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If $$A=\pmatrix{1 &0\\-1&1},$$ show that $$A^2-2A+I_2=0,$$ where $I_{2}$ is the $2x2$ Identity matrix. Hence find $A^{50}$.

We have $$A^2-2A+I_2=A(A-2I_3)+I_=\pmatrix{1 &0\\-1&1}\pmatrix{-1 &0\\-1&-1}+I_2 =-I_2+I_2=0.$$
How can I show the second part?

Bman72
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user181545
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3 Answers3

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Since $(A-I_2)^2 = 0$, by binomial theorem, one has

$$\begin{align}A^{50} &= (I_2 + (A - I_2))^{50}\\ &= I_2 + \binom{50}{1} (A-I_2)^{1} + \binom{50}{2}(A-I_2)^2 + \binom{50}{3}(A-I_2)^3 +\cdots\\ &= I_2 + 50 (A-I_2) + 0 + 0 + \cdots\\ &= \begin{pmatrix}1 & 0\\-50 & 1\end{pmatrix} \end{align} $$

achille hui
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  • +1 This is the neatest way. I was fiddling around with Binomial Theorem, but the insight to write $A = I + (A-I)$ didn't occur to me. By the way, you should remove the subscript of $3$ from the identity matrix since that was an error in the original question (it's meaningless if you think about it). – Deepak Oct 07 '14 at 15:10
  • @Deepak Oops, fixed. – achille hui Oct 07 '14 at 15:14
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The characteristic polynomial of $A$ is

$$\chi_A(t) = \text{det}(A-t\Bbb{1})=t^2-2t+1$$

By the theorem of Caliley Hamilton, you get that $A^2-2A+\Bbb{1}=0$, where $\Bbb{1}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ and $0$ is the zero matrix.

Now we

Claim $$A^n = \begin{pmatrix}1&0\\-n&1\end{pmatrix}$$

Proof

We procede by induction:

If $n=1$ we have $A = \begin{pmatrix}1&0\\-1&0\end{pmatrix}$, that is correct. Assume that the claim is true for $n-1$. We want to prove that it holds also for $n$

\begin{align*} A^n &= A^{n-1}\cdot A\\ &=\begin{pmatrix}1&0\\-(n-1)&1\end{pmatrix}\cdot \begin{pmatrix}1&0\\-1&1\end{pmatrix}\\&=\begin{pmatrix}1&0\\-n+1-1&1\end{pmatrix}=\begin{pmatrix}1&0\\-n&1\end{pmatrix} \end{align*}

Which proves the claim. Now you should be able to evaluate $A^{50}$.

Bman72
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  • I don't quite see how this proof (while good and valid) uses the initially established property that $A^2 -2A + I = 0$. The question does stipulate "Hence..." so it seems that they want you to approach it from that direction. – Deepak Oct 07 '14 at 15:07
  • @Deepak yes you are right. achille hui provided an example of how to use that property, however i think it is good to know how to solve a problem also by different approaches ;). Surely it doesn't hurt. – Bman72 Oct 07 '14 at 15:09
  • Yes, Achille has a nice approach, as I commented under his post. My own answer uses an inductive proof approach like yours, but has the advantage of also using the earlier result, so I guess it might satisfy the marker/examiner better. – Deepak Oct 07 '14 at 15:12
  • @Deepak like i said before i don't think it is bad to have more than an answer. I'm sure the OP is smart enough to choose the one he likes the most. By the way, nice answer yours ;) – Bman72 Oct 07 '14 at 15:15
  • Agreed, and thanks very much. :) – Deepak Oct 07 '14 at 15:29
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You've established that $A^2 -2A + I = 0$

Rearrange, $A^2 = 2A - I$

Let's compute $A^3$:

$A^3 = A^2 A = 2A^2 - IA = 2(2A - I) - A = 3A - 2I$

Similarly, confirm that $A^4 = 4A - 3I$

We can now conjecture that $A^n = nA - (n-1)I, n \geq 2$

We can prove that by induction.

The base case is obvious (work already done for $n=2$).

Assume it's true for some $n=k$, i.e. $A^k = kA - (k-1)I$

Hence $A^{k+1} = A^kA = kA^2 - (k-1)IA = k(2A-I) - (k-1)A = 2kA - kI - kA + A = (k+1)A - kI$

which completes the proof.

Therefore $A^{50} = 50A - 49I = 50\pmatrix{1 &0\\-1&1} - 49\pmatrix{1 &0\\0&1} = \pmatrix{1 &0\\-50&1}$.

Deepak
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