Mixing
Investigate how many endomorphisms $f in L(mathbb R[x]_{3})$ meet the conditions
$begingroup$
Investigate how many endomorphisms $f in L(mathbb R[x]_{3})$ meet the conditions:
(i) $ker f = operatorname{span}(1,x)$,
(ii) $fcirc f=f$,
(iii) $f(x^{2})=1-x+x^{2}$,
(iv) if $f(x^{3})=p$ then $p(1)=p'(1)=0$.
Find a Jordan's matrix and basis for for each such endomorphism
I know that in (ii) $f in L(mathbb R[x]_{3})$ is a projection for a some subspace $U$ along the subspace $V$. But I think it is impossible to such a special observation is crucial in this task because it does not seem so complicated.
At the same time, we did not do anything similar on the lectures, so I have no knowledge about it.
Can you help me?
linear-algebra
edited Feb 2 at 15:36
Bernard
124k741117
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
$endgroup$
add a comment |
$begingroup$
Investigate how many endomorphisms $f in L(mathbb R[x]_{3})$ meet the conditions:
(i) $ker f = operatorname{span}(1,x)$,
(ii) $fcirc f=f$,
(iii) $f(x^{2})=1-x+x^{2}$,
(iv) if $f(x^{3})=p$ then $p(1)=p'(1)=0$.
Find a Jordan's matrix and basis for for each such endomorphism
I know that in (ii) $f in L(mathbb R[x]_{3})$ is a projection for a some subspace $U$ along the subspace $V$. But I think it is impossible to such a special observation is crucial in this task because it does not seem so complicated.
At the same time, we did not do anything similar on the lectures, so I have no knowledge about it.
Can you help me?
linear-algebra
edited Feb 2 at 15:36
Bernard
124k741117
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
$endgroup$
1
$begingroup$
Are (i)-(iv) separate questions, or should you look for endomorphisms which satisfy all of them at the same time?
$endgroup$
– Arthur
Feb 2 at 15:23
$begingroup$
Could you answer to Arthur, please ? If you must meet the 5 conditions, consider a generic element under the form $ax^3+bx^2+cx+d$ then "translate" all your conditions into the "language of $mathbb{R}^n$" ; in particular, accumulate information about the matrix which corresponds to $f$.
$endgroup$
– Jean Marie
Feb 2 at 15:31
$begingroup$
@Arthur I am not sure how to understand the content of this task, but it seems to me that each case is separate
$endgroup$
– VirtualUser
Feb 2 at 16:06
$begingroup$
I don't think so (see my first sentence in my answer.
$endgroup$
– Jean Marie
Feb 2 at 17:04
$begingroup$
I have rectified my answer at its end (no longer two but one solution) taking into account a restriction pinpointed by @egreg
$endgroup$
– Jean Marie
Feb 3 at 11:21
add a comment |
$begingroup$
Investigate how many endomorphisms $f in L(mathbb R[x]_{3})$ meet the conditions:
(i) $ker f = operatorname{span}(1,x)$,
(ii) $fcirc f=f$,
(iii) $f(x^{2})=1-x+x^{2}$,
(iv) if $f(x^{3})=p$ then $p(1)=p'(1)=0$.
Find a Jordan's matrix and basis for for each such endomorphism
I know that in (ii) $f in L(mathbb R[x]_{3})$ is a projection for a some subspace $U$ along the subspace $V$. But I think it is impossible to such a special observation is crucial in this task because it does not seem so complicated.
At the same time, we did not do anything similar on the lectures, so I have no knowledge about it.
Can you help me?
linear-algebra
edited Feb 2 at 15:36
Bernard
124k741117
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
$endgroup$
Investigate how many endomorphisms $f in L(mathbb R[x]_{3})$ meet the conditions:
(i) $ker f = operatorname{span}(1,x)$,
(ii) $fcirc f=f$,
(iii) $f(x^{2})=1-x+x^{2}$,
(iv) if $f(x^{3})=p$ then $p(1)=p'(1)=0$.
Find a Jordan's matrix and basis for for each such endomorphism
I know that in (ii) $f in L(mathbb R[x]_{3})$ is a projection for a some subspace $U$ along the subspace $V$. But I think it is impossible to such a special observation is crucial in this task because it does not seem so complicated.
At the same time, we did not do anything similar on the lectures, so I have no knowledge about it.
Can you help me?
linear-algebra
linear-algebra
edited Feb 2 at 15:36
Bernard
124k741117
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
edited Feb 2 at 15:36
Bernard
124k741117
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
edited Feb 2 at 15:36
Bernard
124k741117
edited Feb 2 at 15:36
Bernard
124k741117
edited Feb 2 at 15:36
Bernard
124k741117
124k741117
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
asked Feb 2 at 15:21
VirtualUserVirtualUser
1,321317
1,321317
1
$begingroup$
Are (i)-(iv) separate questions, or should you look for endomorphisms which satisfy all of them at the same time?
$endgroup$
– Arthur
Feb 2 at 15:23
$begingroup$
Could you answer to Arthur, please ? If you must meet the 5 conditions, consider a generic element under the form $ax^3+bx^2+cx+d$ then "translate" all your conditions into the "language of $mathbb{R}^n$" ; in particular, accumulate information about the matrix which corresponds to $f$.
$endgroup$
– Jean Marie
Feb 2 at 15:31
$begingroup$
@Arthur I am not sure how to understand the content of this task, but it seems to me that each case is separate
$endgroup$
– VirtualUser
Feb 2 at 16:06
$begingroup$
I don't think so (see my first sentence in my answer.
$endgroup$
– Jean Marie
Feb 2 at 17:04
$begingroup$
I have rectified my answer at its end (no longer two but one solution) taking into account a restriction pinpointed by @egreg
$endgroup$
– Jean Marie
Feb 3 at 11:21
add a comment |
1
$begingroup$
Are (i)-(iv) separate questions, or should you look for endomorphisms which satisfy all of them at the same time?
