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Find $B$ if $B=A-1over2 A^2+1over3 A^3 -1over4 A^4+…$
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Let $$
A=beginbmatrix 0 & a & a^2 & a^3 \ 0 & 0 & a & a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$$ and
$B=A-1over2 A^2+1over3 A^3 -1over4 A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ 1over2! B^2+ 1over3! B^3+...$
My attempt:
$i)$
I calculated $A^2$ by multiplying $A$ by itself, then foundnd $A^3$ by multiplying $A$ by $A^2$, ans so on.
Then I noted that $A^n=0$ for $ngeq 4$
$A^2= A.A=beginbmatrix 0 & 0 & a^2 & 2a^3 \ 0 & 0 & 0 & a^2 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^3= A^2.A=beginbmatrix 0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^4=A^3.A =beginbmatrix 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^n=0$ for every $ngeq 4$, so:
$B= beginbmatrix 0 & a & 1over2 a^2 & 1over 3a^3 \ 0 & 0 & a & 1over2 a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$
But is there any way easier than my way ?
And what about $(ii)$ ?
linear-algebra matrices
edited Jul 21 at 5:04
Math Lover
12.4k21232
asked Jul 21 at 3:48
Dima
438214
add a comment |Â
up vote
3
down vote
favorite
Let $$
A=beginbmatrix 0 & a & a^2 & a^3 \ 0 & 0 & a & a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$$ and
$B=A-1over2 A^2+1over3 A^3 -1over4 A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ 1over2! B^2+ 1over3! B^3+...$
My attempt:
$i)$
I calculated $A^2$ by multiplying $A$ by itself, then foundnd $A^3$ by multiplying $A$ by $A^2$, ans so on.
Then I noted that $A^n=0$ for $ngeq 4$
$A^2= A.A=beginbmatrix 0 & 0 & a^2 & 2a^3 \ 0 & 0 & 0 & a^2 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^3= A^2.A=beginbmatrix 0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^4=A^3.A =beginbmatrix 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^n=0$ for every $ngeq 4$, so:
$B= beginbmatrix 0 & a & 1over2 a^2 & 1over 3a^3 \ 0 & 0 & a & 1over2 a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$
But is there any way easier than my way ?
And what about $(ii)$ ?
linear-algebra matrices
edited Jul 21 at 5:04
Math Lover
12.4k21232
asked Jul 21 at 3:48
Dima
438214
1
It's a shame that A & B are matrices. I'm tempted to provide this fake proof: Clearly $log(1+A)=A-frac12A^2+frac13 A^3 - frac14 A^4 + dots = B.$ Then $A=e^B - 1 = B + frac12! B^2 + frac13! B^3 + dots$ as desired.
– Display name
Jul 21 at 3:59
3
@Displayname if I remember correctly, the same idea works for matrices...
– Chris Custer
Jul 21 at 4:22
Not sure if this helps: You can anticipate that $A^4=0$ since $A$ has characteristic equation $lambda^4=0$.
– Ahmed S. Attaalla
Jul 21 at 4:24
there is a problem with the signs here
– Thomas
Jul 21 at 5:10
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $$
A=beginbmatrix 0 & a & a^2 & a^3 \ 0 & 0 & a & a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$$ and
$B=A-1over2 A^2+1over3 A^3 -1over4 A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ 1over2! B^2+ 1over3! B^3+...$
My attempt:
$i)$
I calculated $A^2$ by multiplying $A$ by itself, then foundnd $A^3$ by multiplying $A$ by $A^2$, ans so on.
Then I noted that $A^n=0$ for $ngeq 4$
$A^2= A.A=beginbmatrix 0 & 0 & a^2 & 2a^3 \ 0 & 0 & 0 & a^2 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^3= A^2.A=beginbmatrix 0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^4=A^3.A =beginbmatrix 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^n=0$ for every $ngeq 4$, so:
$B= beginbmatrix 0 & a & 1over2 a^2 & 1over 3a^3 \ 0 & 0 & a & 1over2 a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$
But is there any way easier than my way ?
