I need help to understand a problem in Linear Algebra about calculating a set.
$begingroup$
I'm trying to find the following set (some authors call it "column space"):
Given a matrix $A in mathbb{R} ^{m times n}$, we define:
$$R(A)={b in mathbb{R}^{m} | exists space xin mathbb{R} ^n, b=Ax }.$$
Let $A= left[ begin{array}{ccc}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{array} right]$.
My attempt goes as follows:
I'm trying to find the elements of $R(A)$. Notice that the elements of $R(A)$ are vectors in $mathbb{R} ^3 $. I want to know how those elements are. If I take a generic vector $b=(b_{1},b_{2},b_{3})^T in mathbb{R}^3$, then I'm trying to solve the following equation system:
$$ left[ begin{array}{ccc}
2 & 2 & 0 & 0 &|& b_{1} \
3 & 4 & -1 & 2 &|& b_{2}\
-1 & 1 & -2 & 4 &|& b_{3}end{array} right]$$
I calculated this echelon form for A.
$$ left[ begin{array}{ccc}
1 & -1 & 2 & -1 &|& -b_{3} \
0 & 7 & -7 & 14 &|& b_{2}+3b_{3}\
0 & 0 & 0 & 0 &|& b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3} end{array} right]$$.
(I think it is not necessary to put the steps to find the echelon form, but if you need it, let me know).
If I want the system $Ax=b$ to have solutions, it's clear that we need that $b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3}=0$, which is equivalent to $b_{1}=frac{4b_{2}-2b_{3}}{7}$.
Then my conclussion is that the elements of $R(A)$ must be all of the form $$left[ begin{array}{ccc}
b_{1} \
b_{2} \
b_{3} end{array} right]=left[ begin{array}{ccc}
frac{4b_{2}-2b_{3}}{7} \
b_{2} \
b_{3} end{array} right]$$
because I need that condition if I want the system $Ax=b$ having solution.
The question is, is this the correct way for doing that?
If there's not, which is the steps that I need to follow?
I'm just learning Linear Algebra.
Thanks :)
linear-algebra matrices proof-verification systems-of-equations
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
asked Dec 21 '18 at 19:36
rowcolrowcol
734
$endgroup$
add a comment |
$begingroup$
I'm trying to find the following set (some authors call it "column space"):
Given a matrix $A in mathbb{R} ^{m times n}$, we define:
$$R(A)={b in mathbb{R}^{m} | exists space xin mathbb{R} ^n, b=Ax }.$$
Let $A= left[ begin{array}{ccc}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{array} right]$.
My attempt goes as follows:
I'm trying to find the elements of $R(A)$. Notice that the elements of $R(A)$ are vectors in $mathbb{R} ^3 $. I want to know how those elements are. If I take a generic vector $b=(b_{1},b_{2},b_{3})^T in mathbb{R}^3$, then I'm trying to solve the following equation system:
$$ left[ begin{array}{ccc}
2 & 2 & 0 & 0 &|& b_{1} \
3 & 4 & -1 & 2 &|& b_{2}\
-1 & 1 & -2 & 4 &|& b_{3}end{array} right]$$
I calculated this echelon form for A.
$$ left[ begin{array}{ccc}
1 & -1 & 2 & -1 &|& -b_{3} \
0 & 7 & -7 & 14 &|& b_{2}+3b_{3}\
0 & 0 & 0 & 0 &|& b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3} end{array} right]$$.
(I think it is not necessary to put the steps to find the echelon form, but if you need it, let me know).
If I want the system $Ax=b$ to have solutions, it's clear that we need that $b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3}=0$, which is equivalent to $b_{1}=frac{4b_{2}-2b_{3}}{7}$.
Then my conclussion is that the elements of $R(A)$ must be all of the form $$left[ begin{array}{ccc}
b_{1} \
b_{2} \
b_{3} end{array} right]=left[ begin{array}{ccc}
frac{4b_{2}-2b_{3}}{7} \
b_{2} \
b_{3} end{array} right]$$
because I need that condition if I want the system $Ax=b$ having solution.