$endgroup$
– Arthur
Feb 2 at 15:23
$begingroup$
Could you answer to Arthur, please ? If you must meet the 5 conditions, consider a generic element under the form $ax^3+bx^2+cx+d$ then "translate" all your conditions into the "language of $mathbb{R}^n$" ; in particular, accumulate information about the matrix which corresponds to $f$.
$endgroup$
– Jean Marie
Feb 2 at 15:31
$begingroup$
@Arthur I am not sure how to understand the content of this task, but it seems to me that each case is separate
$endgroup$
– VirtualUser
Feb 2 at 16:06
$begingroup$
I don't think so (see my first sentence in my answer.
$endgroup$
– Jean Marie
Feb 2 at 17:04
$begingroup$
I have rectified my answer at its end (no longer two but one solution) taking into account a restriction pinpointed by @egreg
$endgroup$
– Jean Marie
Feb 3 at 11:21
1
1
$begingroup$
Are (i)-(iv) separate questions, or should you look for endomorphisms which satisfy all of them at the same time?
$endgroup$
– Arthur
Feb 2 at 15:23
$begingroup$
Are (i)-(iv) separate questions, or should you look for endomorphisms which satisfy all of them at the same time?
$endgroup$
– Arthur
Feb 2 at 15:23
$begingroup$
Could you answer to Arthur, please ? If you must meet the 5 conditions, consider a generic element under the form $ax^3+bx^2+cx+d$ then "translate" all your conditions into the "language of $mathbb{R}^n$" ; in particular, accumulate information about the matrix which corresponds to $f$.
$endgroup$
– Jean Marie
Feb 2 at 15:31
$begingroup$
Could you answer to Arthur, please ? If you must meet the 5 conditions, consider a generic element under the form $ax^3+bx^2+cx+d$ then "translate" all your conditions into the "language of $mathbb{R}^n$" ; in particular, accumulate information about the matrix which corresponds to $f$.
$endgroup$
– Jean Marie
Feb 2 at 15:31
$begingroup$
@Arthur I am not sure how to understand the content of this task, but it seems to me that each case is separate
$endgroup$
– VirtualUser
Feb 2 at 16:06
$begingroup$
@Arthur I am not sure how to understand the content of this task, but it seems to me that each case is separate
$endgroup$
– VirtualUser
Feb 2 at 16:06
$begingroup$
I don't think so (see my first sentence in my answer.
$endgroup$
– Jean Marie
Feb 2 at 17:04
$begingroup$
I don't think so (see my first sentence in my answer.
$endgroup$
– Jean Marie
Feb 2 at 17:04
$begingroup$
I have rectified my answer at its end (no longer two but one solution) taking into account a restriction pinpointed by @egreg
$endgroup$
– Jean Marie
Feb 3 at 11:21
$begingroup$
I have rectified my answer at its end (no longer two but one solution) taking into account a restriction pinpointed by @egreg
$endgroup$
– Jean Marie
Feb 3 at 11:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Conditions (i)-...-(v) have to be considered jointly. We are going to show that they determine a single matrix.
We will consider the matrix $M$ describing the transformation with respect to the (canonical) basis ${1,x,x^2,x^3}.$
The columns of such a matrix associated with a linear operation are given by the images of the basis'elements. Let us show in a first step that :
$$M=begin{pmatrix}
0&0&1&d\
0&0&-1&(c-2d)\
0&0&1&(-2c+d)\
0&0&0&cend{pmatrix}.$$
Explanations :
Columns 1 and 2 : they are both zero because condition (i) says that $f(1)=0=0+0x+0x^2+0x^3$ and $f(x)=0=0+0x+0x^2+0x^3$.
Column 3 : (condition (iii)) coefficients of $1-1x+1x^2+0x^3$.
Column 4 : condition (iv) amounts to say that $1$ is a double root of $p$ ; thus the image can be written $p=(x-1)^2(cx+d)=d+(c-2d)x+(-2c+d)x^2+cx^3.$
Now, we must take into account the fact that $M^2=M$ (condition (ii)) with :
$$M^2=begin{pmatrix}
0&0&1&(d - 2c + cd)\
0&0&-1&(2c - d + c^2 - 2cd))\
0&0&1&(c + 1)(d-2c)\
0&0&0&c^2end{pmatrix}.$$
Identifiying $M^2$ and $M$ leads to a first condition $c^2=c$, thus necessarily $c=0$ or $c=1$. Let us consider each of these cases :
if $c=0$, whatever the value of $d$, all other constraints are verified BUT we get in this way a fourth column proportional to the third one ; thus we would have Rank(M)=1 ; this comes in contradiction with the somewhat hidden condition : $dim ker f = 2 implies Rank(f)=4-2=2$. (I am indebted here to the solution by @Egreg : I hadn't seen at first this condition). Thus we must rule out this case.
if $c=1$, we deduce without difficulty that $d=2$ matches all conditions, giving the rank-2 unique solution :
$$M=begin{pmatrix}
0&0&1&2\
0&0&-1&-3\
0&0&1&0\
0&0&0&1end{pmatrix}.$$
I think that the Jordan form will not be difficult. Btw, what are the eigenvalues of a projector ?
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
Can you say, 4 hours later, if this solution aggrees you ? You aren't new on this site : you know it is done for exchanges, It's not just an anonymous booth.
$endgroup$
– Jean Marie
Feb 2 at 20:40
$begingroup$
So If I understand well - you finds matrix of linear transformation in use of given information. You can do this because in each step you use some vectors (there presented as polynomials) which together creates a base of $R[x]_3$. Is that right? After that you have all things to prepare Jordan Matrix, right? Ps: sorry but I am already a bit overtired, I run with the material and I can not cover everything at once
$endgroup$
– VirtualUser
Feb 2 at 20:59
$begingroup$
Thanks for answering. You are excused. Answers : Right. Right (you do it just as if the original problem had been presented in $mathbb{R}^4$).
