Check if functions are independent
$begingroup$
So I recently learned about how to check whether functions are independent.
As far as I understood it one of the methods is to plug in freely chosen values for x and you can calculate the determinate for the matrix that results from that.
$$ f_1(x), f_2(x), f_3(x)$$
$$begin{vmatrix}
f_1(x_1) & f_2(x_1) & f_3(x_1)\
f_1(x_2) & f_2(x_2) & f_3(x_2)\
f_1(x_3) & f_2(x_3) & f_3(x_3)\
end{vmatrix}$$
Where $x_1, x_2, x_3$ are the three values I chose for $x$.
I don't doubt my Professor but for me its still feels kinda "unmathy" if I might call it that way. So I was going to ask whether this a legit strategy or only works in certain scenarios.
Given that I have functions $f_1=cos^2x$, $f_2=sin^2ax$, $f_3=1$.
And I choose $x_1=0$, $x_2=pi/2$ $x_3=pi$
$$begin{vmatrix}
1 & 0 & 1\
0 & sin^2frac{api}{2} & 1\
1 & sin^2api &1\
end{vmatrix}$$
and the determinant would be $sin^2api$, so for each $a in mathbb{Z}$ the three functions would be dependent.
However I could also chose different value for the $x_1$... and I would get a different determinant
linear-algebra independence
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
asked Dec 16 '18 at 11:24
TimTim
63
$endgroup$
add a comment |
$begingroup$
So I recently learned about how to check whether functions are independent.
As far as I understood it one of the methods is to plug in freely chosen values for x and you can calculate the determinate for the matrix that results from that.
$$ f_1(x), f_2(x), f_3(x)$$
$$begin{vmatrix}
f_1(x_1) & f_2(x_1) & f_3(x_1)\
f_1(x_2) & f_2(x_2) & f_3(x_2)\
f_1(x_3) & f_2(x_3) & f_3(x_3)\
end{vmatrix}$$
Where $x_1, x_2, x_3$ are the three values I chose for $x$.
I don't doubt my Professor but for me its still feels kinda "unmathy" if I might call it that way. So I was going to ask whether this a legit strategy or only works in certain scenarios.
Given that I have functions $f_1=cos^2x$, $f_2=sin^2ax$, $f_3=1$.
And I choose $x_1=0$, $x_2=pi/2$ $x_3=pi$
$$begin{vmatrix}
1 & 0 & 1\
0 & sin^2frac{api}{2} & 1\
1 & sin^2api &1\
end{vmatrix}$$
and the determinant would be $sin^2api$, so for each $a in mathbb{Z}$ the three functions would be dependent.
However I could also chose different value for the $x_1$... and I would get a different determinant
linear-algebra independence
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
asked Dec 16 '18 at 11:24
TimTim
63
$endgroup$
$begingroup$
This will find linearly dependent functions each time but it may also suggest that linearly independent functions are linearly dependent. You may have been talking about polynomials only; in the case of checking whether $n$ polynomials of degree at most $n-1$ are linearly independent, this method will always work.
$endgroup$
– Ian
Dec 16 '18 at 11:29
$begingroup$
The criterion is that the functions are inedependent iff for every choice of $x_1,x_2,x_3$ the determinant is zero. If you have second-degree polynomials, as you said, you only need to check it for one triple (ehere all three are distinct, of course).
$endgroup$
– Mindlack
Dec 16 '18 at 11:35
$begingroup$
I edited my question to be more precies
$endgroup$
– Tim
Dec 16 '18 at 12:11
add a comment |
$begingroup$
So I recently learned about how to check whether functions are independent.
As far as I understood it one of the methods is to plug in freely chosen values for x and you can calculate the determinate for the matrix that results from that.
$$ f_1(x), f_2(x), f_3(x)$$
$$begin{vmatrix}
f_1(x_1) & f_2(x_1) & f_3(x_1)\
f_1(x_2) & f_2(x_2) & f_3(x_2)\
f_1(x_3) & f_2(x_3) & f_3(x_3)\
end{vmatrix}$$
Where $x_1, x_2, x_3$ are the three values I chose for $x$.