The question is, is this the correct way for doing that?
If there's not, which is the steps that I need to follow?
I'm just learning Linear Algebra.
Thanks :)
linear-algebra matrices proof-verification systems-of-equations
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
asked Dec 21 '18 at 19:36
rowcolrowcol
734
$endgroup$
$begingroup$
Your process is perfectly fine, and your answer is correct. There are quicker ways of getting the answer in general, but there's nothing wrong with your method.
$endgroup$
– Omnomnomnom
Dec 21 '18 at 20:07
$begingroup$
The set you are looking for is called column space because it is the subvectorspace spanned by the columns of the matrix. You can see this by multiplying your matrix by a generic vector x. This produces a linear combination of the columns of the matrix.
$endgroup$
– Schippmunk
Dec 21 '18 at 20:10
add a comment |
$begingroup$
I'm trying to find the following set (some authors call it "column space"):
Given a matrix $A in mathbb{R} ^{m times n}$, we define:
$$R(A)={b in mathbb{R}^{m} | exists space xin mathbb{R} ^n, b=Ax }.$$
Let $A= left[ begin{array}{ccc}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{array} right]$.
My attempt goes as follows:
I'm trying to find the elements of $R(A)$. Notice that the elements of $R(A)$ are vectors in $mathbb{R} ^3 $. I want to know how those elements are. If I take a generic vector $b=(b_{1},b_{2},b_{3})^T in mathbb{R}^3$, then I'm trying to solve the following equation system:
$$ left[ begin{array}{ccc}
2 & 2 & 0 & 0 &|& b_{1} \
3 & 4 & -1 & 2 &|& b_{2}\
-1 & 1 & -2 & 4 &|& b_{3}end{array} right]$$
I calculated this echelon form for A.
$$ left[ begin{array}{ccc}
1 & -1 & 2 & -1 &|& -b_{3} \
0 & 7 & -7 & 14 &|& b_{2}+3b_{3}\
0 & 0 & 0 & 0 &|& b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3} end{array} right]$$.
(I think it is not necessary to put the steps to find the echelon form, but if you need it, let me know).
If I want the system $Ax=b$ to have solutions, it's clear that we need that $b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3}=0$, which is equivalent to $b_{1}=frac{4b_{2}-2b_{3}}{7}$.
Then my conclussion is that the elements of $R(A)$ must be all of the form $$left[ begin{array}{ccc}
b_{1} \
b_{2} \
b_{3} end{array} right]=left[ begin{array}{ccc}
frac{4b_{2}-2b_{3}}{7} \
b_{2} \
b_{3} end{array} right]$$
because I need that condition if I want the system $Ax=b$ having solution.
The question is, is this the correct way for doing that?
If there's not, which is the steps that I need to follow?
I'm just learning Linear Algebra.
Thanks :)
linear-algebra matrices proof-verification systems-of-equations
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
asked Dec 21 '18 at 19:36
rowcolrowcol
734
$endgroup$
I'm trying to find the following set (some authors call it "column space"):
Given a matrix $A in mathbb{R} ^{m times n}$, we define:
$$R(A)={b in mathbb{R}^{m} | exists space xin mathbb{R} ^n, b=Ax }.$$
Let $A= left[ begin{array}{ccc}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{array} right]$.
My attempt goes as follows:
I'm trying to find the elements of $R(A)$. Notice that the elements of $R(A)$ are vectors in $mathbb{R} ^3 $. I want to know how those elements are. If I take a generic vector $b=(b_{1},b_{2},b_{3})^T in mathbb{R}^3$, then I'm trying to solve the following equation system:
$$ left[ begin{array}{ccc}
2 & 2 & 0 & 0 &|& b_{1} \
3 & 4 & -1 & 2 &|& b_{2}\
-1 & 1 & -2 & 4 &|& b_{3}end{array} right]$$
I calculated this echelon form for A.
$$ left[ begin{array}{ccc}
1 & -1 & 2 & -1 &|& -b_{3} \
0 & 7 & -7 & 14 &|& b_{2}+3b_{3}\
0 & 0 & 0 & 0 &|& b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3} end{array} right]$$.
