Matrix Powers

  • Thread starter shaon0
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  • #1
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Homework Statement


Let B be a matrix with characteristic polynomial λ2-λ√6+3. Evaluate B4.

Homework Equations


Bn=PDnP-1

The Attempt at a Solution


I can find the eigenvalues from the characteristic equation and those would form the diagonal entries of D. But how would I find P, which contains the eigenvectors, if I don't have a matrix?

Side Note: How can I quickly find an inverse for a 2 by 2 matrix. Is it just dividing the 2x2 matrix by it's determinant, then negating the diagonal entries going from a11 to a22 and swapping a12 and a21?
 

Answers and Replies

  • #2
I like Serena
Homework Helper
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Hi shaon0! :smile:

Homework Statement


Let B be a matrix with characteristic polynomial λ2-λ√6+3. Evaluate B4.

Homework Equations


Bn=PDnP-1

The Attempt at a Solution


I can find the eigenvalues from the characteristic equation and those would form the diagonal entries of D. But how would I find P, which contains the eigenvectors, if I don't have a matrix?

Side Note: How can I quickly find an inverse for a 2 by 2 matrix. Is it just dividing the 2x2 matrix by it's determinant, then negating the diagonal entries going from a11 to a22 and swapping a12 and a21?
Let's first see how far we can get before concluding you have too little information shall we?

What did you find for the eigenvalues?
What do you get if you calculate D4?


On your side note: almost, but you need to swap the diagonal entries a11 and a22, and negate a12 and a21.
Check by multiplying your matrix with the supposed inverse. It should yield the identity matrix.
 
  • #3
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Hi shaon0! :smile:



Let's first see how far we can get before concluding you have too little information shall we?

What did you find for the eigenvalues?
What do you get if you calculate D4?


On your side note: almost, but you need to swap the diagonal entries a11 and a22, and negate a12 and a21.
Check by multiplying your matrix with the supposed inverse. It should yield the identity matrix.
Hi I like Serena :);

I've found the answer. Thanks for the help, needed another to look at the problem.
 
Last edited:
  • #4
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You don't really need to calculate the eigenvalues or [itex]D^4[/itex] for this.

Cayley-Hamilton says that

[tex]B^2=\sqrt{6}B-3I[/tex]

Thus

[tex]B^4=B^2*B^2=B^2(\sqrt{6}B-3I)=\sqrt{6}B^3-3B^2=\sqrt{6}B(\sqrt{6}B-3I)-3(\sqrt{6}B-3I)=...[/tex]
 
  • #5
I like Serena
Homework Helper
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Neat! :smile:

Btw, it's also neat to see how Bn=PDnP-1 cancels out. :cool:
 

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