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Diagonalizable matrix

  • Thread starter t_n_p
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  • #1
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Homework Statement



Use diagonalization to find A^9 where A = [0 8;2 0]

The Attempt at a Solution



I perform (A-λI) to find characteristic equation and hence eigenvalues of 4 and -4. These yield eigenvectors of [2;1] and [2;-1] respectively.

Now I know [0 8;2 0][2 2;1 -1] = [2 2;1 1][4 0;0 -4]
Rearranging gives [0 8;2 0] = [2 2;1 -1][4 0;0 -4][2 2;1 -1]^-1

Now to raise to the power of 9 gives, [0 8;2 0]^9 = [2 2;1 -1][4 0;0 -4]^9[2 2;1 -1]^-1

Now I know that any matrix multiplied by its inverse will give the identity matrix, hence this leaves me with [0 8;2 0]^9 = [4 0;0 -4]^9

Therefore by my working, A^9 = [4^9 0; 0 (-4)^9]

The answer suggests [0 2*4^9; .5*4^9 0]

Can someone see where I went wrong?
 

Answers and Replies

  • #2
Defennder
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Now to raise to the power of 9 gives, [0 8;2 0]^9 = [2 2;1 -1][4 0;0 -4]^9[2 2;1 -1]^-1
You're still correct here. Let the matrices on the RHS be labeled P,D^9,P^-1

Now I know that any matrix multiplied by its inverse will give the identity matrix, hence this leaves me with [0 8;2 0]^9 = [4 0;0 -4]^9[/tex]
This is your mistake. You're multiplying PD^9P^-1, not D^9P(P^-1). In matrix multiplication the order matters. You can't randomly switch the order. Just multiply the matrices out normally and you'll get the answer.
 
  • #3
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Ah, so if I have matrices ABC and I wish to perform A*B*C
I must do (A*B)*C.

hmmm, when I did it for a previous question that asked to perform A^6 (where A was the same matrix) I multiplied P and P^-1 to give the inverse and that yielded the correct answer. Just a lucky once off I suppose?
 
  • #4
Defennder
Homework Helper
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Matrix multiplication is associative, meaning ABC = (AB)C = A(BC). But it is not commutative, meaning that ABC not necessarily equals ACB.

I guess you were lucky that time. I didn't try it out and I can't see your working so I can't tell.
 

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