(I think it is not necessary to put the steps to find the echelon form, but if you need it, let me know).
If I want the system $Ax=b$ to have solutions, it's clear that we need that $b_{1}-frac{4}{7}b_{2}+frac{2}{7}b_{3}=0$, which is equivalent to $b_{1}=frac{4b_{2}-2b_{3}}{7}$.
Then my conclussion is that the elements of $R(A)$ must be all of the form $$left[ begin{array}{ccc}
b_{1} \
b_{2} \
b_{3} end{array} right]=left[ begin{array}{ccc}
frac{4b_{2}-2b_{3}}{7} \
b_{2} \
b_{3} end{array} right]$$
because I need that condition if I want the system $Ax=b$ having solution.
The question is, is this the correct way for doing that?
If there's not, which is the steps that I need to follow?
I'm just learning Linear Algebra.
Thanks :)
linear-algebra matrices proof-verification systems-of-equations
linear-algebra matrices proof-verification systems-of-equations
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
asked Dec 21 '18 at 19:36
rowcolrowcol
734
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
asked Dec 21 '18 at 19:36
rowcolrowcol
734
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
edited Dec 21 '18 at 20:06
Omnomnomnom
127k790180
127k790180
asked Dec 21 '18 at 19:36
rowcolrowcol
734
asked Dec 21 '18 at 19:36
rowcolrowcol
734
asked Dec 21 '18 at 19:36
rowcolrowcol
734
734
$begingroup$
Your process is perfectly fine, and your answer is correct. There are quicker ways of getting the answer in general, but there's nothing wrong with your method.
$endgroup$
– Omnomnomnom
Dec 21 '18 at 20:07
$begingroup$
The set you are looking for is called column space because it is the subvectorspace spanned by the columns of the matrix. You can see this by multiplying your matrix by a generic vector x. This produces a linear combination of the columns of the matrix.
$endgroup$
– Schippmunk
Dec 21 '18 at 20:10
add a comment |
$begingroup$
Your process is perfectly fine, and your answer is correct. There are quicker ways of getting the answer in general, but there's nothing wrong with your method.
$endgroup$
– Omnomnomnom
Dec 21 '18 at 20:07
$begingroup$
The set you are looking for is called column space because it is the subvectorspace spanned by the columns of the matrix. You can see this by multiplying your matrix by a generic vector x. This produces a linear combination of the columns of the matrix.
$endgroup$
– Schippmunk
Dec 21 '18 at 20:10
$begingroup$
Your process is perfectly fine, and your answer is correct. There are quicker ways of getting the answer in general, but there's nothing wrong with your method.
$endgroup$
– Omnomnomnom
Dec 21 '18 at 20:07
$begingroup$
Your process is perfectly fine, and your answer is correct. There are quicker ways of getting the answer in general, but there's nothing wrong with your method.
$endgroup$
– Omnomnomnom
Dec 21 '18 at 20:07
$begingroup$
The set you are looking for is called column space because it is the subvectorspace spanned by the columns of the matrix. You can see this by multiplying your matrix by a generic vector x. This produces a linear combination of the columns of the matrix.
$endgroup$
– Schippmunk
Dec 21 '18 at 20:10
$begingroup$
The set you are looking for is called column space because it is the subvectorspace spanned by the columns of the matrix. You can see this by multiplying your matrix by a generic vector x. This produces a linear combination of the columns of the matrix.
