$A^n$ formula not working. Checked arithmetic











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I'm trying to find the general formula for a Matrix raised to a certain power using diagonalization.
My original Matrix A is:
$$
begin{bmatrix}
-3 & 2 \
-12 & 7 \
end{bmatrix}
$$
My eigenvector Matrix,P, is:
$$
begin{bmatrix}
cfrac{1}{2} & cfrac{1}{3} \
1 & 1 \
end{bmatrix}
$$



I found the diagonal Matrix, D to be:
$$
begin{bmatrix}
1 & 0 \
0 & 3 \
end{bmatrix}
$$



And P inverse is:



$$
begin{bmatrix}
6 & -2 \
-6 & 3 \
end{bmatrix}
$$
The final formula i got from multiplying
$PD^n P^-1$



is



$$
begin{bmatrix}
3-6^n & -1+3^n \
6- 18^n & -2 + 9^n \
end{bmatrix}
$$



It checks out in the case of $A^1$ but not $A^0$. Where did I go wrong?






matrices eigenvalues-eigenvectors diagonalization







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  • bmatrix gives square brackets, pmatrix gives parentheses
    – amWhy
    2 days ago












  • @DaniJo: Your eigenvalues / eigenvectors are correct. However, check your multiplication, for example $-2 times 3^n ne -6^n$.
    – Moo
    2 days ago












  • $P, P^{-1},D$ are all fine. The problem is in the last bit.
    – Henno Brandsma
    2 days ago















up vote
0
down vote

favorite












I'm trying to find the general formula for a Matrix raised to a certain power using diagonalization.
My original Matrix A is:
$$
begin{bmatrix}
-3 & 2 \
-12 & 7 \
end{bmatrix}
$$
My eigenvector Matrix,P, is:
$$
begin{bmatrix}
cfrac{1}{2} & cfrac{1}{3} \
1 & 1 \
end{bmatrix}
$$



I found the diagonal Matrix, D to be:
$$
begin{bmatrix}
1 & 0 \
0 & 3 \
end{bmatrix}
$$



And P inverse is:



$$
begin{bmatrix}
6 & -2 \
-6 & 3 \
end{bmatrix}
$$
The final formula i got from multiplying
$PD^n P^-1$



is



$$
begin{bmatrix}
3-6^n & -1+3^n \
6- 18^n & -2 + 9^n \
end{bmatrix}
$$



It checks out in the case of $A^1$ but not $A^0$. Where did I go wrong?






matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question
























  • bmatrix gives square brackets, pmatrix gives parentheses
    – amWhy
    2 days ago












  • @DaniJo: Your eigenvalues / eigenvectors are correct. However, check your multiplication, for example $-2 times 3^n ne -6^n$.
    – Moo
    2 days ago












  • $P, P^{-1},D$ are all fine. The problem is in the last bit.
    – Henno Brandsma
    2 days ago













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm trying to find the general formula for a Matrix raised to a certain power using diagonalization.
My original Matrix A is:
$$
begin{bmatrix}
-3 & 2 \
-12 & 7 \
end{bmatrix}
$$
My eigenvector Matrix,P, is:
$$
begin{bmatrix}
cfrac{1}{2} & cfrac{1}{3} \
1 & 1 \
end{bmatrix}
$$



I found the diagonal Matrix, D to be:
$$
begin{bmatrix}
1 & 0 \
0 & 3 \
end{bmatrix}
$$



And P inverse is:



$$
begin{bmatrix}
6 & -2 \
-6 & 3 \
end{bmatrix}
$$
The final formula i got from multiplying
$PD^n P^-1$



is



$$
begin{bmatrix}
3-6^n & -1+3^n \
6- 18^n & -2 + 9^n \
end{bmatrix}
$$



It checks out in the case of $A^1$ but not $A^0$. Where did I go wrong?






matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question















I'm trying to find the general formula for a Matrix raised to a certain power using diagonalization.
My original Matrix A is:
$$
begin{bmatrix}
-3 & 2 \
-12 & 7 \
end{bmatrix}
$$
My eigenvector Matrix,P, is:
$$
begin{bmatrix}
cfrac{1}{2} & cfrac{1}{3} \
1 & 1 \
end{bmatrix}
$$



I found the diagonal Matrix, D to be:
$$
begin{bmatrix}
1 & 0 \
0 & 3 \
end{bmatrix}
$$



And P inverse is:



$$
begin{bmatrix}
6 & -2 \
-6 & 3 \
end{bmatrix}
$$
The final formula i got from multiplying
$PD^n P^-1$



is



$$
begin{bmatrix}
3-6^n & -1+3^n \
6- 18^n & -2 + 9^n \
end{bmatrix}
$$



It checks out in the case of $A^1$ but not $A^0$. Where did I go wrong?






matrices eigenvalues-eigenvectors diagonalization




matrices eigenvalues-eigenvectors diagonalization






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edited 2 days ago









Bernard

115k637108




115k637108










asked 2 days ago









Dani Jo

41




41












  • bmatrix gives square brackets, pmatrix gives parentheses
    – amWhy
    2 days ago












  • @DaniJo: Your eigenvalues / eigenvectors are correct. However, check your multiplication, for example $-2 times 3^n ne -6^n$.
    – Moo
    2 days ago












  • $P, P^{-1},D$ are all fine. The problem is in the last bit.
    – Henno Brandsma
    2 days ago


















  • bmatrix gives square brackets, pmatrix gives parentheses
    – amWhy
    2 days ago












  • @DaniJo: Your eigenvalues / eigenvectors are correct. However, check your multiplication, for example $-2 times 3^n ne -6^n$.
    – Moo
    2 days ago












  • $P, P^{-1},D$ are all fine. The problem is in the last bit.
    – Henno Brandsma
    2 days ago
















bmatrix gives square brackets, pmatrix gives parentheses
– amWhy
2 days ago






bmatrix gives square brackets, pmatrix gives parentheses
– amWhy
2 days ago














@DaniJo: Your eigenvalues / eigenvectors are correct. However, check your multiplication, for example $-2 times 3^n ne -6^n$.
– Moo
2 days ago






@DaniJo: Your eigenvalues / eigenvectors are correct. However, check your multiplication, for example $-2 times 3^n ne -6^n$.
– Moo
2 days ago














$P, P^{-1},D$ are all fine. The problem is in the last bit.
– Henno Brandsma
2 days ago




$P, P^{-1},D$ are all fine. The problem is in the last bit.
– Henno Brandsma
2 days ago










2 Answers
2






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up vote
1
down vote













Your calculations for $PD^nP^{-1}$ are wrong. Check them.



Remember that, for instance, $2cdot 3^n neq 6^n$.






share|cite|improve this answer




























    up vote
    0
    down vote













    You can simplify the computation by choosing the matrix
    $$
    P=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
    $$
    so that
    $$
    P^{-1}=begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
    $$
    Therefore, with more accurate computations,
    begin{align}
    Pbegin{bmatrix} 1^n & 0 \ 0 & 3^n end{bmatrix}P^{-1}
    &=Pbegin{bmatrix} 1 & 0 \ 0 & 3^n end{bmatrix}
    begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
    \[6px]
    &=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
    begin{bmatrix} 3 &-1 \ -2cdot3^n & 3^n end{bmatrix}
    \[6px]
    &=begin{bmatrix} 3-2cdot3^n & 3^n-1 \ 6(1-3^{n}) & 3^{n+1}-2 end{bmatrix}
    end{align}






    share|cite|improve this answer





















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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

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      up vote
      1
      down vote













      Your calculations for $PD^nP^{-1}$ are wrong. Check them.



      Remember that, for instance, $2cdot 3^n neq 6^n$.






      share|cite|improve this answer

























        up vote
        1
        down vote













        Your calculations for $PD^nP^{-1}$ are wrong. Check them.



