Eigenvalues and Orthogonal Matrices: Proving Properties Without Prefix

[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?

 

[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)?

 

[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) ?

 

[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?

 

[HallsofIvy]

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".

 

[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.