$endgroup$
– Schippmunk
Dec 21 '18 at 20:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Note that you obtained the basis vectors as $begin{bmatrix}4\7\0end{bmatrix},begin{bmatrix}-2\0\7end{bmatrix}$ which is correct. An alternative method to find a basis of $R(A)$ would be to assume $(x,y,z,w)inBbb R^4$ and find its image under $A$:
$begin{bmatrix}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{bmatrix}begin{bmatrix}x\y\z\wend{bmatrix}=begin{bmatrix}2x+2y\3x+4y-z+2w\-x+y-2z+4wend{bmatrix}=xbegin{bmatrix}2\3\-1end{bmatrix}+ybegin{bmatrix}2\4\1end{bmatrix}+zbegin{bmatrix}0\-1\-2end{bmatrix}+wbegin{bmatrix}0\2\4end{bmatrix}$
$therefore R(A)=text{span}Big{begin{bmatrix}2\3\-1end{bmatrix},begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix},begin{bmatrix}0\2\4end{bmatrix}Big}$
Notice that these are just the column vectors of $A^1$. The basis of $R(A)$ is given by the linearly independent vectors in the set; for example, $B=Big{begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix}Big}$. Can you show that the basis you got and $B$ are equivalent?
$^1$Let $T:V^nto W$ be a linear transformation. If
$v_1,v_2,v_3,...,v_n$ are the basis vectors of $V$, then
$R(T)subseteq W$ is spanned by $T(v_1),T(v_2),T(v_3),...,T(v_n)$.
For $V=Bbb R^4$, the most convenient choice of basis is the standard basis $B=Big{begin{bmatrix}1\0\0\0end{bmatrix},begin{bmatrix}0\1\0\0end{bmatrix},begin{bmatrix}0\0\1\0end{bmatrix},begin{bmatrix}0\0\0\1end{bmatrix}Big}$.
This gives $T((1,0,0,0))=Abegin{bmatrix}1\0\0\0end{bmatrix}$ as the first column of the matrix $A$, $T((0,1,0,0))=Abegin{bmatrix}0\1\0\0end{bmatrix}$ as the second column and so on. Thus, the range of $T$ is the column-space of $A$ and is spanned by the column vectors of $A$.
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
$endgroup$
$begingroup$
Thanks! I'm really not able to show that the basis I found and the set B in your answer are equivalent because I'm beginning at Linear Algebra and I'm not study yet the concepts of "span" and "basis"; in fact, I'm learning Linear System Equiations, but why should I know to prove that? thanks :)
$endgroup$
– rowcol
Dec 21 '18 at 20:32
$begingroup$
My apologies, I couldn't judge the level of the question. If you are not yet aware of these terms, your take-away from my answer should be that you solved the question correctly and that you will learn more about faster methods to reach $R(A)$ later. As for the equivalence of our answers, you can see that$$begin{bmatrix}4\7\0end{bmatrix}=2begin{bmatrix}2\4\1end{bmatrix}+begin{bmatrix}0\-1\-2end{bmatrix}\begin{bmatrix}-2\0\7end{bmatrix}=-1begin{bmatrix}2\4\1end{bmatrix}-4begin{bmatrix}0\-1\-2end{bmatrix}$$This means the two bases span the same subspace, that is $R(A)$.
$endgroup$
– Shubham Johri
Dec 21 '18 at 20:43
add a comment |
Your Answer
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1 Answer
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$begingroup$
Note that you obtained the basis vectors as $begin{bmatrix}4\7\0end{bmatrix},begin{bmatrix}-2\0\7end{bmatrix}$ which is correct. An alternative method to find a basis of $R(A)$ would be to assume $(x,y,z,w)inBbb R^4$ and find its image under $A$:
$begin{bmatrix}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{bmatrix}begin{bmatrix}x\y\z\wend{bmatrix}=begin{bmatrix}2x+2y\3x+4y-z+2w\-x+y-2z+4wend{bmatrix}=xbegin{bmatrix}2\3\-1end{bmatrix}+ybegin{bmatrix}2\4\1end{bmatrix}+zbegin{bmatrix}0\-1\-2end{bmatrix}+wbegin{bmatrix}0\2\4end{bmatrix}$
$therefore R(A)=text{span}Big{begin{bmatrix}2\3\-1end{bmatrix},begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix},begin{bmatrix}0\2\4end{bmatrix}Big}$
Notice that these are just the column vectors of $A^1$. The basis of $R(A)$ is given by the linearly independent vectors in the set; for example, $B=Big{begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix}Big}$. Can you show that the basis you got and $B$ are equivalent?
