Mappings to spaces with different numbers of dimensions
The first diagram
first question
second question
I seem to have a good understanding of linear transformation and linear algebra yet I fail to grasp this completely.
Question 1) Please explain how does we get that first equation.
2) Please explain how all the answers to the question 2 are correct.(except for "none of the above"
It would be greatly helpful.
linear-algebra matrices vectors linear-transformations
add a comment |
The first diagram
first question
second question
I seem to have a good understanding of linear transformation and linear algebra yet I fail to grasp this completely.
Question 1) Please explain how does we get that first equation.
2) Please explain how all the answers to the question 2 are correct.(except for "none of the above"
It would be greatly helpful.
linear-algebra matrices vectors linear-transformations
add a comment |
The first diagram
first question
second question
I seem to have a good understanding of linear transformation and linear algebra yet I fail to grasp this completely.
Question 1) Please explain how does we get that first equation.
2) Please explain how all the answers to the question 2 are correct.(except for "none of the above"
It would be greatly helpful.
linear-algebra matrices vectors linear-transformations
The first diagram
first question
second question
I seem to have a good understanding of linear transformation and linear algebra yet I fail to grasp this completely.
Question 1) Please explain how does we get that first equation.
2) Please explain how all the answers to the question 2 are correct.(except for "none of the above"
It would be greatly helpful.
linear-algebra matrices vectors linear-transformations
linear-algebra matrices vectors linear-transformations
asked Dec 9 at 11:52
MaverickEyedea
61
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$hat{mathbf s}$ is the direction the sun is shining, so the shadow of $bf r$ is going to be in that direction from $bf r$. How far is not yet determined, so all that can be said is that it is at $bf r + lambdahat{mathbf s}$ for some yet-to-be determined constant $lambda$.
On the assumption that when you referred to "the first equation", you failed to notice that $bf r' = r + lambdahat s$ is the first equation and were actually referring to the second, the plane representing the ground has equation ${bf hat e_3 cdot v} = 0$ (i.e., the $z$-coordinate of points on the ground is $0$). In particular, since the shadow point $bf r'$ is on the ground, we have ${bf r'cdot hat e}_3 = hat{mathbf e}_3 cdot mathbf r' = 0$. Therefore,
$$(mathbf r + lambdahat{mathbf s})cdothat{mathbf e}_3 = 0\{bf rcdothat e}_3 + lambda {bf hat scdothat e}_3 = 0\{bf rcdothat e}_3 + lambda s_3 = 0$$
where $s_3 = {bf hat scdothat e}_3$ is the $z$-coordinate of $bf hat s$.
For question 2, we also need to express $r_3 = {bf rcdothat e}_3$, the $z$-coordinate of $bf r$. Substituting that back into the equation gives $$r_3 + lambda s_3 = 0\lambda = -frac{r_3}{s_3}$$and so $$mathbf r' = mathbf r - frac{r_3}{s_3}hat{mathbf s}$$
From a coordinate point of view, this is actually 3 equations, one for each coordinate $i$:
$$r_i' = r_i - frac{r_3}{s_3}s_i$$
which is the third choice of question 2. Note that contrary to the way the question reads, this particular expression does not use the Einstein summation convention (no term has repeated variable subscripts). However, in the other answers, the author has chosen to abuse the summation convention to write the simple expression above in convoluted forms (I suppose the idea is just to give you practice interpreting the summation convention).
In the first expression
$$r_i' = r_i - frac{s_i}{s_3}[hat{mathbf e}_3]_jr_j$$ the subscript $j$ is repeated in the second term, so it is summed over all values of $j$. Leaving off the multiple that does not depend on $j$, the convention says
$$ [hat{mathbf e}_3]_jr_j = [hat{mathbf e}_3]_1r_1 + [hat{mathbf e}_3]_2r_2 + [hat{mathbf e}_3]_3r_3 = 0cdot r_1 + 0 cdot r_2 + 1cdot r_3 = r_3$$
since $$hat{mathbf e}_3= begin{bmatrix}0\0\1end{bmatrix}$$
Thus the first expression reduces to $$r_i' = r_i - frac{s_i}{s_3}r_3$$ which is the same as the third.