I don't doubt my Professor but for me its still feels kinda "unmathy" if I might call it that way. So I was going to ask whether this a legit strategy or only works in certain scenarios.
Given that I have functions $f_1=cos^2x$, $f_2=sin^2ax$, $f_3=1$.
And I choose $x_1=0$, $x_2=pi/2$ $x_3=pi$
$$begin{vmatrix}
1 & 0 & 1\
0 & sin^2frac{api}{2} & 1\
1 & sin^2api &1\
end{vmatrix}$$
and the determinant would be $sin^2api$, so for each $a in mathbb{Z}$ the three functions would be dependent.
However I could also chose different value for the $x_1$... and I would get a different determinant
linear-algebra independence
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
asked Dec 16 '18 at 11:24
TimTim
63
$endgroup$
So I recently learned about how to check whether functions are independent.
As far as I understood it one of the methods is to plug in freely chosen values for x and you can calculate the determinate for the matrix that results from that.
$$ f_1(x), f_2(x), f_3(x)$$
$$begin{vmatrix}
f_1(x_1) & f_2(x_1) & f_3(x_1)\
f_1(x_2) & f_2(x_2) & f_3(x_2)\
f_1(x_3) & f_2(x_3) & f_3(x_3)\
end{vmatrix}$$
Where $x_1, x_2, x_3$ are the three values I chose for $x$.
I don't doubt my Professor but for me its still feels kinda "unmathy" if I might call it that way. So I was going to ask whether this a legit strategy or only works in certain scenarios.
Given that I have functions $f_1=cos^2x$, $f_2=sin^2ax$, $f_3=1$.
And I choose $x_1=0$, $x_2=pi/2$ $x_3=pi$
$$begin{vmatrix}
1 & 0 & 1\
0 & sin^2frac{api}{2} & 1\
1 & sin^2api &1\
end{vmatrix}$$
and the determinant would be $sin^2api$, so for each $a in mathbb{Z}$ the three functions would be dependent.
However I could also chose different value for the $x_1$... and I would get a different determinant
linear-algebra independence
linear-algebra independence
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
asked Dec 16 '18 at 11:24
TimTim
63
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
asked Dec 16 '18 at 11:24
TimTim
63
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
edited Dec 16 '18 at 13:19
Davide Giraudo
127k17154268
127k17154268
asked Dec 16 '18 at 11:24
TimTim
63
asked Dec 16 '18 at 11:24
TimTim
63
asked Dec 16 '18 at 11:24
TimTim
63
63
$begingroup$
This will find linearly dependent functions each time but it may also suggest that linearly independent functions are linearly dependent. You may have been talking about polynomials only; in the case of checking whether $n$ polynomials of degree at most $n-1$ are linearly independent, this method will always work.
$endgroup$
– Ian
Dec 16 '18 at 11:29
$begingroup$
The criterion is that the functions are inedependent iff for every choice of $x_1,x_2,x_3$ the determinant is zero. If you have second-degree polynomials, as you said, you only need to check it for one triple (ehere all three are distinct, of course).
$endgroup$
– Mindlack
Dec 16 '18 at 11:35
$begingroup$
I edited my question to be more precies
$endgroup$
– Tim
Dec 16 '18 at 12:11
add a comment |
$begingroup$
This will find linearly dependent functions each time but it may also suggest that linearly independent functions are linearly dependent. You may have been talking about polynomials only; in the case of checking whether $n$ polynomials of degree at most $n-1$ are linearly independent, this method will always work.
$endgroup$
– Ian
Dec 16 '18 at 11:29
$begingroup$
The criterion is that the functions are inedependent iff for every choice of $x_1,x_2,x_3$ the determinant is zero. If you have second-degree polynomials, as you said, you only need to check it for one triple (ehere all three are distinct, of course).
$endgroup$
– Mindlack
Dec 16 '18 at 11:35
$begingroup$
I edited my question to be more precies
$endgroup$
– Tim
Dec 16 '18 at 12:11
$begingroup$
This will find linearly dependent functions each time but it may also suggest that linearly independent functions are linearly dependent. You may have been talking about polynomials only; in the case of checking whether $n$ polynomials of degree at most $n-1$ are linearly independent, this method will always work.
