Calculate the product $A^{10}v$ where $A$ is a $2 times 2$ matrix and $v$ is a vector $[4 ; 4]^T$ [duplicate]












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  • Finding a 2x2 Matrix raised to the power of 1000

    6 answers




Following on from the title, can someone suggest how to proceed with this one.



$$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



and



$$v = [4 ; 4]^T?$$






linear-algebra matrix-equations







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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • One way could be to multiply $A$ by itself nine times then by $v$ from the right. It is pretty straightforward.
    – A.Γ.
    2 days ago


















0















This question already has an answer here:




  • Finding a 2x2 Matrix raised to the power of 1000

    6 answers




Following on from the title, can someone suggest how to proceed with this one.



$$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



and



$$v = [4 ; 4]^T?$$






linear-algebra matrix-equations







share|cite|improve this question















marked as duplicate by amd, stressed out, amWhy linear-algebra
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 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.















  • One way could be to multiply $A$ by itself nine times then by $v$ from the right. It is pretty straightforward.
    – A.Γ.
    2 days ago
















0












0








0








This question already has an answer here:




  • Finding a 2x2 Matrix raised to the power of 1000

    6 answers




Following on from the title, can someone suggest how to proceed with this one.



$$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



and



$$v = [4 ; 4]^T?$$






linear-algebra matrix-equations







share|cite|improve this question
















This question already has an answer here:




  • Finding a 2x2 Matrix raised to the power of 1000

    6 answers




Following on from the title, can someone suggest how to proceed with this one.



$$A=begin{bmatrix}1&1\4&1end{bmatrix}$$



and



$$v = [4 ; 4]^T?$$





This question already has an answer here:




  • Finding a 2x2 Matrix raised to the power of 1000

    6 answers







linear-algebra matrix-equations




linear-algebra matrix-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited 2 days ago









A.Γ.

22.6k32656




22.6k32656










asked 2 days ago









RocketKangaroo

334




334




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 2 days ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • One way could be to multiply $A$ by itself nine times then by $v$ from the right. It is pretty straightforward.
    – A.Γ.
    2 days ago




















  • One way could be to multiply $A$ by itself nine times then by $v$ from the right. It is pretty straightforward.
    – A.Γ.
    2 days ago


















One way could be to multiply $A$ by itself nine times then by $v$ from the right. It is pretty straightforward.
– A.Γ.
2 days ago






One way could be to multiply $A$ by itself nine times then by $v$ from the right. It is pretty straightforward.
– A.Γ.
2 days ago












1 Answer
1






active

oldest

votes


















2














You can




  • calculate $Av$, and then apply $A$ to the result nine times;


  • or, since the eigenvalues of $A$ are distinct, $A$ is diagonalizable: there exist $P$ invertible and $D$ diagonal with $A=PDP^{-1}$. Then $A^{10}=PD^{10}P^{-1}$, where $D^{10}$ is very easy since it's diagonal.







share|cite|improve this answer

















  • 2




    That sounds very inefficient. Each time you multiply two $2times 2$ matrices you are doing 8 products and four sums; so to calculate $A^{10}$ you are doing $9times 8+2=74$ products and $9times 4+1=37$ sums, whereas if you apply $A$ to $v$ and keep doing it, each time you are doing 2 products and one sum, for a total of $10times 2=20$ products and $10times 1=10$ sums.
    – Martin Argerami
    2 days ago






  • 1




    In general, yes. But at this stage, with the 10th power, when you compare with finding the two eigenvalues, two corresponding eigenvectors, then inverting the matrix with the eigenvectors and columns, and then doing the two products in $PD^{10}P^{-1}$, "brute force" is probably less or similar work. It won't as soon as you increase powers.
    – Martin Argerami
    2 days ago








  • 1




    Yes, you are right. Contrary to what @stressedout seems to think, algorithms are not my thing :D
    – Martin Argerami
    2 days ago








  • 1




    If you compute $A^2$ and then $A^4=(A^2)^2$, you can then compute $A^2v$ followed by $A^4(A^2v)$ followed by $A^4(A^4(A^2v))$, with a total of $8+8+4+4+4=28$ multiplications and $4+4+4+2+2=14$ additions.
    – Barry Cipra
    2 days ago






  • 1




    I love how everyone is calculating the number of additions and multiplications necessary to solve this like it is actually an important problem. :P @Martin I never thought algorithms were your thing. But I said that I should never design them. :D
    – stressed out
    2 days ago


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














You can




  • calculate $Av$, and then apply $A$ to the result nine times;


  • or, since the eigenvalues of $A$ are distinct, $A$ is diagonalizable: there exist $P$ invertible and $D$ diagonal with $A=PDP^{-1}$. Then $A^{10}=PD^{10}P^{-1}$, where $D^{10}$ is very easy since it's diagonal.