$^1$Let $T:V^nto W$ be a linear transformation. If
$v_1,v_2,v_3,...,v_n$ are the basis vectors of $V$, then
$R(T)subseteq W$ is spanned by $T(v_1),T(v_2),T(v_3),...,T(v_n)$.
For $V=Bbb R^4$, the most convenient choice of basis is the standard basis $B=Big{begin{bmatrix}1\0\0\0end{bmatrix},begin{bmatrix}0\1\0\0end{bmatrix},begin{bmatrix}0\0\1\0end{bmatrix},begin{bmatrix}0\0\0\1end{bmatrix}Big}$.
This gives $T((1,0,0,0))=Abegin{bmatrix}1\0\0\0end{bmatrix}$ as the first column of the matrix $A$, $T((0,1,0,0))=Abegin{bmatrix}0\1\0\0end{bmatrix}$ as the second column and so on. Thus, the range of $T$ is the column-space of $A$ and is spanned by the column vectors of $A$.
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
$endgroup$
$begingroup$
Thanks! I'm really not able to show that the basis I found and the set B in your answer are equivalent because I'm beginning at Linear Algebra and I'm not study yet the concepts of "span" and "basis"; in fact, I'm learning Linear System Equiations, but why should I know to prove that? thanks :)
$endgroup$
– rowcol
Dec 21 '18 at 20:32
$begingroup$
My apologies, I couldn't judge the level of the question. If you are not yet aware of these terms, your take-away from my answer should be that you solved the question correctly and that you will learn more about faster methods to reach $R(A)$ later. As for the equivalence of our answers, you can see that$$begin{bmatrix}4\7\0end{bmatrix}=2begin{bmatrix}2\4\1end{bmatrix}+begin{bmatrix}0\-1\-2end{bmatrix}\begin{bmatrix}-2\0\7end{bmatrix}=-1begin{bmatrix}2\4\1end{bmatrix}-4begin{bmatrix}0\-1\-2end{bmatrix}$$This means the two bases span the same subspace, that is $R(A)$.
$endgroup$
– Shubham Johri
Dec 21 '18 at 20:43
add a comment |
$begingroup$
Note that you obtained the basis vectors as $begin{bmatrix}4\7\0end{bmatrix},begin{bmatrix}-2\0\7end{bmatrix}$ which is correct. An alternative method to find a basis of $R(A)$ would be to assume $(x,y,z,w)inBbb R^4$ and find its image under $A$:
$begin{bmatrix}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{bmatrix}begin{bmatrix}x\y\z\wend{bmatrix}=begin{bmatrix}2x+2y\3x+4y-z+2w\-x+y-2z+4wend{bmatrix}=xbegin{bmatrix}2\3\-1end{bmatrix}+ybegin{bmatrix}2\4\1end{bmatrix}+zbegin{bmatrix}0\-1\-2end{bmatrix}+wbegin{bmatrix}0\2\4end{bmatrix}$
$therefore R(A)=text{span}Big{begin{bmatrix}2\3\-1end{bmatrix},begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix},begin{bmatrix}0\2\4end{bmatrix}Big}$
Notice that these are just the column vectors of $A^1$. The basis of $R(A)$ is given by the linearly independent vectors in the set; for example, $B=Big{begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix}Big}$. Can you show that the basis you got and $B$ are equivalent?