$endgroup$
– Ian
Dec 16 '18 at 11:29
$begingroup$
This will find linearly dependent functions each time but it may also suggest that linearly independent functions are linearly dependent. You may have been talking about polynomials only; in the case of checking whether $n$ polynomials of degree at most $n-1$ are linearly independent, this method will always work.
$endgroup$
– Ian
Dec 16 '18 at 11:29
$begingroup$
The criterion is that the functions are inedependent iff for every choice of $x_1,x_2,x_3$ the determinant is zero. If you have second-degree polynomials, as you said, you only need to check it for one triple (ehere all three are distinct, of course).
$endgroup$
– Mindlack
Dec 16 '18 at 11:35
$begingroup$
The criterion is that the functions are inedependent iff for every choice of $x_1,x_2,x_3$ the determinant is zero. If you have second-degree polynomials, as you said, you only need to check it for one triple (ehere all three are distinct, of course).
$endgroup$
– Mindlack
Dec 16 '18 at 11:35
$begingroup$
I edited my question to be more precies
$endgroup$
– Tim
Dec 16 '18 at 12:11
$begingroup$
I edited my question to be more precies
$endgroup$
– Tim
Dec 16 '18 at 12:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Possibly the professor lied; more likely he/she didn't say what you thought he/she said.
In fact if that determinant is non-zero then the functions are independent. But that only goes one way - if the determinant is zero that does not imply the functions are dependent.
(As commented, if the determinant is zero for every choice of $x_1,x_2,x_3$ then the functions are dependent. That's possibly not that useful as a "test", since it might take a while to check every one of the infinitely many possibilities.)
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
$begingroup$
But if the functions are independent than they have a non zero determinant right? So if the determinant is zero they must be dependent. Or am I confusing it with the independece of vectors.
$endgroup$
– Tim
Dec 16 '18 at 17:52
$begingroup$
You're confusing something. Say $f(t)=t$ and $g(t)=t^2$. What do you mean by "the determinant of $f$ and $g$", exactly?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 17:57
$begingroup$
Note: If $f(t)=t$ and $g(t)=t^2$ then the determinant of the matrix $begin{bmatrix}f(x_1)&g(x_1)\f(x_2)&g(x_2)end{bmatrix}$ can be zero or non-zero, depending on the choice of $x_1$ and $x_2$..
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:01
$begingroup$
Sorry I meant to ask whether when determinant of the matrix you obtain when plugging in values for $x_1$... .I am really struggling to find the answer for that problem
$endgroup$
– Tim
Dec 16 '18 at 18:38
$begingroup$
That question is answered in the Answer I posted! Let $f(t)=t$ and $g(t)=t^2$. Are $f$ and $g$ dependent? Now let $x_1=0$ and $x_2=1$; what is the determinant of that matrix?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:45
add a comment |
Your Answer
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1 Answer
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$begingroup$
Possibly the professor lied; more likely he/she didn't say what you thought he/she said.
In fact if that determinant is non-zero then the functions are independent. But that only goes one way - if the determinant is zero that does not imply the functions are dependent.
(As commented, if the determinant is zero for every choice of $x_1,x_2,x_3$ then the functions are dependent. That's possibly not that useful as a "test", since it might take a while to check every one of the infinitely many possibilities.)
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
$begingroup$
But if the functions are independent than they have a non zero determinant right? So if the determinant is zero they must be dependent. Or am I confusing it with the independece of vectors.
$endgroup$
– Tim
Dec 16 '18 at 17:52
$begingroup$
You're confusing something. Say $f(t)=t$ and $g(t)=t^2$. What do you mean by "the determinant of $f$ and $g$", exactly?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 17:57
$begingroup$
Note: If $f(t)=t$ and $g(t)=t^2$ then the determinant of the matrix $begin{bmatrix}f(x_1)&g(x_1)\f(x_2)&g(x_2)end{bmatrix}$ can be zero or non-zero, depending on the choice of $x_1$ and $x_2$..
