Let $S$ be the subspace of $Bbb R^3$ spanned by the vector $x= (x_1,x_2,x_3)^T$ and $y= (y_1,y_2,y_3)^T$
$begingroup$
(a)
Let $S$ be the subspace of $Bbb R^3$ Spanned by the vector $x= (x_1,x_2,x_3)^T$ and $y= (y_1,y_2,y_3)^T$, let $A =begin{bmatrix}x_1 &x_2 & x_3 \
y_1 &y_2 &y_3 end{bmatrix}$, show that $S^⊥=N(A)$.
(b)
Find the orthogonal complement of the subspace of $Bbb R^3$ spanned by $(1,2,1)^T$ and $(1,-1,2)^T$.
I'm still so confused by the concept of $S^⊥$, my book defines $Y^⊥= {x∈Bbb R^nmid x^T y=0 text{ for every } y∈Y}$. So to solve (a), I think I have to find $S^T$, I guess it should be $X^T y=0$. Then to find $N(A)$, let $Ax=0$, but I don't know what to do next, so I really need help to solve this kind of question, thanks!
linear-algebra orthogonality orthogonal-polynomials
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
$endgroup$
add a comment |
$begingroup$
(a)
Let $S$ be the subspace of $Bbb R^3$ Spanned by the vector $x= (x_1,x_2,x_3)^T$ and $y= (y_1,y_2,y_3)^T$, let $A =begin{bmatrix}x_1 &x_2 & x_3 \
y_1 &y_2 &y_3 end{bmatrix}$, show that $S^⊥=N(A)$.
(b)
Find the orthogonal complement of the subspace of $Bbb R^3$ spanned by $(1,2,1)^T$ and $(1,-1,2)^T$.
I'm still so confused by the concept of $S^⊥$, my book defines $Y^⊥= {x∈Bbb R^nmid x^T y=0 text{ for every } y∈Y}$. So to solve (a), I think I have to find $S^T$, I guess it should be $X^T y=0$. Then to find $N(A)$, let $Ax=0$, but I don't know what to do next, so I really need help to solve this kind of question, thanks!
linear-algebra orthogonality orthogonal-polynomials
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
$endgroup$
add a comment |
$begingroup$
(a)
Let $S$ be the subspace of $Bbb R^3$ Spanned by the vector $x= (x_1,x_2,x_3)^T$ and $y= (y_1,y_2,y_3)^T$, let $A =begin{bmatrix}x_1 &x_2 & x_3 \
y_1 &y_2 &y_3 end{bmatrix}$, show that $S^⊥=N(A)$.
(b)
Find the orthogonal complement of the subspace of $Bbb R^3$ spanned by $(1,2,1)^T$ and $(1,-1,2)^T$.
I'm still so confused by the concept of $S^⊥$, my book defines $Y^⊥= {x∈Bbb R^nmid x^T y=0 text{ for every } y∈Y}$. So to solve (a), I think I have to find $S^T$, I guess it should be $X^T y=0$. Then to find $N(A)$, let $Ax=0$, but I don't know what to do next, so I really need help to solve this kind of question, thanks!
linear-algebra orthogonality orthogonal-polynomials
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
$endgroup$
(a)
Let $S$ be the subspace of $Bbb R^3$ Spanned by the vector $x= (x_1,x_2,x_3)^T$ and $y= (y_1,y_2,y_3)^T$, let $A =begin{bmatrix}x_1 &x_2 & x_3 \
y_1 &y_2 &y_3 end{bmatrix}$, show that $S^⊥=N(A)$.
(b)
Find the orthogonal complement of the subspace of $Bbb R^3$ spanned by $(1,2,1)^T$ and $(1,-1,2)^T$.
I'm still so confused by the concept of $S^⊥$, my book defines $Y^⊥= {x∈Bbb R^nmid x^T y=0 text{ for every } y∈Y}$. So to solve (a), I think I have to find $S^T$, I guess it should be $X^T y=0$. Then to find $N(A)$, let $Ax=0$, but I don't know what to do next, so I really need help to solve this kind of question, thanks!
linear-algebra orthogonality orthogonal-polynomials
linear-algebra orthogonality orthogonal-polynomials
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
edited Dec 4 '18 at 10:04
Tianlalu
3,08621038
3,08621038
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
asked Dec 4 '18 at 3:29
Shadow ZShadow Z
396
396
add a comment |
add a comment |
1 Answer
1
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$begingroup$
The rows of $A$ are $x$ and $y$, and so $S$ is the row space of A:$$S={alpha x+beta y: alpha,betainmathbb{R}}$$
Therefore, $S^⊥={zinmathbb{R^3: (alpha x+beta y+z)=0}}$ for all $alpha,beta$ ; in particular,
$(x, z) = 0$ and $(y, z) = 0$. But the definition of matrix multiplication says that$$Az=begin{bmatrix} (x,z) \ (y,z)end{bmatrix}=0$$Hence, $zin N(A)$
For the reverse inclusion, suppose that $z in N(A)$. Now $Az = 0$. As above, the definition of matrix multiplication gives $(x, z) = 0 = (y, z)$. Hence, for all
scalars $alpha,beta$ we have$$(alpha x+beta y,z)=alpha(x,z)+beta(y,z)=0$$ So, $zin S^⊥$
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
$endgroup$
$begingroup$
Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0?
