Mixing
I didn't understand 3blue1brown's video on inverse matrices, where a matrix has no inverse if its determinant...
$begingroup$
I was watching 3blue1brown's video on inverse matrices, and I didn't understand what he said about the case where a matrix $A$ has no inverse when $det(A)=0$.
Considering $Ax=v$ where $x$and $v$ are vectors, he made two statements:
1) If a transformation squishes all of plane onto a single line for a $2times 2$ matrix then no transformation can unsquish that line back to a plane. (I didn't understand his explanation based on functions)
2) It is still possible that solution exists
even if inverse doesn't exist when $det(A)=0$ if vector $v$ "lives" somewhere on that line.
I don't understand why no such transformation exists in the first statement.
I also don't understand what vector $v$ "living" on the line means.
Here's the link to the video:
Video by 3blue1brown
linear-algebra linear-transformations
edited Jan 19 at 18:11
Blue
48.7k870156
asked Jan 19 at 17:53
user90596user90596
596
$endgroup$
add a comment |
$begingroup$
I was watching 3blue1brown's video on inverse matrices, and I didn't understand what he said about the case where a matrix $A$ has no inverse when $det(A)=0$.
Considering $Ax=v$ where $x$and $v$ are vectors, he made two statements:
1) If a transformation squishes all of plane onto a single line for a $2times 2$ matrix then no transformation can unsquish that line back to a plane. (I didn't understand his explanation based on functions)
2) It is still possible that solution exists
even if inverse doesn't exist when $det(A)=0$ if vector $v$ "lives" somewhere on that line.
I don't understand why no such transformation exists in the first statement.
I also don't understand what vector $v$ "living" on the line means.
Here's the link to the video:
Video by 3blue1brown
linear-algebra linear-transformations
edited Jan 19 at 18:11
Blue
48.7k870156
asked Jan 19 at 17:53
user90596user90596
596
$endgroup$
add a comment |
$begingroup$
I was watching 3blue1brown's video on inverse matrices, and I didn't understand what he said about the case where a matrix $A$ has no inverse when $det(A)=0$.
Considering $Ax=v$ where $x$and $v$ are vectors, he made two statements:
1) If a transformation squishes all of plane onto a single line for a $2times 2$ matrix then no transformation can unsquish that line back to a plane. (I didn't understand his explanation based on functions)
2) It is still possible that solution exists
even if inverse doesn't exist when $det(A)=0$ if vector $v$ "lives" somewhere on that line.
I don't understand why no such transformation exists in the first statement.
I also don't understand what vector $v$ "living" on the line means.
Here's the link to the video:
Video by 3blue1brown
linear-algebra linear-transformations
edited Jan 19 at 18:11
Blue
48.7k870156
asked Jan 19 at 17:53
user90596user90596
596
$endgroup$
I was watching 3blue1brown's video on inverse matrices, and I didn't understand what he said about the case where a matrix $A$ has no inverse when $det(A)=0$.
Considering $Ax=v$ where $x$and $v$ are vectors, he made two statements:
1) If a transformation squishes all of plane onto a single line for a $2times 2$ matrix then no transformation can unsquish that line back to a plane. (I didn't understand his explanation based on functions)
2) It is still possible that solution exists
even if inverse doesn't exist when $det(A)=0$ if vector $v$ "lives" somewhere on that line.
I don't understand why no such transformation exists in the first statement.
I also don't understand what vector $v$ "living" on the line means.
Here's the link to the video:
Video by 3blue1brown
linear-algebra linear-transformations
linear-algebra linear-transformations
edited Jan 19 at 18:11
Blue
48.7k870156
asked Jan 19 at 17:53
user90596user90596
596
edited Jan 19 at 18:11
Blue
48.7k870156
asked Jan 19 at 17:53
user90596user90596
596
edited Jan 19 at 18:11
Blue
48.7k870156
edited Jan 19 at 18:11
Blue
48.7k870156
edited Jan 19 at 18:11
Blue
48.7k870156
48.7k870156
asked Jan 19 at 17:53
user90596user90596
596
asked Jan 19 at 17:53
user90596user90596
596
asked Jan 19 at 17:53
user90596user90596
596
596
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It might help to look at an example. For instance, consider the matrix
$$
A = pmatrix{1&0\0&0}
$$
The transformation associated with this matrix is
$$
f(x) = Ax = pmatrix{1&0\0&0}pmatrix{x_1\x_2} = pmatrix{x_1\0}
$$
Geometrically, this transformation "squishes" $Bbb R^2$ onto the $x_1$-axis (or the $x$-axis, if you prefer thinking in terms of $x$ and $y$).