$endgroup$
– Jean Marie
Feb 2 at 21:12
$begingroup$
Okay, but I have one more question - to step with column 4. You have seen that it is double root - but let say that I didn't see that. So I have decided to put $ f(x^3) = p(x) = ax^9 +bx^6 + cx^3 + d $ and then check $p(1)=0$ and $p(-1)=0$ And then I get this column $[c+2d,-2c-3d,c,d]^T $ and I don't know if I have take mistake or my approach is not correct (but it seems to be reasonable)
$endgroup$
– VirtualUser
Feb 2 at 21:22
$begingroup$
No : notation $f(x^3)$ is misleading, but shouldn't be interpreted like that: it is the effect of the transformation of $x^3$ by the linear operator$f$. Your intepretation cannot be the right one : we must stay in the vector space of polynomials with degree $leq 3$.
$endgroup$
– Jean Marie
Feb 2 at 21:25
|
show 1 more comment
$begingroup$
We know that $f(1)=0$, $f(x)=0$, $f(x^2)=1-x-x^2$; set $f(x^3)=a+bx+cx^2+dx^3$.
You might be confused by polynomial notation; here it might be convenient to set $v_1=1$, $v_2=x$, $v_3=x^2$ and $v_4=x^3$; then the above statements become more “linear algebra like”:
begin{cases}
f(v_1)=0 \
f(v_2)=0 \
f(v_3)=v_1-v_2+v_3 \
f(v_4)=av_1+bv_2+cv_3+dv_4
end{cases}
Since ${v_1,v_2,v_3,v_4}$ is a basis for $mathbb{R}_3[x]$, the above characterizes the linear map $f$, provided we determine $a,b,c,d$ so that conditions (i)–(v) are satisfied.
The matrix with respect to the basis ${v_1,v_2,v_3,v_4}$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & a \
0 & 0 & -1 & b \
0 & 0 & 1 & c \
0 & 0 & 0 & d
end{bmatrix}
$$
We can compute $A^2$:
$$
A^2=
begin{bmatrix}
0 & 0 & 1 & ad+c \
0 & 0 & -1 & bd-c \
0 & 0 & 1 & cd+c \
0 & 0 & 0 & d^2
end{bmatrix}
$$
Since we need $A=A^2$, we deduce
begin{cases}
ad+c=a \
bd-c=b \
cd+c=c \
d^2=d
end{cases}
that immediately splits into $d=0$ or $d=1$.
Case $d=0$
We get $a=c$ and $b=-c$. Therefore $f(v_4)=c(v_1-v_2+v_3)$. In terms of polynomials, we have $f(v_4)=p$, where $p(x)=c(1-x+x^2)$. We have $p(1)=c$ and condition (v) forces $c=0$. This cannot be the case, because the rank would be $1$, contrary to the requirement that $dimker f=2$.
Case $d=1$
We get $c=0$. Thus $f(v_4)=av_1+bv_2+v_4$ and so $p(x)=a+bx+x^3$. The condition $p(1)=0$ yields $a+b+1=0$; the condition $p'(1)=0$ yields $b+3=0$, so $b=-3$ and $a=2$.
Final discussion
The matrix of $f$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & 2 \
0 & 0 & -1 & -3 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{bmatrix}
$$
The matrix has a double eigenvalue $0$ and a double eigenvalue $1$. Since also the eigenspace relative to $1$ has dimension $2$, the matrix is diagonalizable.
answered Feb 2 at 22:19
egregegreg
186k1486208
$endgroup$
$begingroup$
You are right, case $d=0$ has to be ruled out : my solution (that I have rectified) was an answer to a milder condition : (i) $ ker f supseteq span{1,x}$ .
$endgroup$
– Jean Marie
Feb 3 at 11:02
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Conditions (i)-...-(v) have to be considered jointly. We are going to show that they determine a single matrix.
We will consider the matrix $M$ describing the transformation with respect to the (canonical) basis ${1,x,x^2,x^3}.$
The columns of such a matrix associated with a linear operation are given by the images of the basis'elements. Let us show in a first step that :
$$M=begin{pmatrix}
0&0&1&d\
0&0&-1&(c-2d)\
0&0&1&(-2c+d)\
0&0&0&cend{pmatrix}.$$
Explanations :
Columns 1 and 2 : they are both zero because condition (i) says that $f(1)=0=0+0x+0x^2+0x^3$ and $f(x)=0=0+0x+0x^2+0x^3$.
Column 3 : (condition (iii)) coefficients of $1-1x+1x^2+0x^3$.
Column 4 : condition (iv) amounts to say that $1$ is a double root of $p$ ; thus the image can be written $p=(x-1)^2(cx+d)=d+(c-2d)x+(-2c+d)x^2+cx^3.$
Now, we must take into account the fact that $M^2=M$ (condition (ii)) with :
$$M^2=begin{pmatrix}
0&0&1&(d - 2c + cd)\
0&0&-1&(2c - d + c^2 - 2cd))\
0&0&1&(c + 1)(d-2c)\
0&0&0&c^2end{pmatrix}.$$
Identifiying $M^2$ and $M$ leads to a first condition $c^2=c$, thus necessarily $c=0$ or $c=1$. Let us consider each of these cases :
if $c=0$, whatever the value of $d$, all other constraints are verified BUT we get in this way a fourth column proportional to the third one ; thus we would have Rank(M)=1 ; this comes in contradiction with the somewhat hidden condition : $dim ker f = 2 implies Rank(f)=4-2=2$. (I am indebted here to the solution by @Egreg : I hadn't seen at first this condition). Thus we must rule out this case.
if $c=1$, we deduce without difficulty that $d=2$ matches all conditions, giving the rank-2 unique solution :
$$M=begin{pmatrix}
0&0&1&2\
0&0&-1&-3\
0&0&1&0\
0&0&0&1end{pmatrix}.$$
I think that the Jordan form will not be difficult. Btw, what are the eigenvalues of a projector ?