And what about $(ii)$ ?
linear-algebra matrices
edited Jul 21 at 5:04
Math Lover
12.4k21232
asked Jul 21 at 3:48
Dima
438214
Let $$
A=beginbmatrix 0 & a & a^2 & a^3 \ 0 & 0 & a & a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$$ and
$B=A-1over2 A^2+1over3 A^3 -1over4 A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ 1over2! B^2+ 1over3! B^3+...$
My attempt:
$i)$
I calculated $A^2$ by multiplying $A$ by itself, then foundnd $A^3$ by multiplying $A$ by $A^2$, ans so on.
Then I noted that $A^n=0$ for $ngeq 4$
$A^2= A.A=beginbmatrix 0 & 0 & a^2 & 2a^3 \ 0 & 0 & 0 & a^2 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^3= A^2.A=beginbmatrix 0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^4=A^3.A =beginbmatrix 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 endbmatrix
$
$A^n=0$ for every $ngeq 4$, so:
$B= beginbmatrix 0 & a & 1over2 a^2 & 1over 3a^3 \ 0 & 0 & a & 1over2 a^2 \ 0 & 0 & 0 & a \ 0 & 0 & 0 & 0 endbmatrix
$
But is there any way easier than my way ?
And what about $(ii)$ ?
linear-algebra matrices
edited Jul 21 at 5:04
Math Lover
12.4k21232
asked Jul 21 at 3:48
Dima
438214
edited Jul 21 at 5:04
Math Lover
12.4k21232
edited Jul 21 at 5:04
Math Lover
12.4k21232
edited Jul 21 at 5:04
Math Lover
12.4k21232
12.4k21232
asked Jul 21 at 3:48
Dima
438214
asked Jul 21 at 3:48
Dima
438214
asked Jul 21 at 3:48
Dima
438214
438214
1
It's a shame that A & B are matrices. I'm tempted to provide this fake proof: Clearly $log(1+A)=A-frac12A^2+frac13 A^3 - frac14 A^4 + dots = B.$ Then $A=e^B - 1 = B + frac12! B^2 + frac13! B^3 + dots$ as desired.
– Display name
Jul 21 at 3:59
3
@Displayname if I remember correctly, the same idea works for matrices...
– Chris Custer
Jul 21 at 4:22
Not sure if this helps: You can anticipate that $A^4=0$ since $A$ has characteristic equation $lambda^4=0$.
– Ahmed S. Attaalla
Jul 21 at 4:24
there is a problem with the signs here
– Thomas
Jul 21 at 5:10
add a comment |Â
1
It's a shame that A & B are matrices. I'm tempted to provide this fake proof: Clearly $log(1+A)=A-frac12A^2+frac13 A^3 - frac14 A^4 + dots = B.$ Then $A=e^B - 1 = B + frac12! B^2 + frac13! B^3 + dots$ as desired.
– Display name
Jul 21 at 3:59
3
@Displayname if I remember correctly, the same idea works for matrices...
– Chris Custer
Jul 21 at 4:22
Not sure if this helps: You can anticipate that $A^4=0$ since $A$ has characteristic equation $lambda^4=0$.
– Ahmed S. Attaalla
Jul 21 at 4:24
there is a problem with the signs here
– Thomas
Jul 21 at 5:10
1
1
It's a shame that A & B are matrices. I'm tempted to provide this fake proof: Clearly $log(1+A)=A-frac12A^2+frac13 A^3 - frac14 A^4 + dots = B.$ Then $A=e^B - 1 = B + frac12! B^2 + frac13! B^3 + dots$ as desired.
– Display name
Jul 21 at 3:59
It's a shame that A & B are matrices. I'm tempted to provide this fake proof: Clearly $log(1+A)=A-frac12A^2+frac13 A^3 - frac14 A^4 + dots = B.$ Then $A=e^B - 1 = B + frac12! B^2 + frac13! B^3 + dots$ as desired.
– Display name
Jul 21 at 3:59
3
3
@Displayname if I remember correctly, the same idea works for matrices...
– Chris Custer
Jul 21 at 4:22
@Displayname if I remember correctly, the same idea works for matrices...
– Chris Custer
Jul 21 at 4:22
Not sure if this helps: You can anticipate that $A^4=0$ since $A$ has characteristic equation $lambda^4=0$.
– Ahmed S. Attaalla
Jul 21 at 4:24
Not sure if this helps: You can anticipate that $A^4=0$ since $A$ has characteristic equation $lambda^4=0$.
