Given that $L\in\mathbb{R}^{n\times n}$ be a non singular Lower Triangular matrix. suppose $y\in\mathbb{R}^n$ be a solution of the equation $$ Lx=b; b=(0,0,\dots,0,b_{k+1},\dots,b_n)$$ we have to show
$y=(0,0,\dots,0,y_{k+1},y_{k+2},\dots,y_n)$ in this form.
could anyone help me how to show this? I have checked it for a 3 by 3 lower triangular matrix, and it is true, but how can I generalize it?
Hint: Partitioning $L$ into blocks.
what does hint means?
Let
$$L = \begin{bmatrix} A_{k\times k} & 0_{k\times (n-k)}\\ B_{(n-k)\times k} & C_{(n-k)\times (n-k)}\\ \end{bmatrix} $$
Consider first $k$ rows of your linear system. As soon as $L$ is non singular lower triangular matrix, $A$ is non singular lower triangular matrix as well. Then the system
$$L = \begin{bmatrix} A_{k\times k} & 0_{k\times (n-k)}\\ \end{bmatrix}y = 0 $$ is equal to $$L = \begin{bmatrix} A_{k\times k}\\ \end{bmatrix}y_{1:k} = 0 $$ and have only trivial solution $y_{1:k} = 0$, so therefore $k$ first components of $y$ are essentially zero.
We have $Ly = b$, which can be written using matlab notation as $$ \begin{bmatrix} L_{1:k,1:k} & 0\\ L_{k+1:n,1:k} & L_{k+1:n,k+1:n} \end{bmatrix} \begin{bmatrix} y_{1:k}\\ y_{k+1:n} \end{bmatrix} = \begin{bmatrix} 0\\ b_{k+1:n} \end{bmatrix} $$ So we have \begin{align} L_{1:k,1:k}\,y_{1:k} &= 0\\[1mm] L_{k+1:n,1:k}\,y_{1:k} + L_{k+1:n,k+1:n}\,y_{k+1:n} &= b_{k+1:n} \end{align} Since $L$ is nonsingular it follows that $L_{1:k,1:k}$ is nonsingular and so by the first equation $y_{1:k} = 0$. Thus, the second equation reduces to $$ L_{k+1:n,k+1:n}\,y_{k+1:n} = b_{k+1:n} $$ Therefore, elements the first $k$ elements of $y$ must be zero, and the remaining elements, which may be nonzero, are obtained by solving this linear system.