a solution for $Lx=b$ where $L$ is a non singular lower triangular matrix 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?
 A: 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.
A: 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.
