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

  • Thread starter vish_maths
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  • #1
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Main Question or Discussion Point

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
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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hello vish_maths! :smile:

(try using the X2 button just above the Reply box :wink:)
:
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
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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
AlephZero
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Prove that if ## (A + 2A^{-1})x = 0## then ##x = 0##

Start by multiplying both sides by ##A##, or by ##A^{-1}##.
 
  • #5
tiny-tim
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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
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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
AlephZero
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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
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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
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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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