Proving eigenvalues and diagonalizability

[Yoonique]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >[/color]

Let A be a square matrix of order n such that A^2 = I
a) Prove that if -1 is the only eigenvalue of A, then A= -I
b) Prove that if 1 is the only eigenvalue of A, then A= I
c) Prove that A is diagonalizable.

For part a and b, I consider A=-I and I then showed that the eigenvalues are -1 and 1 respectively. Is it valid by assuming A=-I and I then proving the eigenvalues are -1 and 1?
If A=-I, Av=λv
-v=λv
therefore, λ=-1.

For part c, I have no clue how to even start. I did some research, and I haven't learn minimal polynomial so I can't use it to prove.

 

[fresh_42]

Consider the expression $(t \cdot E_n - A) \cdot (t \cdot E_n + A)$ and the zeros of the characteristic polynomial of $A$ to prove c).
a) and b) are direct consequences of c).

 

[Mark44]

< Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

Let A be a square matrix of order n such that A^2 = I
a) Prove that if -1 is the only eigenvalue of A, then A= -I
b) Prove that if 1 is the only eigenvalue of A, then A= I
c) Prove that A is diagonalizable.

For part a and b, I consider A=-I and I then showed that the eigenvalues are -1 and 1 respectively. Is it valid by assuming A=-I and I then proving the eigenvalues are -1 and 1?

No. You are proving the converse of what you're asked for. For a), assume that -1 is the only eigenvalue of A, then show that A must be -I. Same idea for b).

If A=-I, Av=λv
-v=λv
therefore, λ=-1.

For part c, I have no clue how to even start. I did some research, and I haven't learn minimal polynomial so I can't use it to prove.

 

[Yoonique]

No. You are proving the converse of what you're asked for. For a), assume that -1 is the only eigenvalue of A, then show that A must be -I. Same idea for b).

Av=λv
Av=-v
Av=-Iv
A=-I
Can I deduce A=-I from that?

 

[Yoonique]

Consider the expression $(t \cdot E_n - A) \cdot (t \cdot E_n + A)$ and the zeros of the characteristic polynomial of $A$ to prove c).
a) and b) are direct consequences of c).

I didn't learn anything about that in my course. Can you explain that more explicitly? I know the eigenvalue of A are -1 and 1.

 

[Mark44]

Av=λv
Av=-v
Av=-Iv
A=-I
Can I deduce A=-I from that?

You've omitted what's given, and a step or two. How do you go from equation 2 above to equation 3?

 

[fresh_42]

I didn't learn anything about that in my course. Can you explain that more explicitly? I know the eigenvalue of A are -1 and 1.

$A$ is diagonalizable iff there is a matrix $S$ so that $S^{-1} A S = diag (λ_1, ... , λ_n)$
$diag (λ_1, ... , λ_n)$ denotes a diagonal matrix with entries $λ_1, ... , λ_n$ on its diagonal and 0 elsewhere. $E_n = diag(1, ... , 1)$

The characteristic polynomial $char_A (t)$ of $A$ in $t$ is defined as $det ( t \cdot E_n - A )$. Now
$char_A (t) \cdot det ( t \cdot E_n + A )$
$= det ( t \cdot E_n - A ) \cdot det ( t \cdot E_n + A )$
$= det [( t \cdot E_n - A ) \cdot ( t \cdot E_n + A )]$
$= det (t^2 \cdot E_n - A^2)$
$= det (t^2 \cdot E_n - E_n)$
$= (t^2 - 1)^n$
$= (t-1)^n (t+1)^n$

The $λ_i$ are exactly the zeros of $char_A (t)$. The equation above shows that only $(t ± 1)$ are divisors of $char_A (t)$ and therefore all $λ_i$ must be 1 or -1, proving c) and then a) and b).

I admit there's eventually an easier way to prove this without determinant and characteristic polynomial.
Maybe I'm too used to them that I didn't see it. Which rules about matrices are you expected to use?

