Inverse of upper triangular matrix

  • Thread starter Thread starter topgear
  • Start date Start date
  • Tags Tags
    Inverse Matrix
Click For Summary
An invertible upper triangular matrix A has an inverse A^-1 that is also upper triangular. The discussion highlights a misunderstanding regarding the properties of matrix inverses, particularly that the main diagonal elements do not simply flip and that the off-diagonal elements change signs. An example is provided with a non-upper triangular matrix to illustrate that the initial assumption about the inverse properties is incorrect. The conclusion emphasizes that the properties of upper triangular matrices must be carefully considered when discussing their inverses. Understanding these properties is crucial for accurate mathematical communication.
topgear
Messages
10
Reaction score
0

Homework Statement


Show that if A is an invertible upper triangular matrix, then A^-1 is also an upper triangular matrix


Homework Equations





The Attempt at a Solution


The inverse of something just flips the main diagonal and leaves everything else where is was just changing the signs. How do I say this in math language.
 
Physics news on Phys.org
topgear said:

Homework Statement


Show that if A is an invertible upper triangular matrix, then A^-1 is also an upper triangular matrix


Homework Equations





The Attempt at a Solution


The inverse of something just flips the main diagonal and leaves everything else where is was just changing the signs. How do I say this in math language.

Since it's not true, it's not worth saying in math language.

For example,
A = \begin{bmatrix} 1&1\\1&-1 \end{bmatrix}

It turns out that this matrix is invertible, and its inverse is
A^{-1} = \begin{bmatrix} 1/2&1/2\\1/2&-1/2 \end{bmatrix}

So here is "something" (A) that has an inverse, but I don't see that the main diagonal got flipped and only the signs on the off diagonal got changed. Admittedly, this isn't an upper triangular matrix.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K