Intereting Posts

Enumerating number of solutions to an equation
Understanding how to evaluate $\lim_{x\to\frac\pi2} \frac{2^{-\cos x}-1}{x-\frac\pi2}$
To compute $\tan1-\tan3+\tan5-\cdots+\tan89$, $\tan1+\tan3+\tan5+\cdots+\tan89$
Giving meaning to $R$ (for example) via the evaluation homomorphism
Critical values and critical points of the mapping $z\mapsto z^2 + \bar{z}$
In $\ell^1$ but not in $\ell^2$?
Formula for midpoint in hyperbolic 3-space
Forcing series convergence
Approximation of Riemann integrable function with a continuous function
for a $3 \times 3$ matrix A ,value of $ A^{50} $ is
Number of rooted subtrees of given size in infinite d-regular tree
Maxima and minima of $f(x)=\binom{16-x}{2x-1}+\binom{20-3x}{4x-5}$.
Explicit value for $\sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{1}+\sqrt{2}+\dots+\sqrt{n}}\right)$
Expected value of size of subset
Does a perfect compact metric space have a closed subset homeomorphic to a countable product of 2 point sets?

The inverse of a non-singular lower triangular matrix is lower triangular.

Construct a proof of this fact as follows. Suppose that $L$ is a non-singular lower triangular matrix. If $b \in \mathbb{R^n}$ is such that $b_i = 0$ for $i = 1, . . . , k \leq n$, and $y$ solves $Ly = b$, then $y_i = 0$ for

$i = 1, . . . , k \leq n$. (Hint: partition $L$ by the first $k$ rows and columns.)

Can someone tell me what exactly we are showing here and why it will prove that the inverse of **any** non-singular lower triangular matrix is lower triangular?

- Jordan canonical form of an upper triangular matrix
- Extending a Chebyshev-polynomial determinant identity
- Zero vector of a vector space
- Is the natural map $L^p(X) \otimes L^p(Y) \to L^p(X \times Y)$ injective?
- The probability that two vectors are linearly independent.
- Sequence of positive integers such that their reciprocals are in arithmetic progression

- nilpotent and linear transformation
- Can you use a logarithm coefficient in a linear equation?
- If a Matrix Has Only Zero as an Eigen-Value Then It Is Nilpotent
- Cross product in complex vector spaces
- Probability of making it across a path of $n$ tiles through random walk
- Find a change in variable that will reduce the quadratic form to a sum of squares
- Can a matrix in $\mathbb{R}$ have a minimal polynomial that has coefficients in $\mathbb{C}$?
- Span of Permutation Matrices
- Change basis so that a positive definite matrix $A$ is now seen as $I$.
- Projection of a vector

Let’s write $$L^{-1}=[y_1\:\cdots\:y_n],$$ where each $y_k$ is an $n\times 1$ matrix.

Now, by definition, $$LL^{-1}=I=[e_1\:\cdots\:e_n],$$ where $e_k$ is the $n\times 1$ matrix with a $1$ in the $k$th row and $0$s everywhere else. Observe, though, that $$LL^{-1}=L[y_1\:\cdots\:y_n]=[Ly_1\:\cdots\: Ly_n],$$ so $$Ly_k=e_k\qquad(1\leq k\leq n)$$

By the proposition, since $e_k$ has only $0$s above the $k$th row and $L$ is lower triangular and $Ly_k=e_k$, then $y_k$ has only $0$s above the $k$th row. This is true for all $1\leq k\leq n$, so since $$L^{-1}=[y_1\:\cdots\:y_n],$$ then $L^{-1}$ is lower triangular, too.

$$********$$

Here’s an alternative (but related) approach.