$^1$Let $T:V^nto W$ be a linear transformation. If
$v_1,v_2,v_3,...,v_n$ are the basis vectors of $V$, then
$R(T)subseteq W$ is spanned by $T(v_1),T(v_2),T(v_3),...,T(v_n)$.
For $V=Bbb R^4$, the most convenient choice of basis is the standard basis $B=Big{begin{bmatrix}1\0\0\0end{bmatrix},begin{bmatrix}0\1\0\0end{bmatrix},begin{bmatrix}0\0\1\0end{bmatrix},begin{bmatrix}0\0\0\1end{bmatrix}Big}$.
This gives $T((1,0,0,0))=Abegin{bmatrix}1\0\0\0end{bmatrix}$ as the first column of the matrix $A$, $T((0,1,0,0))=Abegin{bmatrix}0\1\0\0end{bmatrix}$ as the second column and so on. Thus, the range of $T$ is the column-space of $A$ and is spanned by the column vectors of $A$.
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
$endgroup$
$begingroup$
Thanks! I'm really not able to show that the basis I found and the set B in your answer are equivalent because I'm beginning at Linear Algebra and I'm not study yet the concepts of "span" and "basis"; in fact, I'm learning Linear System Equiations, but why should I know to prove that? thanks :)
$endgroup$
– rowcol
Dec 21 '18 at 20:32
$begingroup$
My apologies, I couldn't judge the level of the question. If you are not yet aware of these terms, your take-away from my answer should be that you solved the question correctly and that you will learn more about faster methods to reach $R(A)$ later. As for the equivalence of our answers, you can see that$$begin{bmatrix}4\7\0end{bmatrix}=2begin{bmatrix}2\4\1end{bmatrix}+begin{bmatrix}0\-1\-2end{bmatrix}\begin{bmatrix}-2\0\7end{bmatrix}=-1begin{bmatrix}2\4\1end{bmatrix}-4begin{bmatrix}0\-1\-2end{bmatrix}$$This means the two bases span the same subspace, that is $R(A)$.
$endgroup$
– Shubham Johri
Dec 21 '18 at 20:43
add a comment |
$begingroup$
Note that you obtained the basis vectors as $begin{bmatrix}4\7\0end{bmatrix},begin{bmatrix}-2\0\7end{bmatrix}$ which is correct. An alternative method to find a basis of $R(A)$ would be to assume $(x,y,z,w)inBbb R^4$ and find its image under $A$:
$begin{bmatrix}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{bmatrix}begin{bmatrix}x\y\z\wend{bmatrix}=begin{bmatrix}2x+2y\3x+4y-z+2w\-x+y-2z+4wend{bmatrix}=xbegin{bmatrix}2\3\-1end{bmatrix}+ybegin{bmatrix}2\4\1end{bmatrix}+zbegin{bmatrix}0\-1\-2end{bmatrix}+wbegin{bmatrix}0\2\4end{bmatrix}$
$therefore R(A)=text{span}Big{begin{bmatrix}2\3\-1end{bmatrix},begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix},begin{bmatrix}0\2\4end{bmatrix}Big}$
Notice that these are just the column vectors of $A^1$. The basis of $R(A)$ is given by the linearly independent vectors in the set; for example, $B=Big{begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix}Big}$. Can you show that the basis you got and $B$ are equivalent?
$^1$Let $T:V^nto W$ be a linear transformation. If
$v_1,v_2,v_3,...,v_n$ are the basis vectors of $V$, then
$R(T)subseteq W$ is spanned by $T(v_1),T(v_2),T(v_3),...,T(v_n)$.
For $V=Bbb R^4$, the most convenient choice of basis is the standard basis $B=Big{begin{bmatrix}1\0\0\0end{bmatrix},begin{bmatrix}0\1\0\0end{bmatrix},begin{bmatrix}0\0\1\0end{bmatrix},begin{bmatrix}0\0\0\1end{bmatrix}Big}$.
