Mixing
How can i prove that linear function does not form a vector space basis of $mathcal{C}bigl([0,1]bigr)$ over...
$begingroup$
Prove that the linear function does not form a vector space basis of $mathcal{C}bigl([0,1]bigr)$ over $mathbb R$?
i was trying this question many times but i could not get it.
i was taking the scalar multiplication and addition property
it was showing that this vector form vector space
ie it contain the zero vector
i dont know from where i have to start
if anybody help me , i would be very thankful to him
linear-algebra functional-analysis linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Aug 5 '17 at 15:37
lomberlomber
786420
$endgroup$
add a comment |
$begingroup$
Prove that the linear function does not form a vector space basis of $mathcal{C}bigl([0,1]bigr)$ over $mathbb R$?
i was trying this question many times but i could not get it.
i was taking the scalar multiplication and addition property
it was showing that this vector form vector space
ie it contain the zero vector
i dont know from where i have to start
if anybody help me , i would be very thankful to him
linear-algebra functional-analysis linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Aug 5 '17 at 15:37
lomberlomber
786420
$endgroup$
$begingroup$
whats "the linear function?" do you mean the linear functions? If so, then every linear combination of linear functions is again linear and the function $x^2$ is not linear.
$endgroup$
– Yanko
Aug 5 '17 at 15:41
$begingroup$
Simply, any linear combination of linear functions is weakly monotonic but not every continuous function over $[0,1]$ is weakly monotonic.
$endgroup$
– Jack D'Aurizio
Aug 5 '17 at 15:43
add a comment |
$begingroup$
Prove that the linear function does not form a vector space basis of $mathcal{C}bigl([0,1]bigr)$ over $mathbb R$?
i was trying this question many times but i could not get it.
i was taking the scalar multiplication and addition property
it was showing that this vector form vector space
ie it contain the zero vector
i dont know from where i have to start
if anybody help me , i would be very thankful to him
linear-algebra functional-analysis linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Aug 5 '17 at 15:37
lomberlomber
786420
$endgroup$
Prove that the linear function does not form a vector space basis of $mathcal{C}bigl([0,1]bigr)$ over $mathbb R$?
i was trying this question many times but i could not get it.
i was taking the scalar multiplication and addition property
it was showing that this vector form vector space
ie it contain the zero vector
i dont know from where i have to start
if anybody help me , i would be very thankful to him
linear-algebra functional-analysis linear-transformations
linear-algebra functional-analysis linear-transformations
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Aug 5 '17 at 15:37
lomberlomber
786420
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
asked Aug 5 '17 at 15:37
lomberlomber
786420
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
edited Jan 23 at 9:23
José Carlos Santos
167k22132235
167k22132235
asked Aug 5 '17 at 15:37
lomberlomber
786420
asked Aug 5 '17 at 15:37
lomberlomber
786420
asked Aug 5 '17 at 15:37
lomberlomber
786420
786420
$begingroup$
whats "the linear function?" do you mean the linear functions? If so, then every linear combination of linear functions is again linear and the function $x^2$ is not linear.
$endgroup$
– Yanko
Aug 5 '17 at 15:41
$begingroup$
Simply, any linear combination of linear functions is weakly monotonic but not every continuous function over $[0,1]$ is weakly monotonic.
$endgroup$
– Jack D'Aurizio
Aug 5 '17 at 15:43
add a comment |
$begingroup$
whats "the linear function?" do you mean the linear functions? If so, then every linear combination of linear functions is again linear and the function $x^2$ is not linear.
$endgroup$
– Yanko
Aug 5 '17 at 15:41
$begingroup$
Simply, any linear combination of linear functions is weakly monotonic but not every continuous function over $[0,1]$ is weakly monotonic.
$endgroup$
– Jack D'Aurizio
Aug 5 '17 at 15:43
$begingroup$
whats "the linear function?" do you mean the linear functions? If so, then every linear combination of linear functions is again linear and the function $x^2$ is not linear.
$endgroup$
– Yanko
Aug 5 '17 at 15:41
$begingroup$
whats "the linear function?" do you mean the linear functions? If so, then every linear combination of linear functions is again linear and the function $x^2$ is not linear.
$endgroup$
– Yanko
Aug 5 '17 at 15:41
$begingroup$
Simply, any linear combination of linear functions is weakly monotonic but not every continuous function over $[0,1]$ is weakly monotonic.
$endgroup$
– Jack D'Aurizio
Aug 5 '17 at 15:43
$begingroup$
Simply, any linear combination of linear functions is weakly monotonic but not every continuous function over $[0,1]$ is weakly monotonic.