Observe that a lower triangular matrix is nonsingular if and only if it has all nonzero entries on the diagonal. Let’s proceed by induction on $n$. The base case ($n=1$) is simple, as all scalars are trivially “lower triangular”. Now, let’s suppose that all nonsingular $n\times n$ lower triangular matrices have lower triangular inverses, and let $A$ be any nonsingular $(n+1)\times(n+1)$ lower triangular matrix. In block form, then, we have $$A=\left[\begin{array}{c|c}L & 0_n\\\hline x^T & \alpha\end{array}\right],$$ where $L$ is a nonsingular $n\times n$ lower triangular matrix, $0_n$ is the $n\times 1$ matrix of $0$s, $x$ is some $n\times 1$ matrix, and $\alpha$ is some nonzero scalar. (Can you see why this is true?) Now, in compatible block form, we have $$A^{-1}=\left[\begin{array}{c|c}M & b\\\hline y^T & \beta\end{array}\right],$$ where $M$ is an $n\times n$ matrix, $b,y$ are $n\times 1$ matrices, and $\beta$ some scalar. Letting $I_n$ and $I_{n+1}$ denote the $n\times n$ and $(n+1)\times(n+1)$ identity matrices, respectively, we have $$I_{n+1}=\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right].$$ Hence, $$\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=I_{n+1}=A^{-1}A=\left[\begin{array}{c|c}ML+by^T & M0_n+b\alpha\\\hline x^TM+\alpha y^T & y^T0_n+\beta\alpha\end{array}\right]=\left[\begin{array}{c|c}ML+by^T & \alpha b\\\hline x^TM+\alpha y^T & \beta\alpha\end{array}\right].$$ Since $\alpha$ is a nonzero scalar and $\alpha b=0_n$, then we must have $b=0_n$. Thus, $$A^{-1}=\left[\begin{array}{c|c}M & 0_n\\\hline y^T & \beta\end{array}\right],$$ and $$\left[\begin{array}{c|c}I_n & 0_n\\\hline 0_n^T & 1\end{array}\right]=\left[\begin{array}{c|c}ML & 0_n\\\hline x^TM+\alpha y^T & \beta\alpha\end{array}\right].$$ Since $ML=I_n$, then $M=L^{-1}$, and by inductive hypothesis, we have that $M$ is then lower triangular. Therefore, $$A^{-1}=\left[\begin{array}{c|c}M & 0_n\\\hline y^T & \beta\end{array}\right]$$ is lower triangular, too, as desired.

Suppose you have an invertible lower-triangular matrix $L$. To find its inverse, you must solve the matrix equation $LX = I$, where $I$ denotes the $n$-by-$n$ identity matrix.

Based on how matrix multiplication works, the $i^{\text{th}}$ column of $LX$ is equal to $L$ times the $i^{\text{th}}$ column of $X$. In order for $LX = I$, it must be that the first $i-1$ entries in the $i^{\text{th}}$ column of $LX$ are all zero. The hint is that you can prove that this implies that the first $i-1$ entries in the $i^{\text{th}}$ column of $X$ must all be zero. To do this, you can explicitly write out your calculation, using your assumption that $L$ is lower-triangular. You’ll get a fairly easy system of linear equations to analyze.

In simple form, we can write A = D*(I+L); where A is lower triangular matrix, D is diagonal matrix, I is identity matrix and L is lower triangular with all zeros in diagonal.

Since $A^{-1} = (I+L)^{-1}*D^{-1}$ and inverse of D is simply inverse of diagonal element. And for very large n $L^{-n} = 0$ since it is having only lower triangular elements.

And we can write $(I+L)^{-1} = I – L + L^2 – L^3 + …. (-1)^n*L^n$ which itself is lower triangular matrix.

- How Many Theorems (Tautologies) Exist of 5, 6, 7, 8, and 9 Letters?
- “Probability” of a map being surjective
- Prove that every highly abundant or highly composite number $k$ is a prime distance from the nearest primes $\ne k \pm 1$ on either side
- Why can't the Alpertron solve this Pell-like equation?
- group of units in a topological ring
- Prove:$A B$ and $B A$ has the same characteristic polynomial.
- Every subspace of a compact space is compact?
- How to do well on Math Olympiads
- Inequality with five variables
- In Group theory proofs what is meant by “well defined”
- What's the explanation for these (infinitely many?) Ramanujan-type identities?
- A strange puzzle having two possible solutions
- How do I solve this PDE?
- How to find the CDF and PDF
- Find $\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { k } }{ { n }^{ \frac { 3 }{ 2 } } } } } $