This gives $T((1,0,0,0))=Abegin{bmatrix}1\0\0\0end{bmatrix}$ as the first column of the matrix $A$, $T((0,1,0,0))=Abegin{bmatrix}0\1\0\0end{bmatrix}$ as the second column and so on. Thus, the range of $T$ is the column-space of $A$ and is spanned by the column vectors of $A$.
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
$endgroup$
Note that you obtained the basis vectors as $begin{bmatrix}4\7\0end{bmatrix},begin{bmatrix}-2\0\7end{bmatrix}$ which is correct. An alternative method to find a basis of $R(A)$ would be to assume $(x,y,z,w)inBbb R^4$ and find its image under $A$:
$begin{bmatrix}
2 & 2 & 0 & 0 \
3 & 4 & -1 & 2 \
-1 & 1 & -2 & 4 end{bmatrix}begin{bmatrix}x\y\z\wend{bmatrix}=begin{bmatrix}2x+2y\3x+4y-z+2w\-x+y-2z+4wend{bmatrix}=xbegin{bmatrix}2\3\-1end{bmatrix}+ybegin{bmatrix}2\4\1end{bmatrix}+zbegin{bmatrix}0\-1\-2end{bmatrix}+wbegin{bmatrix}0\2\4end{bmatrix}$
$therefore R(A)=text{span}Big{begin{bmatrix}2\3\-1end{bmatrix},begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix},begin{bmatrix}0\2\4end{bmatrix}Big}$
Notice that these are just the column vectors of $A^1$. The basis of $R(A)$ is given by the linearly independent vectors in the set; for example, $B=Big{begin{bmatrix}2\4\1end{bmatrix},begin{bmatrix}0\-1\-2end{bmatrix}Big}$. Can you show that the basis you got and $B$ are equivalent?
$^1$Let $T:V^nto W$ be a linear transformation. If
$v_1,v_2,v_3,...,v_n$ are the basis vectors of $V$, then
$R(T)subseteq W$ is spanned by $T(v_1),T(v_2),T(v_3),...,T(v_n)$.
For $V=Bbb R^4$, the most convenient choice of basis is the standard basis $B=Big{begin{bmatrix}1\0\0\0end{bmatrix},begin{bmatrix}0\1\0\0end{bmatrix},begin{bmatrix}0\0\1\0end{bmatrix},begin{bmatrix}0\0\0\1end{bmatrix}Big}$.
This gives $T((1,0,0,0))=Abegin{bmatrix}1\0\0\0end{bmatrix}$ as the first column of the matrix $A$, $T((0,1,0,0))=Abegin{bmatrix}0\1\0\0end{bmatrix}$ as the second column and so on. Thus, the range of $T$ is the column-space of $A$ and is spanned by the column vectors of $A$.
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
edited Dec 21 '18 at 20:30
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
answered Dec 21 '18 at 20:07
Shubham JohriShubham Johri
5,122717
5,122717
$begingroup$
Thanks! I'm really not able to show that the basis I found and the set B in your answer are equivalent because I'm beginning at Linear Algebra and I'm not study yet the concepts of "span" and "basis"; in fact, I'm learning Linear System Equiations, but why should I know to prove that? thanks :)
$endgroup$
– rowcol
Dec 21 '18 at 20:32
$begingroup$
My apologies, I couldn't judge the level of the question. If you are not yet aware of these terms, your take-away from my answer should be that you solved the question correctly and that you will learn more about faster methods to reach $R(A)$ later. As for the equivalence of our answers, you can see that$$begin{bmatrix}4\7\0end{bmatrix}=2begin{bmatrix}2\4\1end{bmatrix}+begin{bmatrix}0\-1\-2end{bmatrix}\begin{bmatrix}-2\0\7end{bmatrix}=-1begin{bmatrix}2\4\1end{bmatrix}-4begin{bmatrix}0\-1\-2end{bmatrix}$$This means the two bases span the same subspace, that is $R(A)$.
