Verification: Investigation of a linear map on surjectivity and injectivity
$begingroup$
The linear map I'm investigating is defined like this:
$l_2: P_3(mathbb{R}) rightarrow mathbb{R}, l_2(p):= p'(1)$
And these are my calculations:
Injectivity:
$Let ,, p,q in P_3(mathbb{R}), ,, p(x) := 2x^3 ,, and ,, q(x):= x^3+x^2+x$
$Rightarrow l_2(p) = l_2(q) = 6 ,, but ,, p neq q Rightarrow l_2 ,, is ,, not ,, injective.$
Surjectivity:
$Let ,, p in P_3(mathbb{R}), ,, p(x):= ax^3+bx^2+cx+d ,, and ,, e in mathbb{R}$
Then
$l_2(p) = 3a+2b+c$
$3a+2b+c = e ,, so ,, for ,, example ,, a=frac{e}{3}; b=frac{e}{2} ,, and ,, c=e ,, would,, be ,,one,, possible,, solution,, s.t,, l_2(p) = -e.$
$Rightarrow l_2 ,, is ,, surjective.$
linear-algebra derivatives proof-verification linear-transformations
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
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add a comment |
$begingroup$
The linear map I'm investigating is defined like this:
$l_2: P_3(mathbb{R}) rightarrow mathbb{R}, l_2(p):= p'(1)$
And these are my calculations:
Injectivity:
$Let ,, p,q in P_3(mathbb{R}), ,, p(x) := 2x^3 ,, and ,, q(x):= x^3+x^2+x$
$Rightarrow l_2(p) = l_2(q) = 6 ,, but ,, p neq q Rightarrow l_2 ,, is ,, not ,, injective.$
Surjectivity:
$Let ,, p in P_3(mathbb{R}), ,, p(x):= ax^3+bx^2+cx+d ,, and ,, e in mathbb{R}$
Then
$l_2(p) = 3a+2b+c$
$3a+2b+c = e ,, so ,, for ,, example ,, a=frac{e}{3}; b=frac{e}{2} ,, and ,, c=e ,, would,, be ,,one,, possible,, solution,, s.t,, l_2(p) = -e.$
$Rightarrow l_2 ,, is ,, surjective.$
linear-algebra derivatives proof-verification linear-transformations
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
add a comment |
$begingroup$
The linear map I'm investigating is defined like this:
$l_2: P_3(mathbb{R}) rightarrow mathbb{R}, l_2(p):= p'(1)$
And these are my calculations:
Injectivity:
$Let ,, p,q in P_3(mathbb{R}), ,, p(x) := 2x^3 ,, and ,, q(x):= x^3+x^2+x$
$Rightarrow l_2(p) = l_2(q) = 6 ,, but ,, p neq q Rightarrow l_2 ,, is ,, not ,, injective.$
Surjectivity:
$Let ,, p in P_3(mathbb{R}), ,, p(x):= ax^3+bx^2+cx+d ,, and ,, e in mathbb{R}$
Then
$l_2(p) = 3a+2b+c$
$3a+2b+c = e ,, so ,, for ,, example ,, a=frac{e}{3}; b=frac{e}{2} ,, and ,, c=e ,, would,, be ,,one,, possible,, solution,, s.t,, l_2(p) = -e.$
$Rightarrow l_2 ,, is ,, surjective.$
linear-algebra derivatives proof-verification linear-transformations
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
$endgroup$
The linear map I'm investigating is defined like this:
$l_2: P_3(mathbb{R}) rightarrow mathbb{R}, l_2(p):= p'(1)$
And these are my calculations:
Injectivity:
$Let ,, p,q in P_3(mathbb{R}), ,, p(x) := 2x^3 ,, and ,, q(x):= x^3+x^2+x$
$Rightarrow l_2(p) = l_2(q) = 6 ,, but ,, p neq q Rightarrow l_2 ,, is ,, not ,, injective.$
Surjectivity:
$Let ,, p in P_3(mathbb{R}), ,, p(x):= ax^3+bx^2+cx+d ,, and ,, e in mathbb{R}$
Then
$l_2(p) = 3a+2b+c$
$3a+2b+c = e ,, so ,, for ,, example ,, a=frac{e}{3}; b=frac{e}{2} ,, and ,, c=e ,, would,, be ,,one,, possible,, solution,, s.t,, l_2(p) = -e.$
$Rightarrow l_2 ,, is ,, surjective.$
linear-algebra derivatives proof-verification linear-transformations
linear-algebra derivatives proof-verification linear-transformations
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
asked Jan 23 at 16:55
Fo Young Areal LoFo Young Areal Lo
345
345
add a comment |
add a comment |
2 Answers
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For injectivity, what you have done is correct. However, your examples could have been much simpler. For example, just the constant polynomials $p_1 equiv 0$ and $p_2 equiv 1$ have derivatives identically zero, therefore $l_2(p_1) = l_2(p_2)$, although the two polynomials are different.