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:01
$begingroup$
Sorry I meant to ask whether when determinant of the matrix you obtain when plugging in values for $x_1$... .I am really struggling to find the answer for that problem
$endgroup$
– Tim
Dec 16 '18 at 18:38
$begingroup$
That question is answered in the Answer I posted! Let $f(t)=t$ and $g(t)=t^2$. Are $f$ and $g$ dependent? Now let $x_1=0$ and $x_2=1$; what is the determinant of that matrix?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:45
add a comment |
$begingroup$
Possibly the professor lied; more likely he/she didn't say what you thought he/she said.
In fact if that determinant is non-zero then the functions are independent. But that only goes one way - if the determinant is zero that does not imply the functions are dependent.
(As commented, if the determinant is zero for every choice of $x_1,x_2,x_3$ then the functions are dependent. That's possibly not that useful as a "test", since it might take a while to check every one of the infinitely many possibilities.)
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
$begingroup$
But if the functions are independent than they have a non zero determinant right? So if the determinant is zero they must be dependent. Or am I confusing it with the independece of vectors.
$endgroup$
– Tim
Dec 16 '18 at 17:52
$begingroup$
You're confusing something. Say $f(t)=t$ and $g(t)=t^2$. What do you mean by "the determinant of $f$ and $g$", exactly?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 17:57
$begingroup$
Note: If $f(t)=t$ and $g(t)=t^2$ then the determinant of the matrix $begin{bmatrix}f(x_1)&g(x_1)\f(x_2)&g(x_2)end{bmatrix}$ can be zero or non-zero, depending on the choice of $x_1$ and $x_2$..
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:01
$begingroup$
Sorry I meant to ask whether when determinant of the matrix you obtain when plugging in values for $x_1$... .I am really struggling to find the answer for that problem
$endgroup$
– Tim
Dec 16 '18 at 18:38
$begingroup$
That question is answered in the Answer I posted! Let $f(t)=t$ and $g(t)=t^2$. Are $f$ and $g$ dependent? Now let $x_1=0$ and $x_2=1$; what is the determinant of that matrix?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:45
add a comment |
$begingroup$
Possibly the professor lied; more likely he/she didn't say what you thought he/she said.
In fact if that determinant is non-zero then the functions are independent. But that only goes one way - if the determinant is zero that does not imply the functions are dependent.
(As commented, if the determinant is zero for every choice of $x_1,x_2,x_3$ then the functions are dependent. That's possibly not that useful as a "test", since it might take a while to check every one of the infinitely many possibilities.)
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
$endgroup$
Possibly the professor lied; more likely he/she didn't say what you thought he/she said.
In fact if that determinant is non-zero then the functions are independent. But that only goes one way - if the determinant is zero that does not imply the functions are dependent.
(As commented, if the determinant is zero for every choice of $x_1,x_2,x_3$ then the functions are dependent. That's possibly not that useful as a "test", since it might take a while to check every one of the infinitely many possibilities.)
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
answered Dec 16 '18 at 15:12
David C. UllrichDavid C. Ullrich
61.2k43994
61.2k43994
$begingroup$
But if the functions are independent than they have a non zero determinant right? So if the determinant is zero they must be dependent. Or am I confusing it with the independece of vectors.
$endgroup$
– Tim
Dec 16 '18 at 17:52
$begingroup$
You're confusing something. Say $f(t)=t$ and $g(t)=t^2$. What do you mean by "the determinant of $f$ and $g$", exactly?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 17:57
$begingroup$
Note: If $f(t)=t$ and $g(t)=t^2$ then the determinant of the matrix $begin{bmatrix}f(x_1)&g(x_1)\f(x_2)&g(x_2)end{bmatrix}$ can be zero or non-zero, depending on the choice of $x_1$ and $x_2$..
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:01
$begingroup$
Sorry I meant to ask whether when determinant of the matrix you obtain when plugging in values for $x_1$... .I am really struggling to find the answer for that problem
$endgroup$
– Tim
Dec 16 '18 at 18:38
$begingroup$
That question is answered in the Answer I posted! Let $f(t)=t$ and $g(t)=t^2$. Are $f$ and $g$ dependent? Now let $x_1=0$ and $x_2=1$; what is the determinant of that matrix?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:45
add a comment |
$begingroup$
But if the functions are independent than they have a non zero determinant right? So if the determinant is zero they must be dependent. Or am I confusing it with the independece of vectors.