In the 4th and 5th answers, note that $I$ is the identity matrix, so $$I_{ij}=begin{cases}1& i = j\0&ine jend{cases}$$I'll leave the calculations showing that they also reduce to the 3rd answer to you.
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$hat{mathbf s}$ is the direction the sun is shining, so the shadow of $bf r$ is going to be in that direction from $bf r$. How far is not yet determined, so all that can be said is that it is at $bf r + lambdahat{mathbf s}$ for some yet-to-be determined constant $lambda$.
On the assumption that when you referred to "the first equation", you failed to notice that $bf r' = r + lambdahat s$ is the first equation and were actually referring to the second, the plane representing the ground has equation ${bf hat e_3 cdot v} = 0$ (i.e., the $z$-coordinate of points on the ground is $0$). In particular, since the shadow point $bf r'$ is on the ground, we have ${bf r'cdot hat e}_3 = hat{mathbf e}_3 cdot mathbf r' = 0$. Therefore,
$$(mathbf r + lambdahat{mathbf s})cdothat{mathbf e}_3 = 0\{bf rcdothat e}_3 + lambda {bf hat scdothat e}_3 = 0\{bf rcdothat e}_3 + lambda s_3 = 0$$
where $s_3 = {bf hat scdothat e}_3$ is the $z$-coordinate of $bf hat s$.
For question 2, we also need to express $r_3 = {bf rcdothat e}_3$, the $z$-coordinate of $bf r$. Substituting that back into the equation gives $$r_3 + lambda s_3 = 0\lambda = -frac{r_3}{s_3}$$and so $$mathbf r' = mathbf r - frac{r_3}{s_3}hat{mathbf s}$$
From a coordinate point of view, this is actually 3 equations, one for each coordinate $i$:
$$r_i' = r_i - frac{r_3}{s_3}s_i$$
which is the third choice of question 2. Note that contrary to the way the question reads, this particular expression does not use the Einstein summation convention (no term has repeated variable subscripts). However, in the other answers, the author has chosen to abuse the summation convention to write the simple expression above in convoluted forms (I suppose the idea is just to give you practice interpreting the summation convention).
In the first expression
$$r_i' = r_i - frac{s_i}{s_3}[hat{mathbf e}_3]_jr_j$$ the subscript $j$ is repeated in the second term, so it is summed over all values of $j$. Leaving off the multiple that does not depend on $j$, the convention says
$$ [hat{mathbf e}_3]_jr_j = [hat{mathbf e}_3]_1r_1 + [hat{mathbf e}_3]_2r_2 + [hat{mathbf e}_3]_3r_3 = 0cdot r_1 + 0 cdot r_2 + 1cdot r_3 = r_3$$
since $$hat{mathbf e}_3= begin{bmatrix}0\0\1end{bmatrix}$$
Thus the first expression reduces to $$r_i' = r_i - frac{s_i}{s_3}r_3$$ which is the same as the third.
In the 4th and 5th answers, note that $I$ is the identity matrix, so $$I_{ij}=begin{cases}1& i = j\0&ine jend{cases}$$I'll leave the calculations showing that they also reduce to the 3rd answer to you.
add a comment |
$hat{mathbf s}$ is the direction the sun is shining, so the shadow of $bf r$ is going to be in that direction from $bf r$. How far is not yet determined, so all that can be said is that it is at $bf r + lambdahat{mathbf s}$ for some yet-to-be determined constant $lambda$.
On the assumption that when you referred to "the first equation", you failed to notice that $bf r' = r + lambdahat s$ is the first equation and were actually referring to the second, the plane representing the ground has equation ${bf hat e_3 cdot v} = 0$ (i.e., the $z$-coordinate of points on the ground is $0$). In particular, since the shadow point $bf r'$ is on the ground, we have ${bf r'cdot hat e}_3 = hat{mathbf e}_3 cdot mathbf r' = 0$. Therefore,
$$(mathbf r + lambdahat{mathbf s})cdothat{mathbf e}_3 = 0\{bf rcdothat e}_3 + lambda {bf hat scdothat e}_3 = 0\{bf rcdothat e}_3 + lambda s_3 = 0$$
where $s_3 = {bf hat scdothat e}_3$ is the $z$-coordinate of $bf hat s$.
