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+21 votes

Let A be the $ 2 × 2 $ matrix with elements $a_{11} = a_{12} = a_{21} = +1 $ and $ a_{22} = −1 $ . Then the eigenvalues of the matrix $A^{19}$ are

- $1024$ and $−1024$

- $1024\sqrt{2}$ and $−1024 \sqrt{2}$

- $4 \sqrt{2}$ and $−4 \sqrt{2}$

- $512 \sqrt{2}$ and $−512 \sqrt{2}$

+46 votes

Best answer

Characteristic Equation is $ |A -\mu I| = 0$

$ \begin{vmatrix} 1-\mu & 1\\1 &-1-\mu\end{vmatrix} = 0$

$\implies (1-\mu)(-1-\mu)-1=0$

$\implies -1-\mu+\mu+\mu^{2}-1=0$

$\implies \mu^{2}-2=0$

$\implies \mu=+\sqrt 2 $ and $-\sqrt 2$

According to properties of Eigen values,

eigen values of $A^{19}=$ (eigen value of A)$^{19}$

$=(\sqrt 2)^{19}$ and $(-\sqrt 2)^{19}$

Hence, Ans is **option (D).**

+6 votes

Eigenvalues Of Matrix Powers

Suppose A is a square matrix,λ is an eigenvalue of A, and s≥0 is an integer. Then λ^{s} is an eigenvalue of A^{s}.

0 votes

At the time of solving I just forgot this property & get two method, one method is working fine but other is not,but can't understand why the other method is not working.

**1st Method:**

By using Cayley-hamilton Theorem from the characteristic eqn I get $\lambda ^{2} - 2 = 0$ which is $A^{2} - 2 = 0$

$A^{2} = 2I \Rightarrow ({A^{2}})^9 = 2^{9}I$

i.e $A^{18}=512I$, Now post-multiplied by A on both side $A^{18}.A=512I.A$. which becomes $A^{19}=\begin{bmatrix} 512 &512 \\ 512&-512 \end{bmatrix}$

computing eigen values here we'll get $512\sqrt2$ & $-512\sqrt2$

This method is ok & it's working.

**2nd method:**

finding a pattern to compute A^19 which 1st compute A^2 by A*A , then A^3 by A^2 * A, then A^4 by A^3 * A

by this we'll get a pattern like,

$A^{n} =\begin{bmatrix} n &0 \\ 0 &n \end{bmatrix}$ when n is even

$A^{n} =\begin{bmatrix} n-1 &n-1 \\ n-1 &1-n \end{bmatrix}$ when n is odd

likewise $A^{19} =\begin{bmatrix} 18 &18 \\ 18 &-18 \end{bmatrix}$

but computing eigen values on $A^{19}$ will give $-18\sqrt2$ & $18\sqrt2$

don't know why this 2nd method is not working,Anyone please give a reason why this is not working.

+1

$2^{nd}$ method is not working because your pattern is not correct. Just calculate manually, $A^6$ , $A^{10}$ , , $A^{15}$ etc. your pattern is working for small $n$. if you assume something is correct then check whether it is actually correct or not for every 'n' using induction. Here, using inductive hypothesis, you assume, it is true for 'n', then you have to prove that it will follow the same pattern for n+1, n+2,... then it will be correct.

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