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A) For which $a,b$ is the matrix $A=\begin{bmatrix} a & 0\\ b & b \end{bmatrix}$ invertible?

B) Calculate $A^{1000}$ where $A$ is the above matrix with $a=1$ and $b=2$.

$$$$

I have done the following:

A) $$\begin{vmatrix} a & 0\\ b & b \end{vmatrix} \neq 0 \Rightarrow ab \neq 0 \Rightarrow a \neq 0 \text{ and } b \neq 0$$

B) $$A=\begin{bmatrix} 1 & 0\\ 2 & 2 \end{bmatrix}$$ How can I calculate $A^{1000}$??

Is it maybe $$A^n=\begin{bmatrix} 1 & 0\\ 2 \cdot n & 2^n \end{bmatrix}$$

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1  
why do you think it is that? if you tried say for 3 cases and think that is the case, prove it by induction. –  Lost1 Jun 29 at 15:53
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wow a billion answers to an easy question... –  Lost1 Jun 29 at 16:02

3 Answers 3

Your matrix is diagonalizable, because u have 2 different eigenvalues $\lambda_1=1$ and $\lambda_2=2$. Then you find an invertible matrix $S$, such that $SAS^{-1}=D$ with a diagonal matrix D, which contains your eingenvalues:

Now we obtain the following:

$A^{1000}=(S^{-1}DS)^{1000}=S^{-1}DS*S^{-1}DS*....*S^{-1}DS=S^{-1}D^{1000}S$ (Because $S^{-1}S=I$).

But $D^{1000}=\begin{pmatrix} 1 & 0 \\ 0 & 2^{1000} \\ \end{pmatrix}$. Then it follows that : $A^{1000}=S^{-1}\begin{pmatrix} 1 & 0 \\ 0 & 2^{1000} \\ \end{pmatrix} S$

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it is certainly not diagonalisable, because it is already in (a possible) Jordan normal form and clearly it is not diagonal. –  Lost1 Jun 29 at 15:57
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Thats wrong! If it where in jordan normalform, then there where 2 times "1". It is diagonalizable. See Wolfram Alpha: wolframalpha.com/input/…;. The fact that is has 2 different eigenvalues is sufficient. –  Mesih Jun 29 at 16:00
    
point! (btw i didnt give the downvote), somebody needs to revise linear algebra –  Lost1 Jun 29 at 16:01

You may prove by induction that

$A^n=\begin{bmatrix} 1 & 0\\ 2^{n+1}-2 & 2^n \end{bmatrix}$

but that's out of the blue.


Otherwise, notice that $A=I_2+\begin{bmatrix} 0 & 0\\ 2 & 1 \end{bmatrix}$

Notice that $\forall n, \begin{bmatrix} 0 & 0\\ 2 & 1 \end{bmatrix}^n=\begin{bmatrix} 0 & 0\\ 2 & 1 \end{bmatrix}$

And apply Newton binomial theorem (you're allowed to since $I_2$ commutes with any matrix).

$A^{1000}=I_2+ \begin{bmatrix} 0 & 0\\ 2 & 1 \end{bmatrix}\sum_{k=1}^{1000} \binom{1000}{k}$

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A) looks good

B) Your approach to find a formula for $A^n$ is good. Try to prove by induction that$$A^n=\begin{bmatrix} 1 & 0\\ \sum_{i=1}^{n}2^i & 2^n \end{bmatrix}$$

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