        Remember that, for instance, $2cdot 3^n neq 6^n$.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          Your calculations for $PD^nP^{-1}$ are wrong. Check them.



          Remember that, for instance, $2cdot 3^n neq 6^n$.






          share|cite|improve this answer












          Your calculations for $PD^nP^{-1}$ are wrong. Check them.



          Remember that, for instance, $2cdot 3^n neq 6^n$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Alejandro Nasif Salum

          3,629117




          3,629117






















              up vote
              0
              down vote













              You can simplify the computation by choosing the matrix
              $$
              P=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
              $$
              so that
              $$
              P^{-1}=begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
              $$
              Therefore, with more accurate computations,
              begin{align}
              Pbegin{bmatrix} 1^n & 0 \ 0 & 3^n end{bmatrix}P^{-1}
              &=Pbegin{bmatrix} 1 & 0 \ 0 & 3^n end{bmatrix}
              begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
              \[6px]
              &=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
              begin{bmatrix} 3 &-1 \ -2cdot3^n & 3^n end{bmatrix}
              \[6px]
              &=begin{bmatrix} 3-2cdot3^n & 3^n-1 \ 6(1-3^{n}) & 3^{n+1}-2 end{bmatrix}
              end{align}






              share|cite|improve this answer

























                up vote
                0
                down vote













                You can simplify the computation by choosing the matrix
                $$
                P=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
                $$
                so that
                $$
                P^{-1}=begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
                $$
                Therefore, with more accurate computations,
                begin{align}
                Pbegin{bmatrix} 1^n & 0 \ 0 & 3^n end{bmatrix}P^{-1}
                &=Pbegin{bmatrix} 1 & 0 \ 0 & 3^n end{bmatrix}
                begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
                \[6px]
                &=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
                begin{bmatrix} 3 &-1 \ -2cdot3^n & 3^n end{bmatrix}
                \[6px]
                &=begin{bmatrix} 3-2cdot3^n & 3^n-1 \ 6(1-3^{n}) & 3^{n+1}-2 end{bmatrix}
                end{align}






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You can simplify the computation by choosing the matrix
                  $$
                  P=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
                  $$
                  so that
                  $$
                  P^{-1}=begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
                  $$
                  Therefore, with more accurate computations,
                  begin{align}
                  Pbegin{bmatrix} 1^n & 0 \ 0 & 3^n end{bmatrix}P^{-1}
                  &=Pbegin{bmatrix} 1 & 0 \ 0 & 3^n end{bmatrix}
                  begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
                  \[6px]
                  &=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
                  begin{bmatrix} 3 &-1 \ -2cdot3^n & 3^n end{bmatrix}
                  \[6px]
                  &=begin{bmatrix} 3-2cdot3^n & 3^n-1 \ 6(1-3^{n}) & 3^{n+1}-2 end{bmatrix}
                  end{align}






                  share|cite|improve this answer












                  You can simplify the computation by choosing the matrix
                  $$
                  P=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
                  $$
                  so that
                  $$
                  P^{-1}=begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
                  $$
                  Therefore, with more accurate computations,
                  begin{align}
                  Pbegin{bmatrix} 1^n & 0 \ 0 & 3^n end{bmatrix}P^{-1}
                  &=Pbegin{bmatrix} 1 & 0 \ 0 & 3^n end{bmatrix}
                  begin{bmatrix} 3 & -1 \ -2 & 1 end{bmatrix}
                  \[6px]
                  &=begin{bmatrix} 1 & 1 \ 2 & 3 end{bmatrix}
                  begin{bmatrix} 3 &-1 \ -2cdot3^n & 3^n end{bmatrix}
                  \[6px]
                  &=begin{bmatrix} 3-2cdot3^n & 3^n-1 \ 6(1-3^{n}) & 3^{n+1}-2 end{bmatrix}
                  end{align}







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  egreg

                  173k1383198




                  173k1383198






























                       

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