$endgroup$
– Tim
Dec 16 '18 at 17:52
$begingroup$
You're confusing something. Say $f(t)=t$ and $g(t)=t^2$. What do you mean by "the determinant of $f$ and $g$", exactly?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 17:57
$begingroup$
Note: If $f(t)=t$ and $g(t)=t^2$ then the determinant of the matrix $begin{bmatrix}f(x_1)&g(x_1)\f(x_2)&g(x_2)end{bmatrix}$ can be zero or non-zero, depending on the choice of $x_1$ and $x_2$..
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:01
$begingroup$
Sorry I meant to ask whether when determinant of the matrix you obtain when plugging in values for $x_1$... .I am really struggling to find the answer for that problem
$endgroup$
– Tim
Dec 16 '18 at 18:38
$begingroup$
That question is answered in the Answer I posted! Let $f(t)=t$ and $g(t)=t^2$. Are $f$ and $g$ dependent? Now let $x_1=0$ and $x_2=1$; what is the determinant of that matrix?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:45
$begingroup$
But if the functions are independent than they have a non zero determinant right? So if the determinant is zero they must be dependent. Or am I confusing it with the independece of vectors.
$endgroup$
– Tim
Dec 16 '18 at 17:52
$begingroup$
But if the functions are independent than they have a non zero determinant right? So if the determinant is zero they must be dependent. Or am I confusing it with the independece of vectors.
$endgroup$
– Tim
Dec 16 '18 at 17:52
$begingroup$
You're confusing something. Say $f(t)=t$ and $g(t)=t^2$. What do you mean by "the determinant of $f$ and $g$", exactly?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 17:57
$begingroup$
You're confusing something. Say $f(t)=t$ and $g(t)=t^2$. What do you mean by "the determinant of $f$ and $g$", exactly?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 17:57
$begingroup$
Note: If $f(t)=t$ and $g(t)=t^2$ then the determinant of the matrix $begin{bmatrix}f(x_1)&g(x_1)\f(x_2)&g(x_2)end{bmatrix}$ can be zero or non-zero, depending on the choice of $x_1$ and $x_2$..
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:01
$begingroup$
Note: If $f(t)=t$ and $g(t)=t^2$ then the determinant of the matrix $begin{bmatrix}f(x_1)&g(x_1)\f(x_2)&g(x_2)end{bmatrix}$ can be zero or non-zero, depending on the choice of $x_1$ and $x_2$..
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:01
$begingroup$
Sorry I meant to ask whether when determinant of the matrix you obtain when plugging in values for $x_1$... .I am really struggling to find the answer for that problem
$endgroup$
– Tim
Dec 16 '18 at 18:38
$begingroup$
Sorry I meant to ask whether when determinant of the matrix you obtain when plugging in values for $x_1$... .I am really struggling to find the answer for that problem
$endgroup$
– Tim
Dec 16 '18 at 18:38
$begingroup$
That question is answered in the Answer I posted! Let $f(t)=t$ and $g(t)=t^2$. Are $f$ and $g$ dependent? Now let $x_1=0$ and $x_2=1$; what is the determinant of that matrix?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:45
$begingroup$
That question is answered in the Answer I posted! Let $f(t)=t$ and $g(t)=t^2$. Are $f$ and $g$ dependent? Now let $x_1=0$ and $x_2=1$; what is the determinant of that matrix?
$endgroup$
– David C. Ullrich
Dec 16 '18 at 18:45
add a comment |
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This will find linearly dependent functions each time but it may also suggest that linearly independent functions are linearly dependent. You may have been talking about polynomials only; in the case of checking whether $n$ polynomials of degree at most $n-1$ are linearly independent, this method will always work.
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– Ian
Dec 16 '18 at 11:29
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The criterion is that the functions are inedependent iff for every choice of $x_1,x_2,x_3$ the determinant is zero. If you have second-degree polynomials, as you said, you only need to check it for one triple (ehere all three are distinct, of course).
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– Mindlack
Dec 16 '18 at 11:35
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I edited my question to be more precies
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– Tim
Dec 16 '18 at 12:11