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
Can you say, 4 hours later, if this solution aggrees you ? You aren't new on this site : you know it is done for exchanges, It's not just an anonymous booth.
$endgroup$
– Jean Marie
Feb 2 at 20:40
$begingroup$
So If I understand well - you finds matrix of linear transformation in use of given information. You can do this because in each step you use some vectors (there presented as polynomials) which together creates a base of $R[x]_3$. Is that right? After that you have all things to prepare Jordan Matrix, right? Ps: sorry but I am already a bit overtired, I run with the material and I can not cover everything at once
$endgroup$
– VirtualUser
Feb 2 at 20:59
$begingroup$
Thanks for answering. You are excused. Answers : Right. Right (you do it just as if the original problem had been presented in $mathbb{R}^4$).
$endgroup$
– Jean Marie
Feb 2 at 21:12
$begingroup$
Okay, but I have one more question - to step with column 4. You have seen that it is double root - but let say that I didn't see that. So I have decided to put $ f(x^3) = p(x) = ax^9 +bx^6 + cx^3 + d $ and then check $p(1)=0$ and $p(-1)=0$ And then I get this column $[c+2d,-2c-3d,c,d]^T $ and I don't know if I have take mistake or my approach is not correct (but it seems to be reasonable)
$endgroup$
– VirtualUser
Feb 2 at 21:22
$begingroup$
No : notation $f(x^3)$ is misleading, but shouldn't be interpreted like that: it is the effect of the transformation of $x^3$ by the linear operator$f$. Your intepretation cannot be the right one : we must stay in the vector space of polynomials with degree $leq 3$.
$endgroup$
– Jean Marie
Feb 2 at 21:25
|
show 1 more comment
$begingroup$
Conditions (i)-...-(v) have to be considered jointly. We are going to show that they determine a single matrix.
We will consider the matrix $M$ describing the transformation with respect to the (canonical) basis ${1,x,x^2,x^3}.$
The columns of such a matrix associated with a linear operation are given by the images of the basis'elements. Let us show in a first step that :
$$M=begin{pmatrix}
0&0&1&d\
0&0&-1&(c-2d)\
0&0&1&(-2c+d)\
0&0&0&cend{pmatrix}.$$
Explanations :
Columns 1 and 2 : they are both zero because condition (i) says that $f(1)=0=0+0x+0x^2+0x^3$ and $f(x)=0=0+0x+0x^2+0x^3$.
Column 3 : (condition (iii)) coefficients of $1-1x+1x^2+0x^3$.
Column 4 : condition (iv) amounts to say that $1$ is a double root of $p$ ; thus the image can be written $p=(x-1)^2(cx+d)=d+(c-2d)x+(-2c+d)x^2+cx^3.$
Now, we must take into account the fact that $M^2=M$ (condition (ii)) with :
$$M^2=begin{pmatrix}
0&0&1&(d - 2c + cd)\
0&0&-1&(2c - d + c^2 - 2cd))\
0&0&1&(c + 1)(d-2c)\
0&0&0&c^2end{pmatrix}.$$
Identifiying $M^2$ and $M$ leads to a first condition $c^2=c$, thus necessarily $c=0$ or $c=1$. Let us consider each of these cases :
if $c=0$, whatever the value of $d$, all other constraints are verified BUT we get in this way a fourth column proportional to the third one ; thus we would have Rank(M)=1 ; this comes in contradiction with the somewhat hidden condition : $dim ker f = 2 implies Rank(f)=4-2=2$. (I am indebted here to the solution by @Egreg : I hadn't seen at first this condition). Thus we must rule out this case.
if $c=1$, we deduce without difficulty that $d=2$ matches all conditions, giving the rank-2 unique solution :
$$M=begin{pmatrix}
0&0&1&2\
0&0&-1&-3\
0&0&1&0\
0&0&0&1end{pmatrix}.$$
I think that the Jordan form will not be difficult. Btw, what are the eigenvalues of a projector ?
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
Can you say, 4 hours later, if this solution aggrees you ? You aren't new on this site : you know it is done for exchanges, It's not just an anonymous booth.
$endgroup$
– Jean Marie
Feb 2 at 20:40
$begingroup$
So If I understand well - you finds matrix of linear transformation in use of given information. You can do this because in each step you use some vectors (there presented as polynomials) which together creates a base of $R[x]_3$. Is that right? After that you have all things to prepare Jordan Matrix, right? Ps: sorry but I am already a bit overtired, I run with the material and I can not cover everything at once
$endgroup$
– VirtualUser
Feb 2 at 20:59
$begingroup$
Thanks for answering. You are excused. Answers : Right. Right (you do it just as if the original problem had been presented in $mathbb{R}^4$).
$endgroup$
– Jean Marie
Feb 2 at 21:12
$begingroup$
Okay, but I have one more question - to step with column 4. You have seen that it is double root - but let say that I didn't see that. So I have decided to put $ f(x^3) = p(x) = ax^9 +bx^6 + cx^3 + d $ and then check $p(1)=0$ and $p(-1)=0$ And then I get this column $[c+2d,-2c-3d,c,d]^T $ and I don't know if I have take mistake or my approach is not correct (but it seems to be reasonable)
$endgroup$
– VirtualUser
Feb 2 at 21:22
$begingroup$
No : notation $f(x^3)$ is misleading, but shouldn't be interpreted like that: it is the effect of the transformation of $x^3$ by the linear operator$f$. Your intepretation cannot be the right one : we must stay in the vector space of polynomials with degree $leq 3$.
$endgroup$
– Jean Marie
Feb 2 at 21:25
|
show 1 more comment
$begingroup$
Conditions (i)-...-(v) have to be considered jointly. We are going to show that they determine a single matrix.
We will consider the matrix $M$ describing the transformation with respect to the (canonical) basis ${1,x,x^2,x^3}.$
The columns of such a matrix associated with a linear operation are given by the images of the basis'elements. Let us show in a first step that :
$$M=begin{pmatrix}
0&0&1&d\
0&0&-1&(c-2d)\
0&0&1&(-2c+d)\
0&0&0&cend{pmatrix}.$$
Explanations :
Columns 1 and 2 : they are both zero because condition (i) says that $f(1)=0=0+0x+0x^2+0x^3$ and $f(x)=0=0+0x+0x^2+0x^3$.
