Question

Let A be a square matrix and I is the unit matrix
i) Assume A^2 = 0.
Show that (I - A)^1 = I + A.

ii ) Assume A^3 = 0. Show that (E - A)^-1 = E + A + A^2
.
iii ) Determine the inverse of the matrix
1 2 -1
0 1 3
0 0 1
using one of the above equations .

iv ) Suppose A ^ k = 0 , k ≥ 2. Determine an expression for (E - A) ^ - 1

Answer 1

@Michele_Laino

Answer 2

@Michele_Laino

is i) something like this
\[(I-A)^{-1}=I+A \leftarrow \rightarrow (I-A)*(I-A)^{-1}=(I-A)*(I+A)\]
\[(I-A)*(I-A)^{-1} = I\]
\[I=I^2+I*A-A*I-A^2\]
since A^2=0 and I^2=I we can neglect that, so we have:
\[I = I + I*A-A*I\]

Answer 3

\[I-I=A(I-I) <=> 0 = 0\]

Answer 4

sorry on ii) it should be ii ) Assume A^3 = 0. Show that (I - A)^-1 = I + A + A^2 and NOT ii ) Assume A^3 = 0. Show that (E - A)^-1 = E + A + A^2

Answer 5

we can write this:
\[\Large \begin{gathered}
\left( {I - A} \right){\left( {I - A} \right)^{ - 1}} = \left( {I - A} \right)\left( {I + A} \right) = \hfill \\
\hfill \\
= {I^2} + IA - IA - {A^2} = I - A + A + 0 = I \hfill \\
\end{gathered} \]
and:
\[\Large \begin{gathered}
{\left( {I - A} \right)^{ - 1}}\left( {I - A} \right) = \left( {I + A} \right)\left( {I - A} \right) = \hfill \\
\hfill \\
= {I^2} - IA + IA - {A^2} = I - A + A + 0 = I \hfill \\
\end{gathered} \]
and the proof is complete

Answer 6

I have used this property: \(IA=AI\)

Answer 7

why are you using \((I - A)^{-1} = I + A.\) in the proof?

Answer 8

who?

Answer 9

since the question asks to check that if this Matrix:
\(I+A\) is the inverse of \(I-A\)

Answer 10

and I showed that such statement is true

Answer 11

you only know that \( A^2 = 0\)...you are trying to show that \((I - A)^{-1} = I + A\)

Answer 12

yes! Nevertheless you can use the thesis since the exercise asks for a check

Answer 13

start with \[(I-A)(I+A)=\cdots=I\]

then multiply both sides by \[(I-A)^{-1}\]

Answer 14

Is it okey or not do what I just did? I think I just did the same as Michele, right? but what to do on iii)?

Answer 15

I'll give you a hint...use part (ii) to do part (iii)

Answer 16

so @Zarkon, If I use (I-A)^-1 = E+A and the matrix as A in the equation
\[A=\left[\begin{matrix}1 & 2 &-1\\ 0 & 1&3\\ 0 & 0&1\end{matrix}\right]\]

and

\[I=\left[\begin{matrix}1 & 0 &0\\ 0 & 1&0\\ 0 & 0&1\end{matrix}\right]\]

\[(\left[\begin{matrix}1 & 0 &0\\ 0 & 1&0\\ 0 & 0&1\end{matrix}\right] - \left[\begin{matrix}1 & 2 &-1\\ 0 & 1&3\\ 0 & 0&1\end{matrix}\right])^{-1}=\left[\begin{matrix}1 & 0 &0\\ 0 & 1&0\\ 0 & 0&1\end{matrix}\right] + \left[\begin{matrix}1 & 2 &-1\\ 0 & 1&3\\ 0 & 0&1\end{matrix}\right]\]

\[\left[\begin{matrix}0 & -2 &1\\ 0 & 0&-3\\ 0 & 0&0\end{matrix}\right]^{-1}=\left[\begin{matrix}2 & 2 &-1\\ 0 & 2&3\\ 0 & 0&2\end{matrix}\right]\]

Answer 17

@Michele_Laino check my iii)

Answer 18

What I wanna do next is to multiply the left hand side by the not inverted one and the same to the right hand side, but thats not the inversion

Answer 19

or am I doing completely wrong?

