I am given a matrix $N$ × $N$ symmetric positive definite matrix $B_N$
$B_N = \begin{bmatrix} 2 & -1 & 0 & 0 & \cdots & \cdots\\ -1 & 2 & -1 & 0 & \cdots & \cdots\\ \vdots & \vdots & \ddots & \cdots & \cdots & \cdots \\ \vdots & \vdots & \vdots & \ddots & \cdots & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots & -1 \\ \vdots & \vdots & \vdots & \vdots & -1 & 2 \\ \end{bmatrix}$
$C_N$ is a $N \times (N+1)$ matrix with entries equal to $1$ on the main diagonal and equal to $−1$ on the upper diagonal, and with all the other entries equal to $0$ i.e.
$C_N = \begin{bmatrix} 1 & -1 & 0 & 0 & \cdots & \cdots & \cdots\\ 0 & 1 & -1 & 0 & \cdots & \cdots & \cdots\\ 0 & 0 & 1 & -1 & \cdots & \cdots & \cdots\\ \vdots & \vdots & \vdots & \ddots & \cdots & \cdots & \cdots\\ \vdots & \vdots & \vdots & \vdots & \ddots & -1 & \cdots\\ \vdots & \vdots & \vdots & \vdots & \vdots & 1 & -1 \end{bmatrix}$
I have to show that $B_N = C_N C_N^T$.
I am badly stuck. Any idea how to approach this question.
u =
1 -1 0
0 1 -1
octave-3.2.4.exe:54> transpose(u)*u
ans = 1 -1 0
-1 2 -1
0 -1 1
– 2012-03-14