Column 3 : (condition (iii)) coefficients of $1-1x+1x^2+0x^3$.
Column 4 : condition (iv) amounts to say that $1$ is a double root of $p$ ; thus the image can be written $p=(x-1)^2(cx+d)=d+(c-2d)x+(-2c+d)x^2+cx^3.$
Now, we must take into account the fact that $M^2=M$ (condition (ii)) with :
$$M^2=begin{pmatrix}
0&0&1&(d - 2c + cd)\
0&0&-1&(2c - d + c^2 - 2cd))\
0&0&1&(c + 1)(d-2c)\
0&0&0&c^2end{pmatrix}.$$
Identifiying $M^2$ and $M$ leads to a first condition $c^2=c$, thus necessarily $c=0$ or $c=1$. Let us consider each of these cases :
if $c=0$, whatever the value of $d$, all other constraints are verified BUT we get in this way a fourth column proportional to the third one ; thus we would have Rank(M)=1 ; this comes in contradiction with the somewhat hidden condition : $dim ker f = 2 implies Rank(f)=4-2=2$. (I am indebted here to the solution by @Egreg : I hadn't seen at first this condition). Thus we must rule out this case.
if $c=1$, we deduce without difficulty that $d=2$ matches all conditions, giving the rank-2 unique solution :
$$M=begin{pmatrix}
0&0&1&2\
0&0&-1&-3\
0&0&1&0\
0&0&0&1end{pmatrix}.$$
I think that the Jordan form will not be difficult. Btw, what are the eigenvalues of a projector ?
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
$endgroup$
Conditions (i)-...-(v) have to be considered jointly. We are going to show that they determine a single matrix.
We will consider the matrix $M$ describing the transformation with respect to the (canonical) basis ${1,x,x^2,x^3}.$
The columns of such a matrix associated with a linear operation are given by the images of the basis'elements. Let us show in a first step that :
$$M=begin{pmatrix}
0&0&1&d\
0&0&-1&(c-2d)\
0&0&1&(-2c+d)\
0&0&0&cend{pmatrix}.$$
Explanations :
Columns 1 and 2 : they are both zero because condition (i) says that $f(1)=0=0+0x+0x^2+0x^3$ and $f(x)=0=0+0x+0x^2+0x^3$.
Column 3 : (condition (iii)) coefficients of $1-1x+1x^2+0x^3$.
Column 4 : condition (iv) amounts to say that $1$ is a double root of $p$ ; thus the image can be written $p=(x-1)^2(cx+d)=d+(c-2d)x+(-2c+d)x^2+cx^3.$
Now, we must take into account the fact that $M^2=M$ (condition (ii)) with :
$$M^2=begin{pmatrix}
0&0&1&(d - 2c + cd)\
0&0&-1&(2c - d + c^2 - 2cd))\
0&0&1&(c + 1)(d-2c)\
0&0&0&c^2end{pmatrix}.$$
Identifiying $M^2$ and $M$ leads to a first condition $c^2=c$, thus necessarily $c=0$ or $c=1$. Let us consider each of these cases :
if $c=0$, whatever the value of $d$, all other constraints are verified BUT we get in this way a fourth column proportional to the third one ; thus we would have Rank(M)=1 ; this comes in contradiction with the somewhat hidden condition : $dim ker f = 2 implies Rank(f)=4-2=2$. (I am indebted here to the solution by @Egreg : I hadn't seen at first this condition). Thus we must rule out this case.
if $c=1$, we deduce without difficulty that $d=2$ matches all conditions, giving the rank-2 unique solution :
$$M=begin{pmatrix}
0&0&1&2\
0&0&-1&-3\
0&0&1&0\
0&0&0&1end{pmatrix}.$$
I think that the Jordan form will not be difficult. Btw, what are the eigenvalues of a projector ?
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
edited Feb 5 at 8:41
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
answered Feb 2 at 16:01
Jean MarieJean Marie
31.5k42355
31.5k42355
$begingroup$
Can you say, 4 hours later, if this solution aggrees you ? You aren't new on this site : you know it is done for exchanges, It's not just an anonymous booth.
$endgroup$
– Jean Marie
Feb 2 at 20:40
$begingroup$
So If I understand well - you finds matrix of linear transformation in use of given information. You can do this because in each step you use some vectors (there presented as polynomials) which together creates a base of $R[x]_3$. Is that right? After that you have all things to prepare Jordan Matrix, right? Ps: sorry but I am already a bit overtired, I run with the material and I can not cover everything at once
$endgroup$
– VirtualUser
Feb 2 at 20:59
$begingroup$
Thanks for answering. You are excused. Answers : Right. Right (you do it just as if the original problem had been presented in $mathbb{R}^4$).
$endgroup$
– Jean Marie
Feb 2 at 21:12
$begingroup$
Okay, but I have one more question - to step with column 4. You have seen that it is double root - but let say that I didn't see that. So I have decided to put $ f(x^3) = p(x) = ax^9 +bx^6 + cx^3 + d $ and then check $p(1)=0$ and $p(-1)=0$ And then I get this column $[c+2d,-2c-3d,c,d]^T $ and I don't know if I have take mistake or my approach is not correct (but it seems to be reasonable)
$endgroup$
– VirtualUser
Feb 2 at 21:22
$begingroup$
No : notation $f(x^3)$ is misleading, but shouldn't be interpreted like that: it is the effect of the transformation of $x^3$ by the linear operator$f$. Your intepretation cannot be the right one : we must stay in the vector space of polynomials with degree $leq 3$.
$endgroup$
– Jean Marie
Feb 2 at 21:25
|
show 1 more comment
$begingroup$
Can you say, 4 hours later, if this solution aggrees you ? You aren't new on this site : you know it is done for exchanges, It's not just an anonymous booth.
