Constructing the inverse of random given matrices when a certain condition is levied

In summary: So, the best way to test for invertibility is to try and find a matrix P such that P'P=I and P=A-1. If that holds, then A is invertible. If it doesn't hold, then A isn't invertible.If ##(A^2 + 2I)x = 0## implies ##x = 0##, that means -2 is not an eigenvalue of ##A^2##. That has nothing to do with whether or not A is invertible.
  • #1
vish_maths
61
1
Hello Everyone :)

I have been facing a little difficulty when encountering such kind of problems . i have also written down my line of thinking and approach which i take to solve them. So, please try to give me the correct line of thinking while solving such problems:

1. If A is invertible , then prove that A + 2A^(-1) is invertible

My attempt : A + 2A^(-1) = A A A^-1 + 2A^(-1)
= (A^2 + 2I) A^(-1)

Now, A^(-1) is invertible , which means all i have to do is check for invertibility of
(A^2 + 2I)

What next . i mean i just don't get a clue after this. What is the correct line of thinking ?


2. If A^n = O , then Prove that A - I is invertible.

My attempt : i begin by assuming a matrix exists X so that i arrive back at the given condition

( A - I ) X = I
=> AX - IX = I
=> AX - ( I - A^n )X = I
=> AX - IX = I - A^nX
which solves nothing :(
little low on confidence :(

Thank you
 
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  • #2
hello vish_maths! :smile:

(try using the X2 button just above the Reply box :wink:)
vish_maths said:
:
1. If A is invertible , then prove that A + 2A^(-1) is invertible

try finding the inverse …

it's probably of the form pA + qA-1 :wink:
2. If A^n = O , then Prove that A - I is invertible.

when you can't see a general way, try an easy case first …

in this case, what is (A - 1)-1 if A2 = 0 ? :smile:
 
  • #3


tiny-tim said:
hello vish_maths! :smile:

(try using the X2 button just above the Reply box :wink:)


try finding the inverse …

[/INDENT]

Hello tiny tim :)
thanks for the reply. Is the approach to such solutions more of a hit and trial method ?
 
  • #4


Prove that if ## (A + 2A^{-1})x = 0## then ##x = 0##

Start by multiplying both sides by ##A##, or by ##A^{-1}##.
 
  • #5
actually, now that i look at it again, 1. isn't true …

if we're allowed complex numbers, then it doesn't work for
Code:
1     0
0  √2i
 
  • #6


AlephZero said:
Prove that if ## (A + 2A^{-1})x = 0## then ##x = 0##

Start by multiplying both sides by ##A##, or by ##A^{-1}##.

Hello :)

thanks for the reply. i have already reached that step. Multiplying by A leads to the step
to check for the invertibility of A^2 + 2I

Then the remaining solution is to confirm the null space just consists of the zero vector , but how do u approach towards there :

(A^2 + 2I) X = 0

null space = 0 means : 1. it can be reduced to perfect echelon form
2. all the column vectors and row vectors are linearly independent

what actually bugs me is the link between these two
thanks
 
  • #7


Well, as Tiny Tim said, it is false.

If ##(A^2 + 2I)x = 0## implies ##x = 0##, that means -2 is not an eigenvalue of ##A^2##. That has nothing to do with whether or not A is invertible.

In Tiny Tim's counter-example, the eigenvalues of ##A^2## are 1 and -2.
 
  • #8


AlephZero said:
Well, as Tiny Tim said, it is false.

If ##(A^2 + 2I)x = 0## implies ##x = 0##, that means -2 is not an eigenvalue of ##A^2##. That has nothing to do with whether or not A is invertible.

In Tiny Tim's counter-example, the eigenvalues of ##A^2## are 1 and -2.

thank you , tiny-tim and AlephZero. However, i am facing still issues with questions like if A is invertible , then prove or disprove whether A + A^(-1) is invertible or not.
And the main problem is that problems like testing the invertibility (A^2 + 2I) are located in a serge lang chapter when the concept of determinants or eigen values have not been even touched.

Only the concepts of linear independence and vector spaces have been discussed. So, how to solve such questions with minimal resources ?
 
  • #9


A^-1 = P'(E^-1)P where E is a diagonal matrix composed of the eigenvalues of matrix A. Also, P'P=I and P is compose of the eigenvectors of matrix A.

So for any random matrix A, if any of its eigenvalues are zero, we have an issue. However, back in the real world, any time one of the eigenvalue is very small, there is high instability in computing the inverse (the 'problem' you witnessed).
 
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Related to Constructing the inverse of random given matrices when a certain condition is levied

1. What is the purpose of constructing the inverse of a random matrix?

The inverse of a matrix is used to solve systems of linear equations and perform other mathematical operations. It is also used in various areas of science, such as computer graphics, engineering, and statistics.

2. What does it mean to "levy a condition" on a random matrix?

Levying a condition on a random matrix means imposing a specific requirement or constraint on the matrix. This can be done to simplify the problem or make it more relevant to a particular application.

3. How is the inverse of a random matrix constructed?

The inverse of a random matrix can be constructed using various mathematical methods, such as Gaussian elimination, LU decomposition, or the adjoint method. The specific method used will depend on the properties of the matrix and the desired outcome.

4. Are there any limitations or challenges when constructing the inverse of a random matrix?

Yes, there are limitations and challenges when constructing the inverse of a random matrix. Some matrices may not have an inverse, and others may have a computationally intensive construction process. Additionally, rounding errors and numerical stability can affect the accuracy of the inverse.

5. Can the inverse of a random matrix be used in real-world applications?

Yes, the inverse of a random matrix has many practical applications. It is commonly used in machine learning, data analysis, and finance. It can also be used to solve problems in physics, chemistry, and other scientific fields.

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