Find matrix B

**Arita** · January 31st 2011, 03:33 PM

B is a 2x2 matrix with the property that there is an invertible 2x2 matrix R such that

R^-1BR = [(sqrt 3) 0 ] [ 1.2 -(sqrt 3)/2]
[ 0 (sqrt 3) ] [(sqrt 3)/2 1/2 ]

I know the first part is the scaling matrix and the second on in the rotation matrix.

I'm suppose to find B^6 which is equal to
B^6 = R^-1D^6R

But I dont know how I would find the matrix for R or D.

**Ackbeet** · January 31st 2011, 03:46 PM

Are you saying that $B$ is a matrix such that there exists invertible $R$ such that

$R^{-1}BR=\begin{bmatrix}\sqrt{3}&0\\0&\sqrt{3}\end{bma trix}<br /> \begin{bmatrix}1/2&-\sqrt{3}/2\\\sqrt{3}/2&1/2\end{bmatrix}?$

And you're asked to find $B^{6}?$

**Arita** · January 31st 2011, 04:14 PM

Yea, that is what we were given and then asked to find B^6

**Ackbeet** · January 31st 2011, 04:53 PM

Hmm. Interesting problem. Well, the nice thing about diagonal matrices is that it's easy to find their nth power: just take the nth power of the elements on the main diagonal. Now the problem you've been given is not a typical diagonalization problem. Presumably, you've seen those already, so you know how you can always do this:

If

$B=R^{-1}DR,$ then

$B^{n}=(R^{-1}DR)^{n}=\underbrace{R^{-1}DRR^{-1}DR\dots R^{-1}DR}_{n\times}=R^{-1}\underbrace{DD\dots D}_{n\times}R=R^{-1}D^{n}R.$

I played some "what if" games with your situation. Let's say you've got

$R^{-1}BR=\begin{bmatrix}\sqrt{3}&0\\0&\sqrt{3}\end{bma trix}<br /> \begin{bmatrix}1/2&-\sqrt{3}/2\\\sqrt{3}/2&1/2\end{bmatrix}=C.$

So, $C$ is known. Then

$B=RCR^{-1},$

and the $R^{-1}$ is on the wrong side of $C.$ We want to have

$B=R^{-1}DR.$ If I compare these two, it leads me to ask this question: what if I could find a diagonal $D$ such that

$C=R^{-2}DR^{2}?$ Then

$B=RCR^{-1}=RR^{-2}DR^{2}R^{-1}=R^{-1}DR,$ as desired.

So, can you diagonalize $C=Q^{-1}DQ$ but then, when you find the invertible matrix $Q,$ can you take its square root? There are ways to do that, I think, that aren't overly difficult, especially if you orthogonally diagonalize $C.$

So, give this program a shot: orthogonally diagonalize $C=Q^{-1}DQ.$ What do you get?

**HallsofIvy** · January 31st 2011, 05:08 PM

Are you sure about that? I recommend you go back and reread the problem. I am inclined to suspect that it is R itself that is equal to $\begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$ and that then $R^{-1}BR= \begin{bmatrix}\sqrt{3} & 0 \\ 0 & \sqrt{3}\end{bmatrix}$
which is then the same as saying that $\begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}B= \bmatrix{\sqrt{3} & 0 \\ 0 & \sqrt{3}\end{bmatrix}\begin{bmatrix}\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}$ .

If that is the case, that you have $R^{-1}BR= D$ , where I am using "D" to stand for that diagonal matrix, then you also have $B= RDR^{-1}$ . Now notice that $B^2= (RDR^{-1})(RDR^{-1})= (RD)(R^{-1}R)(DR^{-1})= (RD)(DR^{-1})= RD^2R^{-1}$
$B^3= B(B^2)= (RDR^{=1})(RD^2R^{-1})= RD^3R^{-1}$ , etc.

Further, since D is a diagonal matrix, it is easy to find powers of D:

$\begin{bmatrix}d & 0 \\ 0 & d\end{bmatrix}^n= \begin{bmatrix}d^n & 0 \\ 0 & d^n\end{bmatrix}$ .

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