Answer 20

the inverted A should be
1 -2 7
0 1 -3
0 0 1

Answer 21

I checked that
\[\huge {A^3} \ne O\]

Answer 22

furthermore:
\[\huge {A^2} \ne O\]

Answer 23

what?

Answer 24

if:
\[A = \left( {\begin{array}{*{20}{c}}
1&2&{ - 1} \\
0&1&3 \\
0&0&1
\end{array}} \right) \Rightarrow {A^3} \ne O,\;{A^2} \ne O\]

Answer 25

Let
\[A=\left[\begin{array}{ccc} 0 & 2 & -1 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \\ \end{array}\right]\]

then \[1+A=\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \\ \end{array}\right]\]

Answer 26

Then \[A^3=0\]

Answer 27

ok!

Answer 28

So I have to add 1 to both left and right hand side?

Answer 29

let \[A=-\left[\begin{array}{ccc} 0 & 2 & -1 \\ 0 & 0 & 3 \\ 0 & 0 & 0 \\ \end{array}\right]\]

that will work better

Answer 30

I write this:
\[\begin{gathered}
\left( {\begin{array}{*{20}{c}}
1&2&{ - 1} \\
0&1&3 \\
0&0&1
\end{array}} \right) = I - A \Rightarrow \hfill \\
\hfill \\
\Rightarrow A = I - \left( {\begin{array}{*{20}{c}}
1&2&{ - 1} \\
0&1&3 \\
0&0&1
\end{array}} \right) \hfill \\
\end{gathered} \]

Answer 31

then I-A is your matrix and its inverse is \[I+A+A^2\]

Answer 32

you can do it they I originally said but then the inverse is \[I-A+A^2\]

Answer 33

But I cant just pick a matrix? i am given one in the task?

Answer 34

then the inverse matrix, is:
\[{\left( {\begin{array}{*{20}{c}}
1&2&{ - 1} \\
0&1&3 \\
0&0&1
\end{array}} \right)^{ - 1}} = I + A = I + I - \left( {\begin{array}{*{20}{c}}
1&2&{ - 1} \\
0&1&3 \\
0&0&1
\end{array}} \right)\]

Answer 35

iii ) Determine the inverse of the matrix
1 2 -1
0 1 3
0 0 1
using one of the above equations .

Answer 36

1 -2 7
0 1 -3
0 0 1

Answer 37

yes! I have applied the statement of part ii

Answer 38

sorry my Internet Explorer 11, has crashed

Answer 39

could you explain a little on IV?

Answer 40

yor inverse matrix is correct!

Answer 41

for part iv, we have to apply the Mathematical induction principle

Answer 42

we can say that, if \(A^k=0\) then:
\[\Large {\left( {I - A} \right)^{ - 1}} = I + A + ... + {A^{k - 1}}\]
proof:
we have this computation:
\[\begin{gathered}
{\left( {I - A} \right)^{ - 1}} = I + A + ... + {A^{k - 1}} \hfill \\
\hfill \\
\left( {I - A} \right)\left( {I + A + ... + {A^{k - 1}}} \right) = \hfill \\
\hfill \\
= I + A + ... + {A^{k - 2}} + {A^{k - 1}} + \hfill \\
\hfill \\
- A - {A^2} - ... - {A^{k - 1}} - {A^k} = \hfill \\
\hfill \\
= I - {A^k} = I \hfill \\
\end{gathered} \]
similarly for this part:
\[{\left( {I - A} \right)^{ - 1}}\left( {I - A} \right) = \left( {I + A + ... + {A^{k - 1}}} \right)\left( {I - A} \right)\]