For question 2, we also need to express $r_3 = {bf rcdothat e}_3$, the $z$-coordinate of $bf r$. Substituting that back into the equation gives $$r_3 + lambda s_3 = 0\lambda = -frac{r_3}{s_3}$$and so $$mathbf r' = mathbf r - frac{r_3}{s_3}hat{mathbf s}$$
From a coordinate point of view, this is actually 3 equations, one for each coordinate $i$:
$$r_i' = r_i - frac{r_3}{s_3}s_i$$
which is the third choice of question 2. Note that contrary to the way the question reads, this particular expression does not use the Einstein summation convention (no term has repeated variable subscripts). However, in the other answers, the author has chosen to abuse the summation convention to write the simple expression above in convoluted forms (I suppose the idea is just to give you practice interpreting the summation convention).
In the first expression
$$r_i' = r_i - frac{s_i}{s_3}[hat{mathbf e}_3]_jr_j$$ the subscript $j$ is repeated in the second term, so it is summed over all values of $j$. Leaving off the multiple that does not depend on $j$, the convention says
$$ [hat{mathbf e}_3]_jr_j = [hat{mathbf e}_3]_1r_1 + [hat{mathbf e}_3]_2r_2 + [hat{mathbf e}_3]_3r_3 = 0cdot r_1 + 0 cdot r_2 + 1cdot r_3 = r_3$$
since $$hat{mathbf e}_3= begin{bmatrix}0\0\1end{bmatrix}$$
Thus the first expression reduces to $$r_i' = r_i - frac{s_i}{s_3}r_3$$ which is the same as the third.
In the 4th and 5th answers, note that $I$ is the identity matrix, so $$I_{ij}=begin{cases}1& i = j\0&ine jend{cases}$$I'll leave the calculations showing that they also reduce to the 3rd answer to you.
add a comment |
$hat{mathbf s}$ is the direction the sun is shining, so the shadow of $bf r$ is going to be in that direction from $bf r$. How far is not yet determined, so all that can be said is that it is at $bf r + lambdahat{mathbf s}$ for some yet-to-be determined constant $lambda$.
On the assumption that when you referred to "the first equation", you failed to notice that $bf r' = r + lambdahat s$ is the first equation and were actually referring to the second, the plane representing the ground has equation ${bf hat e_3 cdot v} = 0$ (i.e., the $z$-coordinate of points on the ground is $0$). In particular, since the shadow point $bf r'$ is on the ground, we have ${bf r'cdot hat e}_3 = hat{mathbf e}_3 cdot mathbf r' = 0$. Therefore,
$$(mathbf r + lambdahat{mathbf s})cdothat{mathbf e}_3 = 0\{bf rcdothat e}_3 + lambda {bf hat scdothat e}_3 = 0\{bf rcdothat e}_3 + lambda s_3 = 0$$
where $s_3 = {bf hat scdothat e}_3$ is the $z$-coordinate of $bf hat s$.
For question 2, we also need to express $r_3 = {bf rcdothat e}_3$, the $z$-coordinate of $bf r$. Substituting that back into the equation gives $$r_3 + lambda s_3 = 0\lambda = -frac{r_3}{s_3}$$and so $$mathbf r' = mathbf r - frac{r_3}{s_3}hat{mathbf s}$$
From a coordinate point of view, this is actually 3 equations, one for each coordinate $i$:
$$r_i' = r_i - frac{r_3}{s_3}s_i$$
which is the third choice of question 2. Note that contrary to the way the question reads, this particular expression does not use the Einstein summation convention (no term has repeated variable subscripts). However, in the other answers, the author has chosen to abuse the summation convention to write the simple expression above in convoluted forms (I suppose the idea is just to give you practice interpreting the summation convention).
