed from there? To know whether it's wrong or not you should check whether the proposed set is linearly independent or not.
$endgroup$
– stressed out
Jan 2 at 23:46
1
$begingroup$
There are many ways to obtain a basis for a vector space. A vector space doesn't have a unique basis, so it's possible to find many different bases for the same vector space. If you think your answer works, you should prove that the set you have found is linearly independent and it spans the space. If this condition is satisfied, your answer is correct. Otherwise, it's not. And just because your answer is correct, it won't disprove other answers because as I said, there always exist different bases for the same vector spaces.
$endgroup$
– stressed out
Jan 3 at 1:26
1
$begingroup$
@stressedout got it!
$endgroup$
– JOHN
Jan 3 at 1:30
1
$begingroup$
$U$ is most definitely not $mathbb C^3$. It is a three-dimensional subspace of $mathbb C^5$. Try to get out of the bad habit of calling any old three-dimensional complex vector space “$mathbb C^3$.”
$endgroup$
– amd
Jan 3 at 1:38