The point of an inverse transformation is to undo the effect of $f$. That is, we're looking for a transformation $g(x)$ such that $g(f(x)) = x$. That is, we would like to make a function that satisfies
$$
g(f(x_1,x_2)) = g(x_1,0) = (x_1,x_2)
$$
Intuitively, we see that this is impossible since, after applying $f$, we have lost the information of what $x_2$ was.
More formally and specifically: we know that every point of the form $(1,x_2)$ (i.e. every point of the form $x_1 = 1$) will get "squished" to the point $(1,0)$. So, for $g$ to undo $f$, we would need to have the following be true:
$$
g(f(1,3)) = g(1,0) = (1,3)\
g(f(1,-7)) = g(1,0) = (1,-7)
$$
The problem here is that since $g$ is a function, we can only send the point $(1,0)$ to one output. So, there is no one function $g$ that can give back the coordinates of every point we put in.
Regarding your second point: even though $det(A) = 0$, it is certainly possible for $Ax = v$ to have a solution if $v$ lives in the line (in our case, the $x_1$-axis). For instance, if we have $v = (-4,0)$, then $f(x) = v$ will be true whenever $x = (-4,x_2)$, i.e. whenever $x$ is a point that gets squished to $v$. On the other hand, if we have $v = (-4,2)$, then no $x$ will satisfy $f(x) = v$: $f(x)$ is necessarily on the $x_1$-axis, but $v$ doesn't "live" there.
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Thanks,I understand it now. Could you also explain the second part? Thanks..
$endgroup$
– user90596
Jan 19 at 18:12
$begingroup$
@user90596 see my latest edit
$endgroup$
– Omnomnomnom
Jan 19 at 18:13
add a comment |
$begingroup$
Allow me to offer my two cents. To see how the inverse of any such transformation could not be a function, think about the classic, non-matrixy definition of functions...for any y=f(x) , many x can be sent to the same y (a flat line), but the same x cannot be sent to many y (a vertical line). In matrix speak, this means that many input vectors can be sent to the same output vector, but the same input vector cannot be sent to many output vectors. By living on the line, 3b1b means that there does exist a bijection between vectors that start and end on the line that space is being squished into, because if we limit ourselves to think about these vectors then the effect of the transformation is some simple scaling action.
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It might help to look at an example. For instance, consider the matrix
$$
A = pmatrix{1&0\0&0}
$$
The transformation associated with this matrix is
$$
f(x) = Ax = pmatrix{1&0\0&0}pmatrix{x_1\x_2} = pmatrix{x_1\0}
$$
Geometrically, this transformation "squishes" $Bbb R^2$ onto the $x_1$-axis (or the $x$-axis, if you prefer thinking in terms of $x$ and $y$).
The point of an inverse transformation is to undo the effect of $f$. That is, we're looking for a transformation $g(x)$ such that $g(f(x)) = x$. That is, we would like to make a function that satisfies
$$
g(f(x_1,x_2)) = g(x_1,0) = (x_1,x_2)
$$
Intuitively, we see that this is impossible since, after applying $f$, we have lost the information of what $x_2$ was.
More formally and specifically: we know that every point of the form $(1,x_2)$ (i.e. every point of the form $x_1 = 1$) will get "squished" to the point $(1,0)$. So, for $g$ to undo $f$, we would need to have the following be true:
$$
g(f(1,3)) = g(1,0) = (1,3)\
g(f(1,-7)) = g(1,0) = (1,-7)
$$
The problem here is that since $g$ is a function, we can only send the point $(1,0)$ to one output. So, there is no one function $g$ that can give back the coordinates of every point we put in.
Regarding your second point: even though $det(A) = 0$, it is certainly possible for $Ax = v$ to have a solution if $v$ lives in the line (in our case, the $x_1$-axis). For instance, if we have $v = (-4,0)$, then $f(x) = v$ will be true whenever $x = (-4,x_2)$, i.e. whenever $x$ is a point that gets squished to $v$. On the other hand, if we have $v = (-4,2)$, then no $x$ will satisfy $f(x) = v$: $f(x)$ is necessarily on the $x_1$-axis, but $v$ doesn't "live" there.
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Thanks,I understand it now. Could you also explain the second part? Thanks..
$endgroup$
– user90596
Jan 19 at 18:12
$begingroup$
@user90596 see my latest edit
$endgroup$
– Omnomnomnom
Jan 19 at 18:13
add a comment |
$begingroup$
It might help to look at an example. For instance, consider the matrix
$$
A = pmatrix{1&0\0&0}
$$
The transformation associated with this matrix is
$$
f(x) = Ax = pmatrix{1&0\0&0}pmatrix{x_1\x_2} = pmatrix{x_1\0}
$$
Geometrically, this transformation "squishes" $Bbb R^2$ onto the $x_1$-axis (or the $x$-axis, if you prefer thinking in terms of $x$ and $y$).
