Mixing
Matrix demonstration $A^k$
$begingroup$
Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $
Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$
linear-algebra
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
asked Jan 13 at 0:36
PTSONICPTSONIC
83
$endgroup$
add a comment |
$begingroup$
Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $
Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$
linear-algebra
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
asked Jan 13 at 0:36
PTSONICPTSONIC
83
$endgroup$
$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37
2
$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37
$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40
1
$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46
$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01
add a comment |
$begingroup$
Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $
Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$
linear-algebra
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
asked Jan 13 at 0:36
PTSONICPTSONIC
83
$endgroup$
Given a matrix $A = begin{bmatrix} 7 & 4\ -9 & -5 end{bmatrix}$ $in mathcal{M2times2}, (mathbb{R}) $
Show that $A^k = begin{bmatrix} 1+6k & 4k\ -9k & 1-6k end{bmatrix} $
for every $k in mathbb{N}$
linear-algebra
linear-algebra
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
asked Jan 13 at 0:36
PTSONICPTSONIC
83
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
asked Jan 13 at 0:36
PTSONICPTSONIC
83
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
edited Jan 13 at 0:36
Robert Lewis
46.2k23066
46.2k23066
asked Jan 13 at 0:36
PTSONICPTSONIC
83
asked Jan 13 at 0:36
PTSONICPTSONIC
83
asked Jan 13 at 0:36
PTSONICPTSONIC
83
83
$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37
2
$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37
$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40
1
$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46
$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01
add a comment |
$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37
2
$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37
$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40
1
$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46
$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01
$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37
$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37
2
2
$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37
$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37
$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40
$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40
1
1
$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46
$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46
$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01
$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$
then assuming that for some $k$
$A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$
we find
$A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
$= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$
which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
$endgroup$
add a comment |
$begingroup$
Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$
How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.
answered Jan 13 at 1:26
amdamd
30k21050
$endgroup$
$begingroup$
Very nice indeed, endorsed, +1!!!
$endgroup$
– Robert Lewis
Jan 13 at 1:47
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$
then assuming that for some $k$
$A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$
we find
$A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
$= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$
which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
$endgroup$
add a comment |
$begingroup$
Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$
then assuming that for some $k$
$A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$
we find
$A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
$= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$
which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
$endgroup$
add a comment |
$begingroup$
Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$
then assuming that for some $k$
$A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$
we find
$A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
$= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$
which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
$endgroup$
Try a simple induction on $k$; it is clear that for $k = 1$,
$A^1 = A = begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 1 + 6 cdot 1 & 4 cdot 1 \ -9 cdot 1 & 1 - 6 cdot 1 end{bmatrix}; tag 1$
then assuming that for some $k$
$A^k = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}, tag 2$
we find
$A^{k + 1} = A^kA = begin{bmatrix} 1 + 6 cdot k & 4 cdot k \ -9 cdot k & 1 - 6 cdot k end{bmatrix}begin{bmatrix} 7 & 4 \ -9 & -5 end{bmatrix} = begin{bmatrix} 7 + 42k - 36k & 4 + 24k - 20k \ -63k - 9 + 54k & -36k - 5 + 30k end{bmatrix}$
$= begin{bmatrix} 7 + 6k & 4 + 4k \ -9 -9k & - 5 -6k end{bmatrix} = begin{bmatrix} 1 + 6(k + 1) & 4(k + 1) \ -9(k + 1) & 1 - 6(k + 1) end{bmatrix}, tag 3$
which is simply (2) with $k$ replaced by $k + 1$; we thus infer the formula (2) holds for all $k ge 1$, as desired. $OEDelta$.
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
answered Jan 13 at 0:56
Robert LewisRobert Lewis
46.2k23066
46.2k23066
add a comment |
add a comment |
$begingroup$
Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$
How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.
answered Jan 13 at 1:26
amdamd
30k21050
$endgroup$
$begingroup$
Very nice indeed, endorsed, +1!!!
$endgroup$
– Robert Lewis
Jan 13 at 1:47
add a comment |
$begingroup$
Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$
How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.
answered Jan 13 at 1:26
amdamd
30k21050
$endgroup$
$begingroup$
Very nice indeed, endorsed, +1!!!
$endgroup$
– Robert Lewis
Jan 13 at 1:47
add a comment |
$begingroup$
Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$
How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.
answered Jan 13 at 1:26
amdamd
30k21050
$endgroup$
Let $N = A-I$ and observe that $N^2=0$. Since $N$ and $I$ commute, we can expand via the binomial theorem: $$A^k = (I+N)^k = I+binom k1N+binom k2N^2+dots = I+kN.$$
How did I hit upon this decomposition? I computed that $1$ was the only eigenvalue of $A$, but the only diagonalizable $2times2$ matrices with repeated eigenvalues are multiples of the identity, so $A$ splits into the sum of the identity and a nilpotent matrix. That is, $$A = Pbegin{bmatrix}1&1\0&1end{bmatrix}P^{-1} = I + Pbegin{bmatrix}0&1\0&0end{bmatrix}P^{-1}$$ for some invertible matrix $P$.
answered Jan 13 at 1:26
amdamd
30k21050
edited Jan 16 at 1:10
answered Jan 13 at 1:26
amdamd
30k21050
answered Jan 13 at 1:26
amdamd
30k21050
answered Jan 13 at 1:26
amdamd
30k21050
30k21050
$begingroup$
Very nice indeed, endorsed, +1!!!
$endgroup$
– Robert Lewis
Jan 13 at 1:47
add a comment |
$begingroup$
Very nice indeed, endorsed, +1!!!
$endgroup$
– Robert Lewis
Jan 13 at 1:47
$begingroup$
Very nice indeed, endorsed, +1!!!
$endgroup$
– Robert Lewis
Jan 13 at 1:47
$begingroup$
Very nice indeed, endorsed, +1!!!
$endgroup$
– Robert Lewis
Jan 13 at 1:47
add a comment |
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$begingroup$
What have you tried so far? What approach do you think will work here?
$endgroup$
– user3482749
Jan 13 at 0:37
2
$begingroup$
Have you tried proof by induction?
$endgroup$
– user7530
Jan 13 at 0:37
$begingroup$
yes I thought about the proof by induction but i got lost in the process, I don't know which way is the best, honestly.
$endgroup$
– PTSONIC
Jan 13 at 0:40
1
$begingroup$
Can you start by computing $A^kA$ and simplifying the result? Shows us your work.
$endgroup$
– Git Gud
Jan 13 at 0:46
$begingroup$
yes the result is this matrix I think $begin{bmatrix} 7+6k & 4+4k\ -9-9k & -5-6k end{bmatrix} $
$endgroup$
– PTSONIC
Jan 13 at 1:01