$endgroup$
– Jean Marie
Feb 2 at 20:40
$begingroup$
So If I understand well - you finds matrix of linear transformation in use of given information. You can do this because in each step you use some vectors (there presented as polynomials) which together creates a base of $R[x]_3$. Is that right? After that you have all things to prepare Jordan Matrix, right? Ps: sorry but I am already a bit overtired, I run with the material and I can not cover everything at once
$endgroup$
– VirtualUser
Feb 2 at 20:59
$begingroup$
Thanks for answering. You are excused. Answers : Right. Right (you do it just as if the original problem had been presented in $mathbb{R}^4$).
$endgroup$
– Jean Marie
Feb 2 at 21:12
$begingroup$
Okay, but I have one more question - to step with column 4. You have seen that it is double root - but let say that I didn't see that. So I have decided to put $ f(x^3) = p(x) = ax^9 +bx^6 + cx^3 + d $ and then check $p(1)=0$ and $p(-1)=0$ And then I get this column $[c+2d,-2c-3d,c,d]^T $ and I don't know if I have take mistake or my approach is not correct (but it seems to be reasonable)
$endgroup$
– VirtualUser
Feb 2 at 21:22
$begingroup$
No : notation $f(x^3)$ is misleading, but shouldn't be interpreted like that: it is the effect of the transformation of $x^3$ by the linear operator$f$. Your intepretation cannot be the right one : we must stay in the vector space of polynomials with degree $leq 3$.
$endgroup$
– Jean Marie
Feb 2 at 21:25
$begingroup$
Can you say, 4 hours later, if this solution aggrees you ? You aren't new on this site : you know it is done for exchanges, It's not just an anonymous booth.
$endgroup$
– Jean Marie
Feb 2 at 20:40
$begingroup$
Can you say, 4 hours later, if this solution aggrees you ? You aren't new on this site : you know it is done for exchanges, It's not just an anonymous booth.
$endgroup$
– Jean Marie
Feb 2 at 20:40
$begingroup$
So If I understand well - you finds matrix of linear transformation in use of given information. You can do this because in each step you use some vectors (there presented as polynomials) which together creates a base of $R[x]_3$. Is that right? After that you have all things to prepare Jordan Matrix, right? Ps: sorry but I am already a bit overtired, I run with the material and I can not cover everything at once
$endgroup$
– VirtualUser
Feb 2 at 20:59
$begingroup$
So If I understand well - you finds matrix of linear transformation in use of given information. You can do this because in each step you use some vectors (there presented as polynomials) which together creates a base of $R[x]_3$. Is that right? After that you have all things to prepare Jordan Matrix, right? Ps: sorry but I am already a bit overtired, I run with the material and I can not cover everything at once
$endgroup$
– VirtualUser
Feb 2 at 20:59
$begingroup$
Thanks for answering. You are excused. Answers : Right. Right (you do it just as if the original problem had been presented in $mathbb{R}^4$).
$endgroup$
– Jean Marie
Feb 2 at 21:12
$begingroup$
Thanks for answering. You are excused. Answers : Right. Right (you do it just as if the original problem had been presented in $mathbb{R}^4$).
$endgroup$
– Jean Marie
Feb 2 at 21:12
$begingroup$
Okay, but I have one more question - to step with column 4. You have seen that it is double root - but let say that I didn't see that. So I have decided to put $ f(x^3) = p(x) = ax^9 +bx^6 + cx^3 + d $ and then check $p(1)=0$ and $p(-1)=0$ And then I get this column $[c+2d,-2c-3d,c,d]^T $ and I don't know if I have take mistake or my approach is not correct (but it seems to be reasonable)
$endgroup$
– VirtualUser
Feb 2 at 21:22
$begingroup$
Okay, but I have one more question - to step with column 4. You have seen that it is double root - but let say that I didn't see that. So I have decided to put $ f(x^3) = p(x) = ax^9 +bx^6 + cx^3 + d $ and then check $p(1)=0$ and $p(-1)=0$ And then I get this column $[c+2d,-2c-3d,c,d]^T $ and I don't know if I have take mistake or my approach is not correct (but it seems to be reasonable)
$endgroup$
– VirtualUser
Feb 2 at 21:22
$begingroup$
No : notation $f(x^3)$ is misleading, but shouldn't be interpreted like that: it is the effect of the transformation of $x^3$ by the linear operator$f$. Your intepretation cannot be the right one : we must stay in the vector space of polynomials with degree $leq 3$.
$endgroup$
– Jean Marie
Feb 2 at 21:25
$begingroup$
No : notation $f(x^3)$ is misleading, but shouldn't be interpreted like that: it is the effect of the transformation of $x^3$ by the linear operator$f$. Your intepretation cannot be the right one : we must stay in the vector space of polynomials with degree $leq 3$.
$endgroup$
– Jean Marie
Feb 2 at 21:25
|
show 1 more comment
$begingroup$
We know that $f(1)=0$, $f(x)=0$, $f(x^2)=1-x-x^2$; set $f(x^3)=a+bx+cx^2+dx^3$.
You might be confused by polynomial notation; here it might be convenient to set $v_1=1$, $v_2=x$, $v_3=x^2$ and $v_4=x^3$; then the above statements become more “linear algebra like”:
begin{cases}
f(v_1)=0 \
f(v_2)=0 \
f(v_3)=v_1-v_2+v_3 \
f(v_4)=av_1+bv_2+cv_3+dv_4
end{cases}
Since ${v_1,v_2,v_3,v_4}$ is a basis for $mathbb{R}_3[x]$, the above characterizes the linear map $f$, provided we determine $a,b,c,d$ so that conditions (i)–(v) are satisfied.