$endgroup$
– Shubham Johri
Dec 21 '18 at 20:43
add a comment |
$begingroup$
Thanks! I'm really not able to show that the basis I found and the set B in your answer are equivalent because I'm beginning at Linear Algebra and I'm not study yet the concepts of "span" and "basis"; in fact, I'm learning Linear System Equiations, but why should I know to prove that? thanks :)
$endgroup$
– rowcol
Dec 21 '18 at 20:32
$begingroup$
My apologies, I couldn't judge the level of the question. If you are not yet aware of these terms, your take-away from my answer should be that you solved the question correctly and that you will learn more about faster methods to reach $R(A)$ later. As for the equivalence of our answers, you can see that$$begin{bmatrix}4\7\0end{bmatrix}=2begin{bmatrix}2\4\1end{bmatrix}+begin{bmatrix}0\-1\-2end{bmatrix}\begin{bmatrix}-2\0\7end{bmatrix}=-1begin{bmatrix}2\4\1end{bmatrix}-4begin{bmatrix}0\-1\-2end{bmatrix}$$This means the two bases span the same subspace, that is $R(A)$.
$endgroup$
– Shubham Johri
Dec 21 '18 at 20:43
$begingroup$
Thanks! I'm really not able to show that the basis I found and the set B in your answer are equivalent because I'm beginning at Linear Algebra and I'm not study yet the concepts of "span" and "basis"; in fact, I'm learning Linear System Equiations, but why should I know to prove that? thanks :)
$endgroup$
– rowcol
Dec 21 '18 at 20:32
$begingroup$
Thanks! I'm really not able to show that the basis I found and the set B in your answer are equivalent because I'm beginning at Linear Algebra and I'm not study yet the concepts of "span" and "basis"; in fact, I'm learning Linear System Equiations, but why should I know to prove that? thanks :)
$endgroup$
– rowcol
Dec 21 '18 at 20:32
$begingroup$
My apologies, I couldn't judge the level of the question. If you are not yet aware of these terms, your take-away from my answer should be that you solved the question correctly and that you will learn more about faster methods to reach $R(A)$ later. As for the equivalence of our answers, you can see that$$begin{bmatrix}4\7\0end{bmatrix}=2begin{bmatrix}2\4\1end{bmatrix}+begin{bmatrix}0\-1\-2end{bmatrix}\begin{bmatrix}-2\0\7end{bmatrix}=-1begin{bmatrix}2\4\1end{bmatrix}-4begin{bmatrix}0\-1\-2end{bmatrix}$$This means the two bases span the same subspace, that is $R(A)$.
$endgroup$
– Shubham Johri
Dec 21 '18 at 20:43
$begingroup$
My apologies, I couldn't judge the level of the question. If you are not yet aware of these terms, your take-away from my answer should be that you solved the question correctly and that you will learn more about faster methods to reach $R(A)$ later. As for the equivalence of our answers, you can see that$$begin{bmatrix}4\7\0end{bmatrix}=2begin{bmatrix}2\4\1end{bmatrix}+begin{bmatrix}0\-1\-2end{bmatrix}\begin{bmatrix}-2\0\7end{bmatrix}=-1begin{bmatrix}2\4\1end{bmatrix}-4begin{bmatrix}0\-1\-2end{bmatrix}$$This means the two bases span the same subspace, that is $R(A)$.
$endgroup$
– Shubham Johri
Dec 21 '18 at 20:43
add a comment |
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$begingroup$
Your process is perfectly fine, and your answer is correct. There are quicker ways of getting the answer in general, but there's nothing wrong with your method.
$endgroup$
– Omnomnomnom
Dec 21 '18 at 20:07
$begingroup$
The set you are looking for is called column space because it is the subvectorspace spanned by the columns of the matrix. You can see this by multiplying your matrix by a generic vector x. This produces a linear combination of the columns of the matrix.
$endgroup$
– Schippmunk
Dec 21 '18 at 20:10