# Thread: anyone can help me for this prove matrix

1. ## anyone can help me for this prove matrix

1) Suppose that A is an nxn nonsingular upper triangualr matrix, and let B=A^-1. You may assume that all diagonal elements of A are nonzero.
a) using only the definition of matrix multiplication prove that Bsubnj =0 for all j<n. ( ie, show that all elements in the nth row of B, except for the last element, are zero.)
b) using only the definition of matrix multiplication , prove that B is upper triangular.

Thanks.

2. Originally Posted by happystudent
1) Suppose that A is an nxn nonsingular upper triangualr matrix, and let B=A^-1. You may assume that all diagonal elements of A are nonzero.
a) using only the definition of matrix multiplication prove that Bsubnj =0 for all j<n. ( ie, show that all elements in the nth row of B, except for the last element, are zero.)
b) using only the definition of matrix multiplication , prove that B is upper triangular.

Thanks.
Let the columns of $\bold{A}$ be $\bold{A}_1, .. \bold{A}_n$, then by definition $(\bold{A}_i)_i \ne 0$.

Let the $k$-th row of $\bold{B}$ be $\bold{B}_k$.

Then the $l$-th element in the $k$-th row of $\bold{I}=\bold{B}\bold{A}$ is:

$
\bold{I}_{k,l}=\bold{B}_k\bold{A}_l
$

Now, as $\bold{A}$ is upper trianglar:

$
\bold{I}_{k,1}=\bold{B}_k\bold{A}_1=(\bold{B}_k)_1 (\bold{A}_1)_1
$

and if $k\ne 1$ this is zero and as by definition $(\bold{A}_1)_1 \ne 0$ for $k \ne 1 (\bold{B}_k)_1=0$

Now suppose that for some $j for all $r \le j$

Then:

$
\bold{I}_{k,j+1}=\sum_{r=1}^{n}(\bold{B}_k)_r (\bold{A}_{j+1})_r=\sum_{r=1}^{j+1}(\bold{B}_k)_r (\bold{A}_{j+1})_r
$

because $\bold{A}$ is upper triangular, and:

$
\bold{I}_{k,j+1}=\sum_{r=1}^{n}(\bold{B}_k)_r (\bold{A}_{j+1})_r=\sum_{r=1}^{j+1}(\bold{B}_k)_r (\bold{A}_{j+1})_r=(\bold{B}_k)_{j+1} (\bold{A}_{j+1})_{j+1}
$

because by hypothesis $(\bold{B}_K)_r=0$, for all $r \le j$.

Hence:

$
\bold{I}_{k,j+1}=(\bold{B}_k)_{j+1} (\bold{A}_{j+1})_{j+1}=0
$

So $(\bold{B}_k)_{j+1}=0$

Which completes a proof by induction that $(\bold{B}_k)_j=0$ for all $j, which proves that $\bold{B}$ is upper triangular.

(it is easier to see what is going on doing this element by element along the a row of $\bold{B}$)

RonL