How to prove inverse of Matrix

Question

If you have an invertible upper triangular matrix $M$, how can you prove that $M^{-1}$ is also an upper triangular matrix? I already tried many things but don't know how to prove this. Please help!

Answer 1

Look at the algorithm of invertion. You take yor matrix $A$ and matrix $E$, reduce $A$ to $E$ and perform same conversions to $E$. Then you have $E$ and $A^{-1}$. Note just that $E$ is upper triangular and any reducing transformations do not affect its triangularity.

Answer 2

The co-factor of any element above the diagonal is zero.Resaon :The matrix whose determinant is the co-factor will always be an upper triangular matrix with the determinant zero because either its last row is zero or its first column is zero or one of its diagonal element is zero.This can easily be verified. This will imply that the entries below the diagonal of the inverse of this matrix will all be zero.

Answer 3

An upper triangular $n\times n$ matrix can be written $A=D+N$ with $D$ diagonal, $N$ strictly upper triangular (null diagonal). Then $A$ is invertible if and only if $D$ is invertible. In this case, we have $$ A=D(I_n+D^{-1}N). $$ Now $D^{-1}N$ is strictly upper triangular, so it is nilpotent. Concretely, $(D^{-1}N)^n=0$. So $I_n+D^{-1}N$ is invertible with inverse given by the (finite) Neumann series $$ (I_n+D^{-1}N)^{-1}=\sum_{k=0}^{n-1}(-D^{-1}N)^k. $$ Since upper triangularity is stable under products, we see that $(I_n+D^{-1}N)^{-1}$ is upper triangular, and finally $$ A^{-1}=(I_n+D^{-1}N)^{-1}D^{-1} $$ is upper triangular.

Answer 4

Hint: Use Cramer's rule for the inverse of the matrix.

EDIT: As requested, I give more details:

By Cramer, $$A^{-1} = \frac{1}{\det(A)}\operatorname{Adj}(A),$$ where $\operatorname{Adj}(A)$ is the adjugate matrix of $A$. The entry at the position $(i,j)$ in $\operatorname{Adj}(A)$ is given by $(-1)^{i+j} \det(A_{ji})$, where $A_{ji}$ is the $(n-1)\times (n-1)$ matrix resulting from $A$ by deleting its $j$th row and its $i$th column.

Now $A^{-1}$ is upper triangular iff for all $i > j$ the entry at the position $(i,j)$ is zero. By Cramer, equivalently, the entry $\det(A_{ji})$ at the position $(i,j)$ in $\operatorname{Adj}(A)$ is zero. Now using that $A$ is upper triangular and $i > j$, it is not hard to see that $A_{ji}$ is upper triangular and has a zero on the diagonal, so really, it's determinant is $0$.

Answer 5

Suppose that $M=(a_{ij})$ then since $M$ is upper triangular matrix we have: $$M(e_j)=\sum_{i=1}^j a_{ij}e_i,\tag{1}$$ where $a_{ii}\not=0$ because $M$ is invertible.

Now we prove by induction that $$M^{-1}(e_j)=\sum_{i=1}^j b_{ij}e_i.$$ We have $M^{-1}(e_1)=\frac{1}{a_{11}}e_1$ and suppose the result is true for $j$. For $j+1$: we have from $(1)$: $$e_{j+1}=\sum_{i=1}^{j+1}a_{i,j+1}M^{-1}(e_{i}),$$ so $$M^{-1}(e_{j+1})=\frac{1}{a_{j+1,j+1}}\left(e_{j+1}-\sum_{i=1}^{j}a_{i,j+1}M^{-1}(e_{i})\right)$$ which gives the result taking into account the induction hypothesis.

Answer 6

The easiest way is to prove by mathematical induction. Note that $$ \begin{pmatrix}a&b^T\\0&C\end{pmatrix}^{-1} =\begin{pmatrix}a^{-1}&-\frac1ab^TC^{-1}\\0&C^{-1}\end{pmatrix}. $$