The matrix with respect to the basis ${v_1,v_2,v_3,v_4}$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & a \
0 & 0 & -1 & b \
0 & 0 & 1 & c \
0 & 0 & 0 & d
end{bmatrix}
$$
We can compute $A^2$:
$$
A^2=
begin{bmatrix}
0 & 0 & 1 & ad+c \
0 & 0 & -1 & bd-c \
0 & 0 & 1 & cd+c \
0 & 0 & 0 & d^2
end{bmatrix}
$$
Since we need $A=A^2$, we deduce
begin{cases}
ad+c=a \
bd-c=b \
cd+c=c \
d^2=d
end{cases}
that immediately splits into $d=0$ or $d=1$.
Case $d=0$
We get $a=c$ and $b=-c$. Therefore $f(v_4)=c(v_1-v_2+v_3)$. In terms of polynomials, we have $f(v_4)=p$, where $p(x)=c(1-x+x^2)$. We have $p(1)=c$ and condition (v) forces $c=0$. This cannot be the case, because the rank would be $1$, contrary to the requirement that $dimker f=2$.
Case $d=1$
We get $c=0$. Thus $f(v_4)=av_1+bv_2+v_4$ and so $p(x)=a+bx+x^3$. The condition $p(1)=0$ yields $a+b+1=0$; the condition $p'(1)=0$ yields $b+3=0$, so $b=-3$ and $a=2$.
Final discussion
The matrix of $f$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & 2 \
0 & 0 & -1 & -3 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{bmatrix}
$$
The matrix has a double eigenvalue $0$ and a double eigenvalue $1$. Since also the eigenspace relative to $1$ has dimension $2$, the matrix is diagonalizable.
answered Feb 2 at 22:19
egregegreg
186k1486208
$endgroup$
$begingroup$
You are right, case $d=0$ has to be ruled out : my solution (that I have rectified) was an answer to a milder condition : (i) $ ker f supseteq span{1,x}$ .
$endgroup$
– Jean Marie
Feb 3 at 11:02
add a comment |
$begingroup$
We know that $f(1)=0$, $f(x)=0$, $f(x^2)=1-x-x^2$; set $f(x^3)=a+bx+cx^2+dx^3$.
You might be confused by polynomial notation; here it might be convenient to set $v_1=1$, $v_2=x$, $v_3=x^2$ and $v_4=x^3$; then the above statements become more “linear algebra like”:
begin{cases}
f(v_1)=0 \
f(v_2)=0 \
f(v_3)=v_1-v_2+v_3 \
f(v_4)=av_1+bv_2+cv_3+dv_4
end{cases}
Since ${v_1,v_2,v_3,v_4}$ is a basis for $mathbb{R}_3[x]$, the above characterizes the linear map $f$, provided we determine $a,b,c,d$ so that conditions (i)–(v) are satisfied.
The matrix with respect to the basis ${v_1,v_2,v_3,v_4}$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & a \
0 & 0 & -1 & b \
0 & 0 & 1 & c \
0 & 0 & 0 & d
end{bmatrix}
$$
We can compute $A^2$:
$$
A^2=
begin{bmatrix}
0 & 0 & 1 & ad+c \
0 & 0 & -1 & bd-c \
0 & 0 & 1 & cd+c \
0 & 0 & 0 & d^2
end{bmatrix}
$$
Since we need $A=A^2$, we deduce
begin{cases}
ad+c=a \
bd-c=b \
cd+c=c \
d^2=d
end{cases}
that immediately splits into $d=0$ or $d=1$.
Case $d=0$
We get $a=c$ and $b=-c$. Therefore $f(v_4)=c(v_1-v_2+v_3)$. In terms of polynomials, we have $f(v_4)=p$, where $p(x)=c(1-x+x^2)$. We have $p(1)=c$ and condition (v) forces $c=0$. This cannot be the case, because the rank would be $1$, contrary to the requirement that $dimker f=2$.
Case $d=1$
We get $c=0$. Thus $f(v_4)=av_1+bv_2+v_4$ and so $p(x)=a+bx+x^3$. The condition $p(1)=0$ yields $a+b+1=0$; the condition $p'(1)=0$ yields $b+3=0$, so $b=-3$ and $a=2$.
Final discussion
The matrix of $f$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & 2 \
0 & 0 & -1 & -3 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{bmatrix}
$$
The matrix has a double eigenvalue $0$ and a double eigenvalue $1$. Since also the eigenspace relative to $1$ has dimension $2$, the matrix is diagonalizable.
answered Feb 2 at 22:19
egregegreg
186k1486208
$endgroup$
$begingroup$
You are right, case $d=0$ has to be ruled out : my solution (that I have rectified) was an answer to a milder condition : (i) $ ker f supseteq span{1,x}$ .
$endgroup$
– Jean Marie
Feb 3 at 11:02
add a comment |
$begingroup$
We know that $f(1)=0$, $f(x)=0$, $f(x^2)=1-x-x^2$; set $f(x^3)=a+bx+cx^2+dx^3$.
You might be confused by polynomial notation; here it might be convenient to set $v_1=1$, $v_2=x$, $v_3=x^2$ and $v_4=x^3$; then the above statements become more “linear algebra like”:
begin{cases}
f(v_1)=0 \
f(v_2)=0 \
f(v_3)=v_1-v_2+v_3 \
f(v_4)=av_1+bv_2+cv_3+dv_4
end{cases}
Since ${v_1,v_2,v_3,v_4}$ is a basis for $mathbb{R}_3[x]$, the above characterizes the linear map $f$, provided we determine $a,b,c,d$ so that conditions (i)–(v) are satisfied.
The matrix with respect to the basis ${v_1,v_2,v_3,v_4}$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & a \
0 & 0 & -1 & b \
0 & 0 & 1 & c \
0 & 0 & 0 & d
end{bmatrix}
$$
We can compute $A^2$:
$$
A^2=
begin{bmatrix}
0 & 0 & 1 & ad+c \
0 & 0 & -1 & bd-c \
0 & 0 & 1 & cd+c \
0 & 0 & 0 & d^2
end{bmatrix}
$$
Since we need $A=A^2$, we deduce
begin{cases}
ad+c=a \
bd-c=b \
cd+c=c \
d^2=d
end{cases}
that immediately splits into $d=0$ or $d=1$.
