T/F: Orthogonal matrix has eigenvalues +1, -1

[Mr Davis 97]

Homework Statement

If a 3 x 3 matrix A is diagonalizable with eigenvalues -1, and +1, then it is an orthogonal matrix.

Homework Equations

The Attempt at a Solution

I feel like this question is false, since the true statement is that if a matrix A is orthogonal, then it has a determinant of +1 or -1, which has nothing to do with diagonalozation. However, I don't see how to prove this rigorously. Would the best way just be to search for a counter-example?

 

[micromass]

What do you know about the eigenvectors of an orthogonal matrix?

 

[Mr Davis 97]

I can't seem to recall anything about the eigenvectors of an orthogonal matrix... I looked it up and couldn't find anything either.

 

[fresh_42]

I can't seem to recall anything about the eigenvectors of an orthogonal matrix... I looked it up and couldn't find anything either.

How is an orthogonal matrix defined? And how do eigenvalues behave in that equation?

 

[Mr Davis 97]

How is an orthogonal matrix defined? And how do eigenvalues behave in that equation?

An orthogonal matrix is a square matrix such that $A^{T}A= I$. I don't see how this can be used to analyze its eigenvalues. I can only see that detA = 1 or -1... But those aren't eigenvalues.

 

[fresh_42]

An orthogonal matrix is a square matrix such that $A^{T}A= I$. I don't see how this can be used to analyze its eigenvalues. I can only see that detA = 1 or -1... But those aren't eigenvalues.

Firstly, we have to show (or know), that eigenvalues of $A^ {T}$ are the same as those of $A$.
This is done by the equation $\det (M)=\det(M^T)$ and therewith $\det(\lambda I -A^T)=\det((\lambda I-A^T)^T)= \det(\lambda I^T - (A^T)^T)=\det(\lambda I - A)$

Secondly, for an eigenvector $x$ of $A$ with an eigenvalue $\lambda$ what do we get from $Ax = \lambda x\,$?

Now what does it mean for $A$ to be diagonalizable, and what must be the values of this diagonal?

 

[Mr Davis 97]

Firstly, we have to show (or know), that eigenvalues of $A^ {T}$ are the same as those of $A$.
This is done by the equation $\det (M)=\det(M^T)$ and therewith $\det(\lambda I -A^T)=\det((\lambda I-A^T)^T)= \det(\lambda I^T - (A^T)^T)=\det(\lambda I - A)$

Secondly, for an eigenvector $x$ of $A$ with an eigenvalue $\lambda$ what do we get from $Ax = \lambda x\,$?

Now what does it mean for $A$ to be diagonalizable, and what must be the values of this diagonal?

If A is diagonlizable, then it can be represented in the form A = PDP^-1, where P is a matrix of linearly independent eigenvectors of A, and D is a diagonal matrix with the eigenvalues of A on its diagonal. But I'm not seeing how this definition helps me.

Also, I'm not sure what you mean by what do we get from the $Ax = \lambda x\$. The only thing I see to do with that might be to multiply both sides by the transpose of A and see what happens...

 

[fresh_42]

I meant the consideration of a orthogonal matrix $A$. Being orthogonal leads to $x=I\, x=A^TA\, x = A^T(A(x))=\lambda^2 x$ for an eigenvector $x$ of the (orthogonal) matrix $A$ to the eigenvalue $\lambda$. Thus $\lambda^2 = 1$ and over the real numbers this means $\lambda \in \{-1,+1\}$.

So orthogonal matrices must have eigenvalues $\pm 1$.

Now we have the opposite situation: The eigenvalues are given as $\pm 1$, which is a necessary condition. Checked.
Then we have that $A$ is diagonalizable, i.e. $A=PDP^{-1}$ for some matrix $P$.
The diagonal entries of $D$ also have to be $\pm 1$ by the given condition, which means $D=D^T=D^{-1}$ and $A=A^{-1}$.
Thus to show $A^TA=I$ is equivalent to show $A^T=A$ or $A^T=A^{-1}$.

Here is where I'm stuck. To solve for the corresponding linear equations is rather unpleasant. (I'll answer, if I find the trick.)