$endgroup$
– Jack D'Aurizio
Aug 5 '17 at 15:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Any linear combination of linear functions is still a linear function. Since the function $qcolon[0,1]longrightarrowmathbb R$ defined by $q(x)=x^2$ is not linear, it does not belong to the span of the linear functions.
Besides, the set of all linear functions is not linearly independent.
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
@lomberlego If my answer was useful, then you could mark it as your accepted answer.
$endgroup$
– José Carlos Santos
Aug 5 '17 at 15:48
$begingroup$
when it comes to $C[0,1]$ it is conventional to define a set of functions being a basis if it's span is dense, but a limit of linear functions is again linear...
$endgroup$
– Yanko
Aug 5 '17 at 15:48
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any linear combination of linear functions is still a linear function. Since the function $qcolon[0,1]longrightarrowmathbb R$ defined by $q(x)=x^2$ is not linear, it does not belong to the span of the linear functions.
Besides, the set of all linear functions is not linearly independent.
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
@lomberlego If my answer was useful, then you could mark it as your accepted answer.
$endgroup$
– José Carlos Santos
Aug 5 '17 at 15:48
$begingroup$
when it comes to $C[0,1]$ it is conventional to define a set of functions being a basis if it's span is dense, but a limit of linear functions is again linear...
$endgroup$
– Yanko
Aug 5 '17 at 15:48
add a comment |
$begingroup$
Any linear combination of linear functions is still a linear function. Since the function $qcolon[0,1]longrightarrowmathbb R$ defined by $q(x)=x^2$ is not linear, it does not belong to the span of the linear functions.
Besides, the set of all linear functions is not linearly independent.
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
$begingroup$
@lomberlego If my answer was useful, then you could mark it as your accepted answer.
$endgroup$
– José Carlos Santos
Aug 5 '17 at 15:48
$begingroup$
when it comes to $C[0,1]$ it is conventional to define a set of functions being a basis if it's span is dense, but a limit of linear functions is again linear...
$endgroup$
– Yanko
Aug 5 '17 at 15:48
add a comment |
$begingroup$
Any linear combination of linear functions is still a linear function. Since the function $qcolon[0,1]longrightarrowmathbb R$ defined by $q(x)=x^2$ is not linear, it does not belong to the span of the linear functions.
Besides, the set of all linear functions is not linearly independent.
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
$endgroup$
Any linear combination of linear functions is still a linear function. Since the function $qcolon[0,1]longrightarrowmathbb R$ defined by $q(x)=x^2$ is not linear, it does not belong to the span of the linear functions.
Besides, the set of all linear functions is not linearly independent.
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
answered Aug 5 '17 at 15:44
José Carlos SantosJosé Carlos Santos
167k22132235
167k22132235
$begingroup$
@lomberlego If my answer was useful, then you could mark it as your accepted answer.
$endgroup$
– José Carlos Santos
Aug 5 '17 at 15:48
$begingroup$
when it comes to $C[0,1]$ it is conventional to define a set of functions being a basis if it's span is dense, but a limit of linear functions is again linear...
$endgroup$
– Yanko
Aug 5 '17 at 15:48
add a comment |
$begingroup$
@lomberlego If my answer was useful, then you could mark it as your accepted answer.
$endgroup$
– José Carlos Santos
Aug 5 '17 at 15:48
$begingroup$
when it comes to $C[0,1]$ it is conventional to define a set of functions being a basis if it's span is dense, but a limit of linear functions is again linear...
$endgroup$
– Yanko
Aug 5 '17 at 15:48
$begingroup$
@lomberlego If my answer was useful, then you could mark it as your accepted answer.
$endgroup$
– José Carlos Santos
Aug 5 '17 at 15:48
$begingroup$
@lomberlego If my answer was useful, then you could mark it as your accepted answer.
$endgroup$
– José Carlos Santos
Aug 5 '17 at 15:48
$begingroup$
when it comes to $C[0,1]$ it is conventional to define a set of functions being a basis if it's span is dense, but a limit of linear functions is again linear...
$endgroup$
– Yanko
Aug 5 '17 at 15:48
$begingroup$
when it comes to $C[0,1]$ it is conventional to define a set of functions being a basis if it's span is dense, but a limit of linear functions is again linear...
$endgroup$
– Yanko
Aug 5 '17 at 15:48
add a comment |
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$begingroup$
whats "the linear function?" do you mean the linear functions? If so, then every linear combination of linear functions is again linear and the function $x^2$ is not linear.
$endgroup$
– Yanko
Aug 5 '17 at 15:41
$begingroup$
Simply, any linear combination of linear functions is weakly monotonic but not every continuous function over $[0,1]$ is weakly monotonic.
$endgroup$
– Jack D'Aurizio
Aug 5 '17 at 15:43