TIFR CSE 2012 | Part B | Question: 12

Question

Let $A$ be a matrix such that $A^{k}=0$. What is the inverse of $I - A$?

• A. $0$
• B. $I$
• C. $A$
• D. $1 + A + A^{2} + ...+ A^{k - 1}$
• E. Inverse is not guaranteed to exist.

Comments

Here $A$ is a special kind of matrix called NILPOTENT MATRIX.

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can someone please share an insight as to why $A^{k} = 0$ does not imply A=0. (please do not give examples)

Like when $A^{k} = 0$ $\rightarrow$ A.A.A.A.A……...k times =0

=> $A^{-1}A.A.A….k $ times = $A^{-1}0$

=> eventually we will reach A=0

And then I-A=I.

Where am i going wrong?

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reboot  A inverse doesn’t exist , bcz det(A) = 0

Answer 1

Given $A^{k}=0$

Subtract from $I$ on both sides,

$I-A^{k}=I \qquad \to (1)$
Now $I-A^{k}$  can be written as,

$I-A^{k}=( I - A )(I+A+A^{2}+A^{3}+\ldots +A^{k-1})$ (simplifying RHS we get LHS)

Putting this in $(1),$

$I=( I - A )(I+A+A^{2}+A^{3}+\ldots A^{k-1})$

Now question is $( I - A )^{-1}$ so we multiply with $( I - A )^{-1}$ on both sides,

$( I - A )^{-1} =  (I+A+A^{2}+A^{3}+\ldots +A^{k-1})$

Hence, $(D)$ is the Answer.

Comments

How can we Guarantee that inverse exists??

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I  -  Ak = I

I  -  Ak  = ( I - A ) ( I + A + A2 + A3  +.........+Ak-1 ) 

how second step comes

 

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( I - A ) ( I + A + A2 + A3  +.........+Ak-1 ) 

=( I + A + A2 + A3  +.........+Ak-1 - A- A2 - A3  - ......... - Ak)

=I  -  Ak 

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how u get this step :-

I  = ( I - A ) ( I + A + A² + A³  +.........+A^k-1 ) 

?

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I  -  A^k  = ( I - A ) ( I + A + A² + A³  +.........+A^k-1 ) 

given $A^{k}=0$

I   = ( I - A ) ( I + A + A² + A³  +.........+A^k-1 ) 

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Oh thanks.

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I am trying to understand this question. You seem to have written used the identity:

$ (I - A)^{-1} = (I + A + A^2 + A^3 + ... + A^n) $

But according to this link https://www.physicsforums.com/threads/matrix-inverse-equals-power-series.423897/, this will work only when the right side converges. Since we cannot be sure of the right side converging (I think!), what is your reasoning behind applying it?

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what is A^k-1 - A^k

when is multiplication we add the powers

what incase of subtraction

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$A^{k-1}-A^{k}$

$A^{k} \times A^{-1}-A^{k}$

$A^{k}(A^{-1}-1)$

$A^{k}(\frac{1}{A}-1)$

$A^{k}(\frac{1-A}{A}) $ or $A^{k-1}(1-A)$

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@leen thanku :)

=( I + A + A2 + A3  +.........+Ak-1 - A- A2 - A3  - ......... - Ak)

=I  -  A^k 

then in this everything gets cancelled and what remains is A^k-1 - A^k na

how did that become I-A^k and not (i-A)A^k-1 ?

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A_i_$_h 

$I=( I - A ) ( I + A + A^{2} + A^{3}  +.........+A^{k-1} ) $

$I=( I + A + A^{2} + A^{3}  +.........+A^{k-1} - A- A^{2} - A^{3}  - ......... - A^{k})$

$I = I  -  A^{k} $

which is right but I want something else. I want to find the inverse of $(I-A).$

now if we multiply $(I-A)^{-1}$ both sides  we can easily get inverse of $(I-A).$

$( I - A )^{-1} \times I$ = $( I - A )^{-1} \times ( I - A ) ( I + A + A^{2} + A^{3}  +.........+A^{k-1} ) $

$( I - A )^{-1}= ( I + A + A^{2} + A^{3}  +.........+A^{k-1} ) $

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A_i_$_h 

$( I - A ) ( I + A + A^{2} + A^{3}  +.........+A^{k-1} )$

$( I + A + A^{2} + A^{3}  +.........+A^{k-1} - A - A^{2} - A^{3}  +.........- A^{k} )$

$( I + A + A^{2} + A^{3}  +.......+A^{k-2}+A^{k-1} - A - A^{2} - A^{3}  +.........- A^{k-1} - A^{k} )$

$(I-A^{k})$

You did a mistake . $A^{k-1}$ comes before $A^{K}$.I think your doubt cleared now.

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$1+r+r^{2}+...+r^{k-1} = \frac{r^{k}-1}{r-1}$

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@LeenSharma

I think answer should be E (Inverse is not guaranteed to exist.).

Reason - As $A^{k} = 0$  , so $det (A^{k}) = 0$, this implies $k * det (A) = 0$. so $A$ is singular.

Now $(I  -  A)$ may or may not be singular

https://math.stackexchange.com/questions/1993009/is-the-sum-of-singular-and-nonsingular-matrix-always-a-nonsingular-matrix

If $(I  -  A)$ is non-singular then option D is correct, but if it is singular then inverse does'nt exists. So, option E is correct

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Inverse is guaranteed to exist see the 1st property.

https://en.wikipedia.org/wiki/Nilpotent_matrix#Additional_properties

Answer 2

why inverse is guranteed to exist??
because A^k=0
so for  I-A  det value is 1

how??

lets find (I-A).A^(k-1)

= A^(k-1)-A^k

=A^(k-1)-0

=A^(k-1)

so |(I-A).A^(k-1)| = |A^(k-1)|

=>det(I-A).det(A^(k-1))=det(A^(k-1))

=>det(I-A)=1 PROVED

so for  I-A  inverse always exist..

here A is nilponent with order k.

now we know |M.N|=|M|.|N|

now let inverse of I-A is X

so , (I-A).X=I

=> |(I-A).X|=|I|

=>|I-A|.|X|=1

we showed above |I-A|=1

so, |X|=1

so if X is the answer for the inverse of matrix I-A  then det(X) is 1

check in options,

a) 0 wrong

b) I ; though det is 1, but (I-A).I is certainly not I. ; so wrong

c) A; det is 0; so wrong

e) we know inverse exist ; so wrong

d) only option left; moreover as |A|=1 from 2nd to last term det value is 0; so det of whole is 1(the first term) ANSWER==//