share|cite|improve this answer

















  • 2




    That sounds very inefficient. Each time you multiply two $2times 2$ matrices you are doing 8 products and four sums; so to calculate $A^{10}$ you are doing $9times 8+2=74$ products and $9times 4+1=37$ sums, whereas if you apply $A$ to $v$ and keep doing it, each time you are doing 2 products and one sum, for a total of $10times 2=20$ products and $10times 1=10$ sums.
    – Martin Argerami
    2 days ago






  • 1




    In general, yes. But at this stage, with the 10th power, when you compare with finding the two eigenvalues, two corresponding eigenvectors, then inverting the matrix with the eigenvectors and columns, and then doing the two products in $PD^{10}P^{-1}$, "brute force" is probably less or similar work. It won't as soon as you increase powers.
    – Martin Argerami
    2 days ago








  • 1




    Yes, you are right. Contrary to what @stressedout seems to think, algorithms are not my thing :D
    – Martin Argerami
    2 days ago








  • 1




    If you compute $A^2$ and then $A^4=(A^2)^2$, you can then compute $A^2v$ followed by $A^4(A^2v)$ followed by $A^4(A^4(A^2v))$, with a total of $8+8+4+4+4=28$ multiplications and $4+4+4+2+2=14$ additions.
    – Barry Cipra
    2 days ago






  • 1




    I love how everyone is calculating the number of additions and multiplications necessary to solve this like it is actually an important problem. :P @Martin I never thought algorithms were your thing. But I said that I should never design them. :D
    – stressed out
    2 days ago
















2














You can




  • calculate $Av$, and then apply $A$ to the result nine times;


  • or, since the eigenvalues of $A$ are distinct, $A$ is diagonalizable: there exist $P$ invertible and $D$ diagonal with $A=PDP^{-1}$. Then $A^{10}=PD^{10}P^{-1}$, where $D^{10}$ is very easy since it's diagonal.







share|cite|improve this answer

















  • 2




    That sounds very inefficient. Each time you multiply two $2times 2$ matrices you are doing 8 products and four sums; so to calculate $A^{10}$ you are doing $9times 8+2=74$ products and $9times 4+1=37$ sums, whereas if you apply $A$ to $v$ and keep doing it, each time you are doing 2 products and one sum, for a total of $10times 2=20$ products and $10times 1=10$ sums.
    – Martin Argerami
    2 days ago






  • 1




    In general, yes. But at this stage, with the 10th power, when you compare with finding the two eigenvalues, two corresponding eigenvectors, then inverting the matrix with the eigenvectors and columns, and then doing the two products in $PD^{10}P^{-1}$, "brute force" is probably less or similar work. It won't as soon as you increase powers.
    – Martin Argerami
    2 days ago








  • 1




    Yes, you are right. Contrary to what @stressedout seems to think, algorithms are not my thing :D
    – Martin Argerami
    2 days ago








  • 1




    If you compute $A^2$ and then $A^4=(A^2)^2$, you can then compute $A^2v$ followed by $A^4(A^2v)$ followed by $A^4(A^4(A^2v))$, with a total of $8+8+4+4+4=28$ multiplications and $4+4+4+2+2=14$ additions.
    – Barry Cipra
    2 days ago






  • 1




    I love how everyone is calculating the number of additions and multiplications necessary to solve this like it is actually an important problem. :P @Martin I never thought algorithms were your thing. But I said that I should never design them. :D
    – stressed out
    2 days ago














2












2








2






You can




  • calculate $Av$, and then apply $A$ to the result nine times;


  • or, since the eigenvalues of $A$ are distinct, $A$ is diagonalizable: there exist $P$ invertible and $D$ diagonal with $A=PDP^{-1}$. Then $A^{10}=PD^{10}P^{-1}$, where $D^{10}$ is very easy since it's diagonal.







share|cite|improve this answer












You can




  • calculate $Av$, and then apply $A$ to the result nine times;


  • or, since the eigenvalues of $A$ are distinct, $A$ is diagonalizable: there exist $P$ invertible and $D$ diagonal with $A=PDP^{-1}$. Then $A^{10}=PD^{10}P^{-1}$, where $D^{10}$ is very easy since it's diagonal.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









Martin Argerami

124k1176174




124k1176174








  • 2




    That sounds very inefficient. Each time you multiply two $2times 2$ matrices you are doing 8 products and four sums; so to calculate $A^{10}$ you are doing $9times 8+2=74$ products and $9times 4+1=37$ sums, whereas if you apply $A$ to $v$ and keep doing it, each time you are doing 2 products and one sum, for a total of $10times 2=20$ products and $10times 1=10$ sums.
    – Martin Argerami
    2 days ago






  • 1




    In general, yes. But at this stage, with the 10th power, when you compare with finding the two eigenvalues, two corresponding eigenvectors, then inverting the matrix with the eigenvectors and columns, and then doing the two products in $PD^{10}P^{-1}$, "brute force" is probably less or similar work. It won't as soon as you increase powers.
    – Martin Argerami
    2 days ago








  • 1




    Yes, you are right. Contrary to what @stressedout seems to think, algorithms are not my thing :D
    – Martin Argerami
    2 days ago