 

[Yoonique]

You've omitted what's given, and a step or two. How do you go from equation 2 above to equation 3?

I multiply both sides by identity matrix so it IAv is still Av.

 

[Mark44]

$A$ is diagonalizable iff there is a matrix $S$ so that $S^{-1} A S = diag (λ_1, ... , λ_n)$
$diag (λ_1, ... , λ_n)$ denotes a diagonal matrix with entries $λ_1, ... , λ_n$ on its diagonal and 0 elsewhere. $E_n = diag(1, ... , 1)$

I don't believe that the OP is this far along in his/her studies.
Try to keep your hints aligned with what the person you're attempting to help can handle.

The characteristic polynomial $char_A (t)$ of $A$ in $t$ is defined as $det ( t \cdot E_n - A )$. Now
## det ( t \cdot E_n - A ) \ cdot det ( t \cdot E_n + A )
= det [( t \cdot E_n - A ) \ cdot ( t \cdot E_n + A )]
= det (t^2 \cdot E_n - A^2)
= det (t^2 \cdot E_n - E_n)
= (t^2 - 1)^n
= (t-1)^n (t+1)^n

 

[Mark44]

I multiply both sides by identity matrix so it IAv is still Av.

Yes, that's clear. What I really meant was how did you go from equation 3 to equation 4?

 

[Yoonique]

Yes, that's clear. What I really meant was how did you go from equation 3 to equation 4?

I'm not sure whether it is right. But I compared them like coefficients.

 

[Mark44]

I'm not sure whether it is right. But I compared them like coefficients.

The trouble is, that isn't guaranteed to work, when you're doing matrix multiplication. For example, it's possible for AB = 0, where neither A nor B is the zero matrix.
For example:
$A = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}$

So starting here: Av=-Iv, do you have any other ideas? You know the v is an eigenvector, right? Is there a property of eigenvectors you can use?

 

[Yoonique]

The trouble is, that isn't guaranteed to work, when you're doing matrix multiplication. For example, it's possible for AB = 0, where neither A nor B is the zero matrix.
For example:
$A = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}$

So starting here: Av=-Iv, do you have any other ideas? You know the v is an eigenvector, right? Is there a property of eigenvectors you can use?

The only properties I know are that they are linearly independent and they form a basis for R^n.

 

[Mark44]

The only properties I know are that they are linearly independent and they form a basis for R^n.

"They" - for each problem part, you have only one eigenvector.
Can any old vector be an eigenvector, or are there any restrictions?

 

[Yoonique]

If I know that the rank(I-A) + rank(I+A) = n, then does it mean that the dimension of eigenspace of A is n, therefore there are n eigenvectors so A is diagonalizable.

 

[Yoonique]

"They" - for each problem part, you have only one eigenvector.
Can any old vector be an eigenvector, or are there any restrictions?

Eigenvector cannot be the zero vector?

 

[Mark44]

Eigenvector cannot be the zero vector?

Yes!
OK, now starting from here -- Av = -Iv, can you get another equation and conclude that A = -I?

 

[Yoonique]

Yes!
OK, now starting from here -- Av = -Iv, can you get another equation and conclude that A = -I?

Another equation? (A+I)v=0?

 

[Mark44]

Another equation? (A+I)v=0?

That's what I was steering you toward. With what you know about v, and what you know about A (from post #1), why can you conclude that A + I must be the zero matrix?

Note that because of the way matrix multiplication works, it's possible for Bx = 0, with $\vec{x} \ne \vec{0}$ and B not equal to the zero matrix. For example, with $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

 

[fresh_42]

Try to keep your hints aligned with what the person you're attempting to help can handle.

mea culpa. You're right.

 

[Yoonique]

That's what I was steering you toward. With what you know about v, and what you know about A (from post #1), why can you conclude that A + I must be the zero matrix?

Note that because of the way matrix multiplication works, it's possible for Bx = 0, with $\vec{x} \ne \vec{0}$ and B not equal to the zero matrix. For example, with $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and $\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

I have no clue for this :/

 

[Yoonique]

I have no clue for this :/

Oh wait. I think I have an idea. Does it mean that rank(A+I)=0?

