if $C=BAB'(B' $transpose of $B)$ then find $B(C^{100})B'$

Question

Let $$A=\begin {bmatrix}1&1\\0&1\\ \end {bmatrix} $$ and

$$B= \begin {bmatrix}\frac{3}{2}&\frac{1}{2}\\\frac{-1}{2}&\frac{3}{2}\\ \end {bmatrix}$$

now if $$C=BAB'$$ $(B'$ transpose of $B)$ then find $B(C^{100})B'$.

pls tell me what is the trick behind to find $C^{100}$

Answer 1

With

$B= \begin {bmatrix} \dfrac{3}{2} & \dfrac{1}{2}\\ -\dfrac{1}{2} & \dfrac{3}{2}\\ \end {bmatrix}, \tag 1$

we have

$B^T = \begin {bmatrix} \dfrac{3}{2} & -\dfrac{1}{2}\\ \dfrac{1}{2} & \dfrac{3}{2}\\ \end {bmatrix}, \tag 2$

whence

$B^T B = \begin {bmatrix} \dfrac{3}{2} & -\dfrac{1}{2} \\ \dfrac{1}{2} & \dfrac{3}{2}\\ \end {bmatrix} \begin {bmatrix} \dfrac{3}{2} & \dfrac{1}{2}\\ -\dfrac{1}{2} & \dfrac{3}{2}\\ \end {bmatrix} = \begin {bmatrix} \dfrac{5}{2} & 0 \\ 0 & \dfrac{5}{2}\\ \end {bmatrix} = \dfrac{5}{2}I, \tag 3$

and also,

$BB^T = \dfrac{5}{2}I, \tag 4$

which follows from (3) by first right multiplying by $B^{-1}$ and then left multiplying by $B$. With

$C = BAB^T, \tag 5$

we have

$C^2 = BAB^TBAB^T = BA \left (\dfrac{5}{2}I \right )AB^T = \dfrac{5}{2}BA^2B^T, \tag 6$

whence

$C^3 = \dfrac{5}{2}BA^2B^T BAB^T = \dfrac{5}{2} BA^2 \left (\dfrac{5}{2}I\right) AB^T = \left (\dfrac{5}{2} \right )^2 BA^3B^T;\tag 7$

based on (6) and (7) we conjecture that the general pattern is

$C^n = \left (\dfrac{5}{2} \right )^{n - 1} BA^nB^T,\tag 8$

which we prove by induction: with

$C^k = \left (\dfrac{5}{2} \right )^{k - 1} BA^kB^T,\tag 9$

we have

$C^{k + 1} = C^kC = \left (\dfrac{5}{2} \right )^{k - 1} BA^kB^T BAB^T$ $= \left (\dfrac{5}{2} \right )^{k - 1} BA^k \left (\dfrac{5}{2}I \right )AB^T = \left (\dfrac{5}{2} \right )^k BA^{k + 1}B^T,\tag{10}$

whence (8) binds for all $n \ge 1$; to compute $A^n$ we note that

$A = I + N, \tag{11}$

where

$N = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}; \tag{12}$

we see that

$N^2 = 0; \tag{13}$

since $NI = IN$ we may apply the bionomial theorem to

$A^n = (I + N)^n; \tag{14}$

since $N^m = 0$ for $m \ge 2$, we only retain the first two terms of the expansion:

$A^n = (I + N)^n = I + nN = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}; \tag{15}$

thus, from (8),

$C^n = \left (\dfrac{5}{2} \right )^{n - 1} B(I + nN)B^T$ $= \left (\dfrac{5}{2} \right )^{n - 1} (BB^T + nBNB^T) = \left (\dfrac{5}{2} \right )^{n - 1}\left ( \dfrac{5}{2}I + nBNB^T \right ); \tag {16}$

to finish the calculation we need

$BNB^T = \begin {bmatrix} \dfrac{3}{2} & \dfrac{1}{2}\\ -\dfrac{1}{2} & \dfrac{3}{2}\\ \end {bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin {bmatrix} \dfrac{3}{2} & -\dfrac{1}{2}\\ \dfrac{1}{2} & \dfrac{3}{2}\\ \end {bmatrix}$ $= \begin {bmatrix} \dfrac{3}{2} & \dfrac{1}{2}\\ -\dfrac{1}{2} & \dfrac{3}{2}\\ \end {bmatrix} \begin{bmatrix} \dfrac{1}{2} & \dfrac{3}{2} \\0 & 0 \end{bmatrix} = \begin{bmatrix} \dfrac{3}{4} & \dfrac{9}{4} \\ -\dfrac{1}{4} & -\dfrac{3}{4} \end{bmatrix}; \tag{17}$

we can assemble these findings to conclude that

$C^n = \left (\dfrac{5}{2} \right )^nI + \left (\dfrac{5}{2} \right )^{n - 1} n \begin{bmatrix} \dfrac{3}{4} & \dfrac{9}{4} \\ -\dfrac{1}{4} & -\dfrac{3}{4} \end{bmatrix}$ $= \left ( \dfrac{5}{2} \right )^{n - 1} \left (\begin{bmatrix} \dfrac{5}{2} & 0 \\ 0 & \dfrac{5}{2} \end{bmatrix} + \begin{bmatrix} \dfrac{3}{4}n & \dfrac{9}{4}n \\ -\dfrac{1}{4}n & -\dfrac{3}{4}n \end{bmatrix} \right ); \tag{18}$

finally,

$C^n = \left ( \dfrac{5}{2} \right )^{n - 1} \begin{bmatrix} \dfrac{5}{2} + \dfrac{3}{4}n & \dfrac{9}{4}n \\ -\dfrac{1}{4}n & \dfrac{5}{2} - \dfrac{3}{4}n \end{bmatrix}. \tag{20}$

Answer 2

Hint: calculate explicitly $BB'$ - you will get that is is $\alpha I$ where $\alpha$ is a scalar and $I$ is the identity matrix. Then use $C^2=BA(B'B)AB'=BA(\alpha I)AB'=\alpha\cdot BA^2B'$. What happens for other powers?

Answer 3

Note that $det(B)= 5/2 $

$$B= \begin {bmatrix}\frac{3}{2}&\frac{1}{2}\\\frac{-1}{2}&\frac{3}{2}\\ \end {bmatrix}\implies B'=\begin {bmatrix}\frac{3}{2}&\frac{-1}{2}\\\frac{1}{2}&\frac{3}{2}\\ \end{bmatrix}= (5/2)B^{-1} $$

$$C=BAB'= (5/2)BAB^{-1} \implies C^{100}=(5/2)^{100}BA^{100}B^{-1}$$