Help for finding Bases of Sets
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I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
asked 16 hours ago
Danny
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up vote
-1
down vote
favorite
I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
asked 16 hours ago
Danny
11
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
16 hours ago
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
12 hours ago
Thank you for the advice.
– Danny
8 hours ago
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
asked 16 hours ago
Danny
11
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I'm struggling on a linear algebra assignment where I have to find the basis of the set: $U={left[begin{array}{c}a \b \ cend{array}right]| a,b,cin mathbb R, a=2b+3c}subseteqmathbb R^{3}$. I don't really understand how to do this when I look at other examples. The question first required me to prove that $U$ is a subspace of $mathbb R^{3}$ (I used the subspace test). Now for this basis portion, I know that I need to use a system of equations to find a spanning set and show its linear independence. I got stuck after thinking of how to form a system using only the given information of $a-2b-3c=0$. How would I use linear combinations to find the basis of $U$? Any help would be appreciated. Thanks! (I'm a first year engineering student who is new to vector (sub)spaces, just to give context on my current knowledge).
linear-algebra
linear-algebra
asked 16 hours ago
Danny
11
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 16 hours ago
Danny
11
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 15 hours ago
asked 16 hours ago
Danny
11
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 16 hours ago
Danny
11
asked 16 hours ago
Danny
11
11
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Danny is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
16 hours ago
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
12 hours ago
Thank you for the advice.
– Danny
8 hours ago
add a comment |
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
16 hours ago
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
12 hours ago
Thank you for the advice.
– Danny
8 hours ago
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
16 hours ago
Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
16 hours ago
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
12 hours ago
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
12 hours ago
Thank you for the advice.
– Danny
8 hours ago
Thank you for the advice.
– Danny
8 hours ago
add a comment |
1 Answer
1
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There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered 10 hours ago
gandalf61
7,052522
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
8 hours ago
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
7 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered 10 hours ago
gandalf61
7,052522
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
8 hours ago
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
7 hours ago
add a comment |
up vote
0
down vote
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered 10 hours ago
gandalf61
7,052522
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
8 hours ago
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
7 hours ago
add a comment |
up vote
0
down vote
up vote
0
down vote
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered 10 hours ago
gandalf61
7,052522
There are many choices for a basis for $U$, but one choice becomes fairly obvious if you think of a typical member of $U$ as follows:
$begin{pmatrix}2b+3c\b\cend{pmatrix} = begin{pmatrix}2b\b\0end{pmatrix} + begin{pmatrix}3c\0\cend{pmatrix} = bbegin{pmatrix}2\1\0end{pmatrix} + cbegin{pmatrix}3\0\1end{pmatrix}$
answered 10 hours ago
gandalf61
7,052522
answered 10 hours ago
gandalf61
7,052522
answered 10 hours ago
gandalf61
7,052522
answered 10 hours ago
gandalf61
7,052522
7,052522
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
8 hours ago
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
7 hours ago
add a comment |
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
8 hours ago
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
7 hours ago
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
8 hours ago
I see, so I just had to show that a is linearly dependent of b and c and form a linear combination to form the matrix. Thank you.
– Danny
8 hours ago
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
7 hours ago
Well, you already know $a$ is linearly dependent on $b$ and $c$ because you are told that $a=2b+3c$. And now you also know that any vector in U is a linear combination of the two vectors $(2,1,0)^T$ and $(3,0,1)^T$.
– gandalf61
7 hours ago
add a comment |
Danny is a new contributor. Be nice, and check out our Code of Conduct.
Danny is a new contributor. Be nice, and check out our Code of Conduct.
Danny is a new contributor. Be nice, and check out our Code of Conduct.
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Try to view $U$ as the null space of some matrix $A$, then use the standard method for finding a basis for $mathrm{Nul}A$.
– Morgan Rodgers
16 hours ago
Alternatively, find a nonzero vector in $U$. Then find another one, but not a scalar multiple of the first one. Then argue from geometry that those two vectors form a basis for $U$.
– Gerry Myerson
12 hours ago
Thank you for the advice.
– Danny
8 hours ago