Problem with idempotent matrices

[Samme013]

OK so i can prove that the given inverse is actually the inverse but i can not prove that I+A is non singular without using the given inverse so how do i go about doing that?(I have done part a)
Thanks in advance.View attachment 3586

 

[Ackbach]

I'm confused. Is there a reason you can't use the fact that $(I+A)(I-\tfrac12 A)=I$ in order to show that $I+A$ is invertible?

 

[Samme013]

I'm confused. Is there a reason you can't use the fact that $(I+A)(I-\tfrac12 A)=I$ in order to show that $I+A$ is invertible?

I am not sure if i can use that fact and i can't really ask but it is kind of trivial if i can use it.Is it even possible to show it without using that?

 

[Ackbach]

I am not sure if i can use that fact and i can't really ask but it is kind of trivial if i can use it.Is it even possible to show it without using that?

It might be. You might be able to do something clever with determinants. Let me marshall the facts such as we know them:
\begin{align*}
&\text{Idempotent matrices have determinant equal to } 0 \text{ or } 1. \text{ Why?} \\
A^2&=A \\
(\det(A))^2&=\det(A) \\
(\det(A))^2-\det(A)&=0 \\
\det(A)(\det(A)-1)&=0 \\
\det(A)&\in\{0,1\} \\
\therefore \; \det(I-A)&\in\{0,1\}. \\
&\text{Idempotent matrices are either the identity matrix, or are singular.} \\
&\text{Why? Assume } A \text{ is non-singular. Then } \\
A^2&=A \\
A^{-1}AA&=A^{-1}A \\
IA&=I \\
A&=I.
\end{align*}
Now we could take two cases: 1. $A=I$. 2. $\det(A)=0$.

For the first case, $I+A=I+I=2I$, and the determinant is $2^n\not=0$, where $A$ is $n\times n$.

For the second case, we have $\det(A)=0$. What if you formed the product
$$(I+A)(I+A)=I^2+IA+AI+A^2=I+2A+A=I+3A.$$
Not sure where that leads you. You could also form
$$(I-A)(I+A)=I^2+IA-AI-A^2=I+A-A-A=I-A.$$
Just seeing here:
$$(I-A)(I-A)=I^2-IA-AI+A^2=I-A-A+A=I-A.$$
Evidently, $(I-A)(I-A)=(I-A)(I+A)$, but I don't see where you can go further. There's no guarantee that $I-A$ is invertible (indeed, unless $I-A=I$, it won't be!), so you can't cancel anything, and the determinants won't give you anything here, either.

I think that using the formula for the inverse the problem asks you to show is just fine. If you simply compute $(I+A)(I-\tfrac12 A)=I$, you have just shown that $I+A$ is invertible. The problem asks you to show those two things. It does not tell you in what order to do them.

 

[Samme013]

It might be. You might be able to do something clever with determinants. Let me marshall the facts such as we know them:
\begin{align*}
&\text{Idempotent matrices have determinant equal to } 0 \text{ or } 1. \text{ Why?} \\
A^2&=A \\
(\det(A))^2&=\det(A) \\
(\det(A))^2-\det(A)&=0 \\
\det(A)(\det(A)-1)&=0 \\
\det(A)&\in\{0,1\} \\
\therefore \; \det(I-A)&\in\{0,1\}. \\
&\text{Idempotent matrices are either the identity matrix, or are singular.} \\
&\text{Why? Assume } A \text{ is non-singular. Then } \\
A^2&=A \\
A^{-1}AA&=A^{-1}A \\
IA&=I \\
A&=I.
\end{align*}
Now we could take two cases: 1. $A=I$. 2. $\det(A)=0$.

For the first case, $I+A=I+I=2I$, and the determinant is $2^n\not=0$, where $A$ is $n\times n$.

For the second case, we have $\det(A)=0$. What if you formed the product
$$(I+A)(I+A)=I^2+IA+AI+A^2=I+2A+A=I+3A.$$
Not sure where that leads you. You could also form
$$(I-A)(I+A)=I^2+IA-AI-A^2=I+A-A-A=I-A.$$
Just seeing here:
$$(I-A)(I-A)=I^2-IA-AI+A^2=I-A-A+A=I-A.$$
Evidently, $(I-A)(I-A)=(I-A)(I+A)$, but I don't see where you can go further. There's no guarantee that $I-A$ is invertible (indeed, unless $I-A=I$, it won't be!), so you can't cancel anything, and the determinants won't give you anything here, either.

I think that using the formula for the inverse the problem asks you to show is just fine. If you simply compute $(I+A)(I-\tfrac12 A)=I$, you have just shown that $I+A$ is invertible. The problem asks you to show those two things. It does not tell you in what order to do them.

The exercise is listed under a chapter which is before determinants are introduced( i have no clue what they are yet) so most likely it is not expected to come up wit that solution.I will definitely come back and take a look at this when i get to the next chapter though, thanks for helping !

 

[Ackbach]

The exercise is listed under a chapter which is before determinants are introduced( i have no clue what they are yet) so most likely it is not expected to come up wit that solution.I will definitely come back and take a look at this when i get to the next chapter though, thanks for helping !

You're very welcome!