In the first expression
$$r_i' = r_i - frac{s_i}{s_3}[hat{mathbf e}_3]_jr_j$$ the subscript $j$ is repeated in the second term, so it is summed over all values of $j$. Leaving off the multiple that does not depend on $j$, the convention says
$$ [hat{mathbf e}_3]_jr_j = [hat{mathbf e}_3]_1r_1 + [hat{mathbf e}_3]_2r_2 + [hat{mathbf e}_3]_3r_3 = 0cdot r_1 + 0 cdot r_2 + 1cdot r_3 = r_3$$
since $$hat{mathbf e}_3= begin{bmatrix}0\0\1end{bmatrix}$$
Thus the first expression reduces to $$r_i' = r_i - frac{s_i}{s_3}r_3$$ which is the same as the third.
In the 4th and 5th answers, note that $I$ is the identity matrix, so $$I_{ij}=begin{cases}1& i = j\0&ine jend{cases}$$I'll leave the calculations showing that they also reduce to the 3rd answer to you.
$hat{mathbf s}$ is the direction the sun is shining, so the shadow of $bf r$ is going to be in that direction from $bf r$. How far is not yet determined, so all that can be said is that it is at $bf r + lambdahat{mathbf s}$ for some yet-to-be determined constant $lambda$.
On the assumption that when you referred to "the first equation", you failed to notice that $bf r' = r + lambdahat s$ is the first equation and were actually referring to the second, the plane representing the ground has equation ${bf hat e_3 cdot v} = 0$ (i.e., the $z$-coordinate of points on the ground is $0$). In particular, since the shadow point $bf r'$ is on the ground, we have ${bf r'cdot hat e}_3 = hat{mathbf e}_3 cdot mathbf r' = 0$. Therefore,
$$(mathbf r + lambdahat{mathbf s})cdothat{mathbf e}_3 = 0\{bf rcdothat e}_3 + lambda {bf hat scdothat e}_3 = 0\{bf rcdothat e}_3 + lambda s_3 = 0$$
where $s_3 = {bf hat scdothat e}_3$ is the $z$-coordinate of $bf hat s$.
For question 2, we also need to express $r_3 = {bf rcdothat e}_3$, the $z$-coordinate of $bf r$. Substituting that back into the equation gives $$r_3 + lambda s_3 = 0\lambda = -frac{r_3}{s_3}$$and so $$mathbf r' = mathbf r - frac{r_3}{s_3}hat{mathbf s}$$
From a coordinate point of view, this is actually 3 equations, one for each coordinate $i$:
$$r_i' = r_i - frac{r_3}{s_3}s_i$$
which is the third choice of question 2. Note that contrary to the way the question reads, this particular expression does not use the Einstein summation convention (no term has repeated variable subscripts). However, in the other answers, the author has chosen to abuse the summation convention to write the simple expression above in convoluted forms (I suppose the idea is just to give you practice interpreting the summation convention).
In the first expression
$$r_i' = r_i - frac{s_i}{s_3}[hat{mathbf e}_3]_jr_j$$ the subscript $j$ is repeated in the second term, so it is summed over all values of $j$. Leaving off the multiple that does not depend on $j$, the convention says
$$ [hat{mathbf e}_3]_jr_j = [hat{mathbf e}_3]_1r_1 + [hat{mathbf e}_3]_2r_2 + [hat{mathbf e}_3]_3r_3 = 0cdot r_1 + 0 cdot r_2 + 1cdot r_3 = r_3$$
since $$hat{mathbf e}_3= begin{bmatrix}0\0\1end{bmatrix}$$
Thus the first expression reduces to $$r_i' = r_i - frac{s_i}{s_3}r_3$$ which is the same as the third.
In the 4th and 5th answers, note that $I$ is the identity matrix, so $$I_{ij}=begin{cases}1& i = j\0&ine jend{cases}$$I'll leave the calculations showing that they also reduce to the 3rd answer to you.
answered Dec 9 at 20:28
Paul Sinclair
19.3k21441
19.3k21441
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