$endgroup$
– Shadow Z
Dec 4 '18 at 3:59
$begingroup$
@ShadowZ Since, $S$ is the row space of $A$
$endgroup$
– Key Flex
Dec 4 '18 at 4:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The rows of $A$ are $x$ and $y$, and so $S$ is the row space of A:$$S={alpha x+beta y: alpha,betainmathbb{R}}$$
Therefore, $S^⊥={zinmathbb{R^3: (alpha x+beta y+z)=0}}$ for all $alpha,beta$ ; in particular,
$(x, z) = 0$ and $(y, z) = 0$. But the definition of matrix multiplication says that$$Az=begin{bmatrix} (x,z) \ (y,z)end{bmatrix}=0$$Hence, $zin N(A)$
For the reverse inclusion, suppose that $z in N(A)$. Now $Az = 0$. As above, the definition of matrix multiplication gives $(x, z) = 0 = (y, z)$. Hence, for all
scalars $alpha,beta$ we have$$(alpha x+beta y,z)=alpha(x,z)+beta(y,z)=0$$ So, $zin S^⊥$
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
$endgroup$
$begingroup$
Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0?
$endgroup$
– Shadow Z
Dec 4 '18 at 3:59
$begingroup$
@ShadowZ Since, $S$ is the row space of $A$
$endgroup$
– Key Flex
Dec 4 '18 at 4:49
add a comment |
$begingroup$
The rows of $A$ are $x$ and $y$, and so $S$ is the row space of A:$$S={alpha x+beta y: alpha,betainmathbb{R}}$$
Therefore, $S^⊥={zinmathbb{R^3: (alpha x+beta y+z)=0}}$ for all $alpha,beta$ ; in particular,
$(x, z) = 0$ and $(y, z) = 0$. But the definition of matrix multiplication says that$$Az=begin{bmatrix} (x,z) \ (y,z)end{bmatrix}=0$$Hence, $zin N(A)$
For the reverse inclusion, suppose that $z in N(A)$. Now $Az = 0$. As above, the definition of matrix multiplication gives $(x, z) = 0 = (y, z)$. Hence, for all
scalars $alpha,beta$ we have$$(alpha x+beta y,z)=alpha(x,z)+beta(y,z)=0$$ So, $zin S^⊥$
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
$endgroup$
$begingroup$
Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0?
$endgroup$
– Shadow Z
Dec 4 '18 at 3:59
$begingroup$
@ShadowZ Since, $S$ is the row space of $A$
$endgroup$
– Key Flex
Dec 4 '18 at 4:49
add a comment |
$begingroup$
The rows of $A$ are $x$ and $y$, and so $S$ is the row space of A:$$S={alpha x+beta y: alpha,betainmathbb{R}}$$
Therefore, $S^⊥={zinmathbb{R^3: (alpha x+beta y+z)=0}}$ for all $alpha,beta$ ; in particular,
$(x, z) = 0$ and $(y, z) = 0$. But the definition of matrix multiplication says that$$Az=begin{bmatrix} (x,z) \ (y,z)end{bmatrix}=0$$Hence, $zin N(A)$
For the reverse inclusion, suppose that $z in N(A)$. Now $Az = 0$. As above, the definition of matrix multiplication gives $(x, z) = 0 = (y, z)$. Hence, for all
scalars $alpha,beta$ we have$$(alpha x+beta y,z)=alpha(x,z)+beta(y,z)=0$$ So, $zin S^⊥$
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
$endgroup$
The rows of $A$ are $x$ and $y$, and so $S$ is the row space of A:$$S={alpha x+beta y: alpha,betainmathbb{R}}$$
Therefore, $S^⊥={zinmathbb{R^3: (alpha x+beta y+z)=0}}$ for all $alpha,beta$ ; in particular,
$(x, z) = 0$ and $(y, z) = 0$. But the definition of matrix multiplication says that$$Az=begin{bmatrix} (x,z) \ (y,z)end{bmatrix}=0$$Hence, $zin N(A)$
For the reverse inclusion, suppose that $z in N(A)$. Now $Az = 0$. As above, the definition of matrix multiplication gives $(x, z) = 0 = (y, z)$. Hence, for all
scalars $alpha,beta$ we have$$(alpha x+beta y,z)=alpha(x,z)+beta(y,z)=0$$ So, $zin S^⊥$
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
answered Dec 4 '18 at 3:37
Key FlexKey Flex
7,78461232
7,78461232
$begingroup$
Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0?
$endgroup$
– Shadow Z
Dec 4 '18 at 3:59
$begingroup$
@ShadowZ Since, $S$ is the row space of $A$
$endgroup$
– Key Flex
Dec 4 '18 at 4:49
add a comment |
$begingroup$
Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0?
$endgroup$
– Shadow Z
Dec 4 '18 at 3:59
$begingroup$
@ShadowZ Since, $S$ is the row space of $A$
$endgroup$
– Key Flex
Dec 4 '18 at 4:49
$begingroup$
Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0?
$endgroup$
– Shadow Z
Dec 4 '18 at 3:59
$begingroup$
Thanks though! I wonder why S⊥ can be written as αx+βy+z=0 without using its definition: x^Ty=0?
$endgroup$
– Shadow Z
Dec 4 '18 at 3:59
$begingroup$
@ShadowZ Since, $S$ is the row space of $A$
$endgroup$
– Key Flex
Dec 4 '18 at 4:49
$begingroup$
@ShadowZ Since, $S$ is the row space of $A$
$endgroup$
– Key Flex
Dec 4 '18 at 4:49
add a comment |
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