# One more EigenValues proof

1. Jan 15, 2008

### Bertrandkis

Question 1
Let A be an nxn matrix such that $$(A-I)^{2}=O$$ where $$O$$ is the zero matrix
Prove that if $${\lambda}$$ is an eigen value of A then $${\lambda}=1$$
My attempt
If $$(A-I)^{2}=O$$ then $$A=I$$ (1)
if $${\lambda}$$ is an eigen value of A then $$Ax={\lambda}x$$ (2)
replace (1) in (2) $$Ix={\lambda}x$$ , but $$Ix=x$$ therefore $${\lambda}=1$$

Question 2
If A is an orthogonal Matrix, then prove that det(A)=+-1
My attempt
if A is orthogonal then $$AA^{T}=I$$ and $$A^{-1}=A^{T}$$
therefore $$AA^{-1}=I$$ and $$det(AA^{-1})=1$$ . Where does the + - comes from?

2. Jan 15, 2008

### Dick

1) You can't say A=I. It's not necessarily true. You do know (A-I)(A-I)x=0. Now let x be an eigenvector...
2) You've got det(A*A^T)=1. What can you say about the relationship between det(AB) and det(A) and det(B)? How about between det(A) and det(A^T)?

3. Jan 16, 2008

### Bertrandkis

Dick
1)How do you know that (A-I)(A-I)x=0? Why not $$(A-I)(A-I)x={\lambda}x$$.
If (A-I)(A-I)x=0 and x being the eigen vector, this suggests that $${\lambda}=0$$.

2)I know that $$det(A^{T})=det(A)$$ but how does that prove that
$$det(A)=+-1)$$ ?

4. Jan 16, 2008

### Dick

The problem says (A-I)^2=0. Operating on a vector means (A-I)(A-I)x=0. (A-I)x=Ax-Ix. If $$Ax={\lambda}x$$ that's $$(\lambda-1)x$$. Now apply the other (A-I). At the end you have a NUMBER times a nonzero vector equaling zero. So the number is zero.

det(A*A^T)=1=det(A^T)*det(A)=det(A)^2. Solving that equation for det(A) is just like solving x^2=1 for x. What are the solutions?

Last edited: Jan 16, 2008
5. Jan 16, 2008

### HallsofIvy

Staff Emeritus
Dick's original point was simply that you cannot, from M2= 0, with M a matrix and 0 the zero matrix, conclude that M= 0 because there are "zero-divisors" in the algebra of matrices. For example,
$$M= \left(\begin{array}{cc} 1 & 1 \\ -1 & -1 \end{array}\left)$$
has the property that M2= 0.

Of course, saying M2= 0 means that M2x= 0x= 0 for any vector x. Separate that into M(Mx)= 0 and solve "twice".

6. Jan 16, 2008

### Bertrandkis

for the first question I came up with this: $$0=(A-I)^2=A^2-2A+I.$$ so if $$Ax=\lambda x,$$ for some
vector $$x \neq 0,$$ then $$A^2x=\lambda^2x,$$ and $$0=(\lambda^2 - 2\lambda + 1)x=(\lambda - 1)^2x.$$ thus $$\lambda = 1.$$
This looks very right, doesn't it?

For question 2, I figue that x^2=1 yields x=+-1.
Thanks To all.