– Ahmed S. Attaalla
Jul 21 at 4:24
there is a problem with the signs here
– Thomas
Jul 21 at 5:10
there is a problem with the signs here
– Thomas
Jul 21 at 5:10
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
$B=ln(I+A)$, and so $e^B=I+Aimplies ii)$...
You asked if there was a better way of doing it... I believe the answer is to notice that you have the power series for $ln$ and $exp$ respectively; and then use the fact (probably covered in your course) that those series can be applied to matrices. See for instance .
answered Jul 21 at 4:14
Chris Custer
5,4282622
This may be a bit cryptic but it's actually correct.
– mathreadler
Jul 21 at 5:37
I do not think the OP is able to satisfy the condition $||A||<1$.
– loup blanc
Jul 21 at 15:43
By choosing $a$ small, we get $midmid I+A-Imidmid =midmid Amidmidltepsilon $.
– Chris Custer
Jul 21 at 16:30
add a comment |Â
up vote
1
down vote
You are on the right track. For (ii), just compute $B^2$, $B^3$, $B^4$, and observe that $B^4 = O$. Afterwards, verify that
$$A = B + frac12!B^2 + frac13! B^3.$$
For your information,
$$B^2 = beginbmatrix0 & 0 & a^2 & a^3 \ 0 & 0 & 0 & a^2 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$ and
$$B^3 = beginbmatrix0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix.$$
answered Jul 21 at 4:32
Math Lover
12.4k21232
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$B=ln(I+A)$, and so $e^B=I+Aimplies ii)$...
You asked if there was a better way of doing it... I believe the answer is to notice that you have the power series for $ln$ and $exp$ respectively; and then use the fact (probably covered in your course) that those series can be applied to matrices. See for instance .
answered Jul 21 at 4:14
Chris Custer
5,4282622
This may be a bit cryptic but it's actually correct.
– mathreadler
Jul 21 at 5:37
I do not think the OP is able to satisfy the condition $||A||<1$.
– loup blanc
Jul 21 at 15:43
By choosing $a$ small, we get $midmid I+A-Imidmid =midmid Amidmidltepsilon $.
– Chris Custer
Jul 21 at 16:30
add a comment |Â
up vote
2
down vote
$B=ln(I+A)$, and so $e^B=I+Aimplies ii)$...
You asked if there was a better way of doing it... I believe the answer is to notice that you have the power series for $ln$ and $exp$ respectively; and then use the fact (probably covered in your course) that those series can be applied to matrices. See for instance .
answered Jul 21 at 4:14
Chris Custer
5,4282622
This may be a bit cryptic but it's actually correct.
– mathreadler
Jul 21 at 5:37
I do not think the OP is able to satisfy the condition $||A||<1$.
– loup blanc
Jul 21 at 15:43
By choosing $a$ small, we get $midmid I+A-Imidmid =midmid Amidmidltepsilon $.
– Chris Custer
Jul 21 at 16:30
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$B=ln(I+A)$, and so $e^B=I+Aimplies ii)$...
You asked if there was a better way of doing it... I believe the answer is to notice that you have the power series for $ln$ and $exp$ respectively; and then use the fact (probably covered in your course) that those series can be applied to matrices. See for instance .
answered Jul 21 at 4:14
Chris Custer
5,4282622
$B=ln(I+A)$, and so $e^B=I+Aimplies ii)$...
You asked if there was a better way of doing it... I believe the answer is to notice that you have the power series for $ln$ and $exp$ respectively; and then use the fact (probably covered in your course) that those series can be applied to matrices. See for instance .
answered Jul 21 at 4:14
Chris Custer
5,4282622
edited Jul 21 at 5:45
answered Jul 21 at 4:14
Chris Custer
5,4282622
answered Jul 21 at 4:14
Chris Custer
5,4282622
answered Jul 21 at 4:14
Chris Custer
5,4282622
5,4282622
This may be a bit cryptic but it's actually correct.
– mathreadler
Jul 21 at 5:37
I do not think the OP is able to satisfy the condition $||A||<1$.
– loup blanc
Jul 21 at 15:43
By choosing $a$ small, we get $midmid I+A-Imidmid =midmid Amidmidltepsilon $.