The point of an inverse transformation is to undo the effect of $f$. That is, we're looking for a transformation $g(x)$ such that $g(f(x)) = x$. That is, we would like to make a function that satisfies
$$
g(f(x_1,x_2)) = g(x_1,0) = (x_1,x_2)
$$
Intuitively, we see that this is impossible since, after applying $f$, we have lost the information of what $x_2$ was.
More formally and specifically: we know that every point of the form $(1,x_2)$ (i.e. every point of the form $x_1 = 1$) will get "squished" to the point $(1,0)$. So, for $g$ to undo $f$, we would need to have the following be true:
$$
g(f(1,3)) = g(1,0) = (1,3)\
g(f(1,-7)) = g(1,0) = (1,-7)
$$
The problem here is that since $g$ is a function, we can only send the point $(1,0)$ to one output. So, there is no one function $g$ that can give back the coordinates of every point we put in.
Regarding your second point: even though $det(A) = 0$, it is certainly possible for $Ax = v$ to have a solution if $v$ lives in the line (in our case, the $x_1$-axis). For instance, if we have $v = (-4,0)$, then $f(x) = v$ will be true whenever $x = (-4,x_2)$, i.e. whenever $x$ is a point that gets squished to $v$. On the other hand, if we have $v = (-4,2)$, then no $x$ will satisfy $f(x) = v$: $f(x)$ is necessarily on the $x_1$-axis, but $v$ doesn't "live" there.
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
$begingroup$
Thanks,I understand it now. Could you also explain the second part? Thanks..
$endgroup$
– user90596
Jan 19 at 18:12
$begingroup$
@user90596 see my latest edit
$endgroup$
– Omnomnomnom
Jan 19 at 18:13
add a comment |
$begingroup$
It might help to look at an example. For instance, consider the matrix
$$
A = pmatrix{1&0\0&0}
$$
The transformation associated with this matrix is
$$
f(x) = Ax = pmatrix{1&0\0&0}pmatrix{x_1\x_2} = pmatrix{x_1\0}
$$
Geometrically, this transformation "squishes" $Bbb R^2$ onto the $x_1$-axis (or the $x$-axis, if you prefer thinking in terms of $x$ and $y$).
The point of an inverse transformation is to undo the effect of $f$. That is, we're looking for a transformation $g(x)$ such that $g(f(x)) = x$. That is, we would like to make a function that satisfies
$$
g(f(x_1,x_2)) = g(x_1,0) = (x_1,x_2)
$$
Intuitively, we see that this is impossible since, after applying $f$, we have lost the information of what $x_2$ was.
More formally and specifically: we know that every point of the form $(1,x_2)$ (i.e. every point of the form $x_1 = 1$) will get "squished" to the point $(1,0)$. So, for $g$ to undo $f$, we would need to have the following be true:
$$
g(f(1,3)) = g(1,0) = (1,3)\
g(f(1,-7)) = g(1,0) = (1,-7)
$$
The problem here is that since $g$ is a function, we can only send the point $(1,0)$ to one output. So, there is no one function $g$ that can give back the coordinates of every point we put in.
Regarding your second point: even though $det(A) = 0$, it is certainly possible for $Ax = v$ to have a solution if $v$ lives in the line (in our case, the $x_1$-axis). For instance, if we have $v = (-4,0)$, then $f(x) = v$ will be true whenever $x = (-4,x_2)$, i.e. whenever $x$ is a point that gets squished to $v$. On the other hand, if we have $v = (-4,2)$, then no $x$ will satisfy $f(x) = v$: $f(x)$ is necessarily on the $x_1$-axis, but $v$ doesn't "live" there.
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
$endgroup$
It might help to look at an example. For instance, consider the matrix
$$
A = pmatrix{1&0\0&0}
$$
The transformation associated with this matrix is
$$
f(x) = Ax = pmatrix{1&0\0&0}pmatrix{x_1\x_2} = pmatrix{x_1\0}
$$
Geometrically, this transformation "squishes" $Bbb R^2$ onto the $x_1$-axis (or the $x$-axis, if you prefer thinking in terms of $x$ and $y$).
The point of an inverse transformation is to undo the effect of $f$. That is, we're looking for a transformation $g(x)$ such that $g(f(x)) = x$. That is, we would like to make a function that satisfies
$$
g(f(x_1,x_2)) = g(x_1,0) = (x_1,x_2)
$$
Intuitively, we see that this is impossible since, after applying $f$, we have lost the information of what $x_2$ was.
