prove that 2 vectors is linearly independent in the vector space of vectors of length 2 with entries of...
I have the following vectors
$V_1=(e^t,te^t), V_2=(1,t)$
now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.
when I'm writing the formula:
$C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.
I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.
Would appreciate some thoughts.
linear-algebra vector-spaces
asked Dec 9 at 15:47
chemist
144
add a comment |
I have the following vectors
$V_1=(e^t,te^t), V_2=(1,t)$
now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.
when I'm writing the formula:
$C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.
I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.
Would appreciate some thoughts.
linear-algebra vector-spaces
asked Dec 9 at 15:47
chemist
144
add a comment |
I have the following vectors
$V_1=(e^t,te^t), V_2=(1,t)$
now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.
when I'm writing the formula:
$C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.
I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.
Would appreciate some thoughts.
linear-algebra vector-spaces
asked Dec 9 at 15:47
chemist
144
I have the following vectors
$V_1=(e^t,te^t), V_2=(1,t)$
now I want to prove that this two vectors are linearly independent in the vector space of vectors of length 2 with entries of real-valued functions.
when I'm writing the formula:
$C_1*V_1^T+C_2*V_2^T=0$ and writing it in matrix notation I can't get to prove that the coefficients needs to be 0.
I do note that if I'm fixing t for a given value then they are linearly dependent which is somewhat weird because i know i need to prove that for general t they are not.
Would appreciate some thoughts.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Dec 9 at 15:47
chemist
144
asked Dec 9 at 15:47
chemist
144
asked Dec 9 at 15:47
chemist
144
asked Dec 9 at 15:47
chemist
144
asked Dec 9 at 15:47
chemist
144
144
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You get the linear equations:
$$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$
The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:
$$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$
Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.
answered Dec 9 at 16:00
DonAntonio
177k1491225
so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
– chemist
Dec 9 at 16:16
I can't see anything "robust" in this proof. It is just going by the very simple definition ...
– DonAntonio
Dec 9 at 16:23
Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
– chemist
Dec 9 at 16:36
@chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
– DonAntonio
Dec 9 at 17:33
add a comment |
Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.
answered Dec 9 at 20:27
Christian Blatter
172k7112325
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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votes
You get the linear equations:
$$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$
The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:
$$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$
Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.
answered Dec 9 at 16:00
DonAntonio
177k1491225
so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
– chemist
Dec 9 at 16:16
I can't see anything "robust" in this proof. It is just going by the very simple definition ...
– DonAntonio
Dec 9 at 16:23
Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
– chemist
Dec 9 at 16:36
@chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
– DonAntonio
Dec 9 at 17:33
add a comment |
You get the linear equations:
$$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$
The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:
$$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$
Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.
answered Dec 9 at 16:00
DonAntonio
177k1491225
so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
– chemist
Dec 9 at 16:16
I can't see anything "robust" in this proof. It is just going by the very simple definition ...
– DonAntonio
Dec 9 at 16:23
Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
– chemist
Dec 9 at 16:36
@chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
– DonAntonio
Dec 9 at 17:33
add a comment |
You get the linear equations:
$$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$
The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:
$$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$
Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.
answered Dec 9 at 16:00
DonAntonio
177k1491225
You get the linear equations:
$$begin{cases}I&c_1e^t+c_2=0\{}\II&c_1te^t+c_2t=0end{cases}$$
The above two equations are true for any calue of $;tinBbb R;$ (I'm assuming the functions are defined on the whole real line), so take some particular values:
$$t=0stackrel Iimplies c_1+c_2=0implies c_2=-c_1stackrel {II}implies c_1tleft(e^t-1right)=0$$
Take now $;t=1;$ and you get $;c_1(e-1)=0implies c_1=0;$ and thus also $;c_2=0;$ and we're done.
answered Dec 9 at 16:00
DonAntonio
177k1491225
answered Dec 9 at 16:00
DonAntonio
177k1491225
answered Dec 9 at 16:00
DonAntonio
177k1491225
answered Dec 9 at 16:00
DonAntonio
177k1491225
177k1491225
so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
– chemist
Dec 9 at 16:16
I can't see anything "robust" in this proof. It is just going by the very simple definition ...
– DonAntonio
Dec 9 at 16:23
Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
– chemist
Dec 9 at 16:36
@chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
– DonAntonio
Dec 9 at 17:33
add a comment |
so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
– chemist
Dec 9 at 16:16
I can't see anything "robust" in this proof. It is just going by the very simple definition ...
– DonAntonio
Dec 9 at 16:23
Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
– chemist
Dec 9 at 16:36
@chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
– DonAntonio
Dec 9 at 17:33
so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
– chemist
Dec 9 at 16:16
so this is a valid proof, isn't any robust method for this kind of cases? i did it but for some reason it seems to simple for a proof
– chemist
Dec 9 at 16:16
I can't see anything "robust" in this proof. It is just going by the very simple definition ...
– DonAntonio
Dec 9 at 16:23
I can't see anything "robust" in this proof. It is just going by the very simple definition ...
– DonAntonio
Dec 9 at 16:23
Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
– chemist
Dec 9 at 16:36
Yeah i get that, but what if there were like k vectors, this would be the only way cause if i can represent in a matrix and isolate the part that depends on t, then its just matrix reduction
– chemist
Dec 9 at 16:36
@chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
– DonAntonio
Dec 9 at 17:33
@chemist I don't quite follow you. The above is a proof. Maybe there are other ones though I suspect not a simple and elementary as this one. If there were $;k;$ one could probably do the same, although then it would probably be a little more difficult.
– DonAntonio
Dec 9 at 17:33
add a comment |
Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.
answered Dec 9 at 20:27
Christian Blatter
172k7112325
add a comment |
Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.
answered Dec 9 at 20:27
Christian Blatter
172k7112325
add a comment |
Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.
answered Dec 9 at 20:27
Christian Blatter
172k7112325
Linear dependence of ${bf v}_1$ and ${bf v}_2$ in the vector space $X$ of functions ${bf x}:>tmapstobigl(x_1(t),x_2(t)bigr)$ means that there are fixed numbers $lambda$, $mu in{mathbb R}$, not both zero, such that $$lambda {bf v}_1(t)+mu {bf v}_2(t)={bf 0}qquadforall> t .tag{1}$$
Inspecting the first component of $(1)$ we obtain the condition $lambda e^t+mu=0$. This condition cannot be satisfied for all $t$ with fixed numbers $lambda$, $mu$, other than $lambda=mu=0$. It follows that the two given vectors are linearly independent in $X$.
answered Dec 9 at 20:27
Christian Blatter
172k7112325
answered Dec 9 at 20:27
Christian Blatter
172k7112325
answered Dec 9 at 20:27
Christian Blatter
172k7112325
answered Dec 9 at 20:27
Christian Blatter
172k7112325
172k7112325
add a comment |
add a comment |
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