 

[Mark44]

Oh wait. I think I have an idea. Does it mean that rank(A+I)=0?

Yes, but I'm not sure that is helpful.

You are given that $A^2 = I$. You definitely want to use that information.

 

[fresh_42]

Oh wait. I think I have an idea. Does it mean that rank(A+I)=0?

If you find it, tell me. I'm blocked, too. What Mark wrote was that you cannot conclude from (A+I)v = 0 that v = 0 or A+I=0 (see his example with Ax=0). The difficulty is to show that it holds anyway. Somewhere $A^2 = 1$ or (A+I)(A-I)=0 should be used.

 

[Yoonique]

If you find it, tell me. I'm blocked, too. What Mark wrote was that you cannot conclude from (A+I)v = 0 that v = 0 or A+I=0 (see his example with Ax=0). The difficulty is to show that it holds anyway. Somewhere $A^2 = 1$ or (A+I)(A-I)=0 should be used.

(A+I)v=(A+I)(A-I)? I don't know how can it help me to conclude A+I=0

 

[Mark44]

Somewhere $A^2 = 1$ or (A+I)(A-I)=0 should be used.

That would be $A^2 = I$ rather than $A^2 = 1$.
If $A^2 = I$, then $|A^2| = ?$

 

[Yoonique]

That would be $A^2 = I$ rather than $A^2 = 1$.
If $A^2 = I$, then $|A^2| = ?$

I assume |A^2| means determinant? Then determinant of A is 1 or -1. If it means length then it is 1.

 

[Mark44]

I assume |A^2| means determinant?

Yes, that's standard notation for the matrix determinant.

Then determinant of A is 1 or -1. If it means length then it is 1.

The determinant doesn't measure the "length" of a matrix.
It's useful to determine whether a matrix has an inverse ($|A| \ne 0$) or doesn't have an inverse ($|A| = 0$).

However, the important fact here is that $A^2 = I$. If the product of two square matrices is I, then the two matrices are inverses of each other. What does this fact imply about A?

BTW, if it seems like I've changed course on this, I have. I was thinking that if $A^2 = I$, then $|A^2| = 1$, from which it follows that $|A| = \pm 1$ which is true. I was then thinking that it must be the case that A = I or A = -I, which does not necessarily follow. A counterexample is this matrix:
$B = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$
For this matrix, |B| = 1 for any value of $\theta$, but $B \ne I$, in general.

 

[Yoonique]

Yes, that's standard notation for the matrix determinant.
The determinant doesn't measure the "length" of a matrix.
It's useful to determine whether a matrix has an inverse ($|A| \ne 0$) or doesn't have an inverse ($|A| = 0$).

However, the important fact here is that $A^2 = I$. If the product of two square matrices is I, then the two matrices are inverses of each other. What does this fact imply about A?

BTW, if it seems like I've changed course on this, I have. I was thinking that if $A^2 = I$, then $|A^2| = 1$, from which it follows that $|A| = \pm 1$ which is true. I was then thinking that it must be the case that A = I or A = -I, which does not necessarily follow. A counterexample is this matrix:
$B = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$
For this matrix, |B| = 1 for any value of $\theta$, but $B \ne I$, in general.

A is invertible?

 

[fresh_42]

A is invertible?

Yes.
$A^2 = A \cdot A = 1$ (or I or $E_n$ or whatever the unity matrix may be denoted) means that $A$ is it's own inverse.
Just as $(-1) \cdot (-1) = 1$.
Let us first assume all eigenvalues are +1.

By multiplying we see that $(A^2 - 1) = (A + 1)(A - 1) = 0$.
What does this mean for $im(A - 1) = \{ v | v = ( A - 1)w$ for a $w \}$ and $ker(A + 1) = \{ v | (A + 1)v = 0 \}$?
Can you calculate $ker(A + 1) = \{ v | (A + 1)v = 0 \}$?