More formally and specifically: we know that every point of the form $(1,x_2)$ (i.e. every point of the form $x_1 = 1$) will get "squished" to the point $(1,0)$. So, for $g$ to undo $f$, we would need to have the following be true:
$$
g(f(1,3)) = g(1,0) = (1,3)\
g(f(1,-7)) = g(1,0) = (1,-7)
$$
The problem here is that since $g$ is a function, we can only send the point $(1,0)$ to one output. So, there is no one function $g$ that can give back the coordinates of every point we put in.
Regarding your second point: even though $det(A) = 0$, it is certainly possible for $Ax = v$ to have a solution if $v$ lives in the line (in our case, the $x_1$-axis). For instance, if we have $v = (-4,0)$, then $f(x) = v$ will be true whenever $x = (-4,x_2)$, i.e. whenever $x$ is a point that gets squished to $v$. On the other hand, if we have $v = (-4,2)$, then no $x$ will satisfy $f(x) = v$: $f(x)$ is necessarily on the $x_1$-axis, but $v$ doesn't "live" there.
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
edited Jan 19 at 18:10
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
answered Jan 19 at 18:05
OmnomnomnomOmnomnomnom
128k791184
128k791184
$begingroup$
Thanks,I understand it now. Could you also explain the second part? Thanks..
$endgroup$
– user90596
Jan 19 at 18:12
$begingroup$
@user90596 see my latest edit
$endgroup$
– Omnomnomnom
Jan 19 at 18:13
add a comment |
$begingroup$
Thanks,I understand it now. Could you also explain the second part? Thanks..
$endgroup$
– user90596
Jan 19 at 18:12
$begingroup$
@user90596 see my latest edit
$endgroup$
– Omnomnomnom
Jan 19 at 18:13
$begingroup$
Thanks,I understand it now. Could you also explain the second part? Thanks..
$endgroup$
– user90596
Jan 19 at 18:12
$begingroup$
Thanks,I understand it now. Could you also explain the second part? Thanks..
$endgroup$
– user90596
Jan 19 at 18:12
$begingroup$
@user90596 see my latest edit
$endgroup$
– Omnomnomnom
Jan 19 at 18:13
$begingroup$
@user90596 see my latest edit
$endgroup$
– Omnomnomnom
Jan 19 at 18:13
add a comment |
$begingroup$
Allow me to offer my two cents. To see how the inverse of any such transformation could not be a function, think about the classic, non-matrixy definition of functions...for any y=f(x) , many x can be sent to the same y (a flat line), but the same x cannot be sent to many y (a vertical line). In matrix speak, this means that many input vectors can be sent to the same output vector, but the same input vector cannot be sent to many output vectors. By living on the line, 3b1b means that there does exist a bijection between vectors that start and end on the line that space is being squished into, because if we limit ourselves to think about these vectors then the effect of the transformation is some simple scaling action.
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
$endgroup$
add a comment |
$begingroup$
Allow me to offer my two cents. To see how the inverse of any such transformation could not be a function, think about the classic, non-matrixy definition of functions...for any y=f(x) , many x can be sent to the same y (a flat line), but the same x cannot be sent to many y (a vertical line). In matrix speak, this means that many input vectors can be sent to the same output vector, but the same input vector cannot be sent to many output vectors. By living on the line, 3b1b means that there does exist a bijection between vectors that start and end on the line that space is being squished into, because if we limit ourselves to think about these vectors then the effect of the transformation is some simple scaling action.
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
$endgroup$
add a comment |
$begingroup$
Allow me to offer my two cents. To see how the inverse of any such transformation could not be a function, think about the classic, non-matrixy definition of functions...for any y=f(x) , many x can be sent to the same y (a flat line), but the same x cannot be sent to many y (a vertical line). In matrix speak, this means that many input vectors can be sent to the same output vector, but the same input vector cannot be sent to many output vectors. By living on the line, 3b1b means that there does exist a bijection between vectors that start and end on the line that space is being squished into, because if we limit ourselves to think about these vectors then the effect of the transformation is some simple scaling action.
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
$endgroup$
Allow me to offer my two cents. To see how the inverse of any such transformation could not be a function, think about the classic, non-matrixy definition of functions...for any y=f(x) , many x can be sent to the same y (a flat line), but the same x cannot be sent to many y (a vertical line). In matrix speak, this means that many input vectors can be sent to the same output vector, but the same input vector cannot be sent to many output vectors. By living on the line, 3b1b means that there does exist a bijection between vectors that start and end on the line that space is being squished into, because if we limit ourselves to think about these vectors then the effect of the transformation is some simple scaling action.
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
answered Jan 19 at 18:40
Math EnthusiastMath Enthusiast
1168
1168
add a comment |
add a comment |
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