Case $d=0$
We get $a=c$ and $b=-c$. Therefore $f(v_4)=c(v_1-v_2+v_3)$. In terms of polynomials, we have $f(v_4)=p$, where $p(x)=c(1-x+x^2)$. We have $p(1)=c$ and condition (v) forces $c=0$. This cannot be the case, because the rank would be $1$, contrary to the requirement that $dimker f=2$.
Case $d=1$
We get $c=0$. Thus $f(v_4)=av_1+bv_2+v_4$ and so $p(x)=a+bx+x^3$. The condition $p(1)=0$ yields $a+b+1=0$; the condition $p'(1)=0$ yields $b+3=0$, so $b=-3$ and $a=2$.
Final discussion
The matrix of $f$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & 2 \
0 & 0 & -1 & -3 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{bmatrix}
$$
The matrix has a double eigenvalue $0$ and a double eigenvalue $1$. Since also the eigenspace relative to $1$ has dimension $2$, the matrix is diagonalizable.
answered Feb 2 at 22:19
egregegreg
186k1486208
$endgroup$
We know that $f(1)=0$, $f(x)=0$, $f(x^2)=1-x-x^2$; set $f(x^3)=a+bx+cx^2+dx^3$.
You might be confused by polynomial notation; here it might be convenient to set $v_1=1$, $v_2=x$, $v_3=x^2$ and $v_4=x^3$; then the above statements become more “linear algebra like”:
begin{cases}
f(v_1)=0 \
f(v_2)=0 \
f(v_3)=v_1-v_2+v_3 \
f(v_4)=av_1+bv_2+cv_3+dv_4
end{cases}
Since ${v_1,v_2,v_3,v_4}$ is a basis for $mathbb{R}_3[x]$, the above characterizes the linear map $f$, provided we determine $a,b,c,d$ so that conditions (i)–(v) are satisfied.
The matrix with respect to the basis ${v_1,v_2,v_3,v_4}$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & a \
0 & 0 & -1 & b \
0 & 0 & 1 & c \
0 & 0 & 0 & d
end{bmatrix}
$$
We can compute $A^2$:
$$
A^2=
begin{bmatrix}
0 & 0 & 1 & ad+c \
0 & 0 & -1 & bd-c \
0 & 0 & 1 & cd+c \
0 & 0 & 0 & d^2
end{bmatrix}
$$
Since we need $A=A^2$, we deduce
begin{cases}
ad+c=a \
bd-c=b \
cd+c=c \
d^2=d
end{cases}
that immediately splits into $d=0$ or $d=1$.
Case $d=0$
We get $a=c$ and $b=-c$. Therefore $f(v_4)=c(v_1-v_2+v_3)$. In terms of polynomials, we have $f(v_4)=p$, where $p(x)=c(1-x+x^2)$. We have $p(1)=c$ and condition (v) forces $c=0$. This cannot be the case, because the rank would be $1$, contrary to the requirement that $dimker f=2$.
Case $d=1$
We get $c=0$. Thus $f(v_4)=av_1+bv_2+v_4$ and so $p(x)=a+bx+x^3$. The condition $p(1)=0$ yields $a+b+1=0$; the condition $p'(1)=0$ yields $b+3=0$, so $b=-3$ and $a=2$.
Final discussion
The matrix of $f$ is
$$
A=begin{bmatrix}
0 & 0 & 1 & 2 \
0 & 0 & -1 & -3 \
0 & 0 & 1 & 0 \
0 & 0 & 0 & 1
end{bmatrix}
$$
The matrix has a double eigenvalue $0$ and a double eigenvalue $1$. Since also the eigenspace relative to $1$ has dimension $2$, the matrix is diagonalizable.
answered Feb 2 at 22:19
egregegreg
186k1486208
answered Feb 2 at 22:19
egregegreg
186k1486208
answered Feb 2 at 22:19
egregegreg
186k1486208
answered Feb 2 at 22:19
egregegreg
186k1486208
186k1486208
$begingroup$
You are right, case $d=0$ has to be ruled out : my solution (that I have rectified) was an answer to a milder condition : (i) $ ker f supseteq span{1,x}$ .
$endgroup$
– Jean Marie
Feb 3 at 11:02
add a comment |
$begingroup$
You are right, case $d=0$ has to be ruled out : my solution (that I have rectified) was an answer to a milder condition : (i) $ ker f supseteq span{1,x}$ .
$endgroup$
– Jean Marie
Feb 3 at 11:02
$begingroup$
You are right, case $d=0$ has to be ruled out : my solution (that I have rectified) was an answer to a milder condition : (i) $ ker f supseteq span{1,x}$ .
$endgroup$
– Jean Marie
Feb 3 at 11:02
$begingroup$
You are right, case $d=0$ has to be ruled out : my solution (that I have rectified) was an answer to a milder condition : (i) $ ker f supseteq span{1,x}$ .
$endgroup$
– Jean Marie
Feb 3 at 11:02
add a comment |
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Are (i)-(iv) separate questions, or should you look for endomorphisms which satisfy all of them at the same time?
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– Arthur
Feb 2 at 15:23
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Could you answer to Arthur, please ? If you must meet the 5 conditions, consider a generic element under the form $ax^3+bx^2+cx+d$ then "translate" all your conditions into the "language of $mathbb{R}^n$" ; in particular, accumulate information about the matrix which corresponds to $f$.
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– Jean Marie
Feb 2 at 15:31
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@Arthur I am not sure how to understand the content of this task, but it seems to me that each case is separate
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– VirtualUser
Feb 2 at 16:06
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I don't think so (see my first sentence in my answer.
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– Jean Marie
Feb 2 at 17:04
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I have rectified my answer at its end (no longer two but one solution) taking into account a restriction pinpointed by @egreg
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– Jean Marie
Feb 3 at 11:21