Similarly for surjectivity, you are okay, but simply note that if $p_c(x) := cx$, then $p_c'(1)=l_2(p_c)=c$.
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
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add a comment |
$begingroup$
Yes, this is correct. An easier way to see the lack of injectivity is that for any $fin mathcal{P}_3(mathbb{R})$,
$$f'(1)=frac{d}{dx}(f(x)+c)bigg|_{x=1}$$
for $c$ a constant. Surjectivity can be constructed using just the degree $1$ functions $g_a(x)=ax$. For instance, given any $ain mathbb{R}$, $g_a'(1)=a$.
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
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2 Answers
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2 Answers
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$begingroup$
For injectivity, what you have done is correct. However, your examples could have been much simpler. For example, just the constant polynomials $p_1 equiv 0$ and $p_2 equiv 1$ have derivatives identically zero, therefore $l_2(p_1) = l_2(p_2)$, although the two polynomials are different.
Similarly for surjectivity, you are okay, but simply note that if $p_c(x) := cx$, then $p_c'(1)=l_2(p_c)=c$.
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
add a comment |
$begingroup$
For injectivity, what you have done is correct. However, your examples could have been much simpler. For example, just the constant polynomials $p_1 equiv 0$ and $p_2 equiv 1$ have derivatives identically zero, therefore $l_2(p_1) = l_2(p_2)$, although the two polynomials are different.
Similarly for surjectivity, you are okay, but simply note that if $p_c(x) := cx$, then $p_c'(1)=l_2(p_c)=c$.
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
add a comment |
$begingroup$
For injectivity, what you have done is correct. However, your examples could have been much simpler. For example, just the constant polynomials $p_1 equiv 0$ and $p_2 equiv 1$ have derivatives identically zero, therefore $l_2(p_1) = l_2(p_2)$, although the two polynomials are different.
Similarly for surjectivity, you are okay, but simply note that if $p_c(x) := cx$, then $p_c'(1)=l_2(p_c)=c$.
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
$endgroup$
For injectivity, what you have done is correct. However, your examples could have been much simpler. For example, just the constant polynomials $p_1 equiv 0$ and $p_2 equiv 1$ have derivatives identically zero, therefore $l_2(p_1) = l_2(p_2)$, although the two polynomials are different.
Similarly for surjectivity, you are okay, but simply note that if $p_c(x) := cx$, then $p_c'(1)=l_2(p_c)=c$.
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
answered Jan 23 at 17:03
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.9k33477
38.9k33477
add a comment |
add a comment |
$begingroup$
Yes, this is correct. An easier way to see the lack of injectivity is that for any $fin mathcal{P}_3(mathbb{R})$,
$$f'(1)=frac{d}{dx}(f(x)+c)bigg|_{x=1}$$
for $c$ a constant. Surjectivity can be constructed using just the degree $1$ functions $g_a(x)=ax$. For instance, given any $ain mathbb{R}$, $g_a'(1)=a$.
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Yes, this is correct. An easier way to see the lack of injectivity is that for any $fin mathcal{P}_3(mathbb{R})$,
$$f'(1)=frac{d}{dx}(f(x)+c)bigg|_{x=1}$$
for $c$ a constant. Surjectivity can be constructed using just the degree $1$ functions $g_a(x)=ax$. For instance, given any $ain mathbb{R}$, $g_a'(1)=a$.
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Yes, this is correct. An easier way to see the lack of injectivity is that for any $fin mathcal{P}_3(mathbb{R})$,
$$f'(1)=frac{d}{dx}(f(x)+c)bigg|_{x=1}$$
for $c$ a constant. Surjectivity can be constructed using just the degree $1$ functions $g_a(x)=ax$. For instance, given any $ain mathbb{R}$, $g_a'(1)=a$.
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
Yes, this is correct. An easier way to see the lack of injectivity is that for any $fin mathcal{P}_3(mathbb{R})$,
$$f'(1)=frac{d}{dx}(f(x)+c)bigg|_{x=1}$$
for $c$ a constant. Surjectivity can be constructed using just the degree $1$ functions $g_a(x)=ax$. For instance, given any $ain mathbb{R}$, $g_a'(1)=a$.
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 23 at 17:01
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
10.5k41741
add a comment |
add a comment |
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