– Chris Custer
Jul 21 at 16:30
add a comment |Â
This may be a bit cryptic but it's actually correct.
– mathreadler
Jul 21 at 5:37
I do not think the OP is able to satisfy the condition $||A||<1$.
– loup blanc
Jul 21 at 15:43
By choosing $a$ small, we get $midmid I+A-Imidmid =midmid Amidmidltepsilon $.
– Chris Custer
Jul 21 at 16:30
This may be a bit cryptic but it's actually correct.
– mathreadler
Jul 21 at 5:37
This may be a bit cryptic but it's actually correct.
– mathreadler
Jul 21 at 5:37
I do not think the OP is able to satisfy the condition $||A||<1$.
– loup blanc
Jul 21 at 15:43
I do not think the OP is able to satisfy the condition $||A||<1$.
– loup blanc
Jul 21 at 15:43
By choosing $a$ small, we get $midmid I+A-Imidmid =midmid Amidmidltepsilon $.
– Chris Custer
Jul 21 at 16:30
By choosing $a$ small, we get $midmid I+A-Imidmid =midmid Amidmidltepsilon $.
– Chris Custer
Jul 21 at 16:30
add a comment |Â
up vote
1
down vote
You are on the right track. For (ii), just compute $B^2$, $B^3$, $B^4$, and observe that $B^4 = O$. Afterwards, verify that
$$A = B + frac12!B^2 + frac13! B^3.$$
For your information,
$$B^2 = beginbmatrix0 & 0 & a^2 & a^3 \ 0 & 0 & 0 & a^2 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$ and
$$B^3 = beginbmatrix0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix.$$
answered Jul 21 at 4:32
Math Lover
12.4k21232
add a comment |Â
up vote
1
down vote
You are on the right track. For (ii), just compute $B^2$, $B^3$, $B^4$, and observe that $B^4 = O$. Afterwards, verify that
$$A = B + frac12!B^2 + frac13! B^3.$$
For your information,
$$B^2 = beginbmatrix0 & 0 & a^2 & a^3 \ 0 & 0 & 0 & a^2 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$ and
$$B^3 = beginbmatrix0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix.$$
answered Jul 21 at 4:32
Math Lover
12.4k21232
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You are on the right track. For (ii), just compute $B^2$, $B^3$, $B^4$, and observe that $B^4 = O$. Afterwards, verify that
$$A = B + frac12!B^2 + frac13! B^3.$$
For your information,
$$B^2 = beginbmatrix0 & 0 & a^2 & a^3 \ 0 & 0 & 0 & a^2 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$ and
$$B^3 = beginbmatrix0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix.$$
answered Jul 21 at 4:32
Math Lover
12.4k21232
You are on the right track. For (ii), just compute $B^2$, $B^3$, $B^4$, and observe that $B^4 = O$. Afterwards, verify that
$$A = B + frac12!B^2 + frac13! B^3.$$
For your information,
$$B^2 = beginbmatrix0 & 0 & a^2 & a^3 \ 0 & 0 & 0 & a^2 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix,$$ and
$$B^3 = beginbmatrix0 & 0 & 0 & a^3 \ 0 & 0 & 0 & 0 \0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0endbmatrix.$$
answered Jul 21 at 4:32
Math Lover
12.4k21232
edited Jul 21 at 5:02
answered Jul 21 at 4:32
Math Lover
12.4k21232
answered Jul 21 at 4:32
Math Lover
12.4k21232
answered Jul 21 at 4:32
Math Lover
12.4k21232
12.4k21232
add a comment |Â
add a comment |Â
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1
It's a shame that A & B are matrices. I'm tempted to provide this fake proof: Clearly $log(1+A)=A-frac12A^2+frac13 A^3 - frac14 A^4 + dots = B.$ Then $A=e^B - 1 = B + frac12! B^2 + frac13! B^3 + dots$ as desired.
– Display name
Jul 21 at 3:59
3
@Displayname if I remember correctly, the same idea works for matrices...
– Chris Custer
Jul 21 at 4:22
Not sure if this helps: You can anticipate that $A^4=0$ since $A$ has characteristic equation $lambda^4=0$.
– Ahmed S. Attaalla
Jul 21 at 4:24
there is a problem with the signs here
– Thomas
Jul 21 at 5:10