  • 1




    If you compute $A^2$ and then $A^4=(A^2)^2$, you can then compute $A^2v$ followed by $A^4(A^2v)$ followed by $A^4(A^4(A^2v))$, with a total of $8+8+4+4+4=28$ multiplications and $4+4+4+2+2=14$ additions.
    – Barry Cipra
    2 days ago






  • 1




    I love how everyone is calculating the number of additions and multiplications necessary to solve this like it is actually an important problem. :P @Martin I never thought algorithms were your thing. But I said that I should never design them. :D
    – stressed out
    2 days ago














  • 2




    That sounds very inefficient. Each time you multiply two $2times 2$ matrices you are doing 8 products and four sums; so to calculate $A^{10}$ you are doing $9times 8+2=74$ products and $9times 4+1=37$ sums, whereas if you apply $A$ to $v$ and keep doing it, each time you are doing 2 products and one sum, for a total of $10times 2=20$ products and $10times 1=10$ sums.
    – Martin Argerami
    2 days ago






  • 1




    In general, yes. But at this stage, with the 10th power, when you compare with finding the two eigenvalues, two corresponding eigenvectors, then inverting the matrix with the eigenvectors and columns, and then doing the two products in $PD^{10}P^{-1}$, "brute force" is probably less or similar work. It won't as soon as you increase powers.
    – Martin Argerami
    2 days ago








  • 1




    Yes, you are right. Contrary to what @stressedout seems to think, algorithms are not my thing :D
    – Martin Argerami
    2 days ago








  • 1




    If you compute $A^2$ and then $A^4=(A^2)^2$, you can then compute $A^2v$ followed by $A^4(A^2v)$ followed by $A^4(A^4(A^2v))$, with a total of $8+8+4+4+4=28$ multiplications and $4+4+4+2+2=14$ additions.
    – Barry Cipra
    2 days ago






  • 1




    I love how everyone is calculating the number of additions and multiplications necessary to solve this like it is actually an important problem. :P @Martin I never thought algorithms were your thing. But I said that I should never design them. :D
    – stressed out
    2 days ago








2




2




That sounds very inefficient. Each time you multiply two $2times 2$ matrices you are doing 8 products and four sums; so to calculate $A^{10}$ you are doing $9times 8+2=74$ products and $9times 4+1=37$ sums, whereas if you apply $A$ to $v$ and keep doing it, each time you are doing 2 products and one sum, for a total of $10times 2=20$ products and $10times 1=10$ sums.
– Martin Argerami
2 days ago




That sounds very inefficient. Each time you multiply two $2times 2$ matrices you are doing 8 products and four sums; so to calculate $A^{10}$ you are doing $9times 8+2=74$ products and $9times 4+1=37$ sums, whereas if you apply $A$ to $v$ and keep doing it, each time you are doing 2 products and one sum, for a total of $10times 2=20$ products and $10times 1=10$ sums.
– Martin Argerami
2 days ago




1




1




In general, yes. But at this stage, with the 10th power, when you compare with finding the two eigenvalues, two corresponding eigenvectors, then inverting the matrix with the eigenvectors and columns, and then doing the two products in $PD^{10}P^{-1}$, "brute force" is probably less or similar work. It won't as soon as you increase powers.
– Martin Argerami
2 days ago






In general, yes. But at this stage, with the 10th power, when you compare with finding the two eigenvalues, two corresponding eigenvectors, then inverting the matrix with the eigenvectors and columns, and then doing the two products in $PD^{10}P^{-1}$, "brute force" is probably less or similar work. It won't as soon as you increase powers.
– Martin Argerami
2 days ago






1




1




Yes, you are right. Contrary to what @stressedout seems to think, algorithms are not my thing :D
– Martin Argerami
2 days ago






Yes, you are right. Contrary to what @stressedout seems to think, algorithms are not my thing :D
– Martin Argerami
2 days ago






1




1




If you compute $A^2$ and then $A^4=(A^2)^2$, you can then compute $A^2v$ followed by $A^4(A^2v)$ followed by $A^4(A^4(A^2v))$, with a total of $8+8+4+4+4=28$ multiplications and $4+4+4+2+2=14$ additions.
– Barry Cipra
2 days ago




If you compute $A^2$ and then $A^4=(A^2)^2$, you can then compute $A^2v$ followed by $A^4(A^2v)$ followed by $A^4(A^4(A^2v))$, with a total of $8+8+4+4+4=28$ multiplications and $4+4+4+2+2=14$ additions.
– Barry Cipra
2 days ago




1




1




I love how everyone is calculating the number of additions and multiplications necessary to solve this like it is actually an important problem. :P @Martin I never thought algorithms were your thing. But I said that I should never design them. :D
– stressed out
2 days ago




I love how everyone is calculating the number of additions and multiplications necessary to solve this like it is actually an important problem. :P @Martin I never thought algorithms were your thing. But I said that I should never design them. :D
– stressed out
2 days ago



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