2
$\begingroup$

In the questions compute $\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6$ and $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$, how would you solve these?

  • 0
    Are you starting a course in linear algebra? – 2017-02-17
  • 0
    For the one on the left, have you learned how to find the eigenvalues of a matrix? For the one on the right, I suggest you start multiplying the first few iterations, and look for a pattern. – 2017-02-17
  • 0
    I don't know eigenvalues :( – 2017-02-17
  • 3
    Then for the one on the left brute force multiply it 3 times (it is easier than it looks). You will get something surprisingly nice. Now square that. – 2017-02-17

3 Answers 3

4

Since first part is answered by Doug M , for the second bit we can approach by the method of induction.

We consider this matrix ,$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n}$

Lets check for n=2, $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$.

Lets check for n=3, $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}=\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$.

I think we got a pattern! So, our hypthesis is $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{n} = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ .To prove our hypothesis we use first principle of mathematical induction.

Let us assume that this form is true for $n = k$ that is multiplying $k$ times which gives us $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k} =\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ Now if we prove it for $n=k+1$ then it's true for all $n \geq 1$ So, consider $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ Now $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$ from our assumption, so $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{k+1} = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}.\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1) \\ 0 & 1 \end{pmatrix}$ from our first case. Hence this holds for any $n \geq 1$,so as a particular case of yours,for $n=99$,this case also holds that is $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99} = \begin{pmatrix} 1 & 99 \\ 0 & 1 \end{pmatrix}$.

Hope this helps!

  • 0
    Please check your work for the square of the matrix. I think the entry in the upper right corner is wrong. – 2017-02-17
  • 0
    you are right i will edit this – 2017-02-17
  • 0
    did an edit @EthanBolker. – 2017-02-17
1

$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}= 2 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6} \\ \sin\frac {\pi}{6} & \cos \frac {\pi}{6} \end{pmatrix}$

$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6= 2^6 \begin{pmatrix} \cos \frac {\pi}{6} & -\sin\frac {\pi}{6} \\ \sin\frac {\pi}{6} & \cos \frac {\pi}{6} \end{pmatrix}^6$

It is worth the exercise to see what happens when you multiply matrices that can be put into this form.

0

The complex numbers are isomorphic to the set of matrices of the form $$ a+bi \sim \begin{pmatrix} a & -b \\ b & a \end{pmatrix} =a I + b J, \text{ where } J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, J^2=-I $$ Your first matrix corresponds to $z=\sqrt3 +i$.

Now $\dfrac{iz}{2}=\dfrac{-1+\sqrt3}{2}$ is a third root of unity and so $\left(\dfrac{iz}{2}\right)^6=1$.

Therefore $z^6=-64$ and the corresponding matrix is $-64I$.

The second matrix can be written $A=I+N$ where $N^2=0$.

Therefore, $A^k=I+kN$ by the binomial theorem.