Eigenvalues and Diagonalization

I have a few miscellaneous questions here that I started to do but am unsure about; any clarification would be great.

The first is "Let A be the matrix of the linear transformation T. Without writing A, find an eigenvalue of A and describe the eigenspace. T is the transformation on R3 that rotates points about some line through the origin."

My original assumption was that A would not have any eigenvalues because it is a rotation. But someone told me that because A is in R3 and not R2, any point on the line of rotation would be an eigenvalue because it would be the same after the transformation. Could someone confirm this or perhaps clarify it? Do rotation transforms only not contain eigenvalues in R2?

The second involved two True/False questions, and the statements are as follows: (Assume A, B,P, and D are n x n matrices)

1. If AP = PD, with D diagonal, then the nonzero columns of P must be eigenvectors of A.

2. If A is invertible, then A is diagonalizable.

At first I was inclined to think that 1 is true because it corresponds to a theorum in my my textbook that says if A = (P)(D)(P inverse), then A is similar to D and that this means the nonzero columns of P are in fact eigenvectors of A. But this only holds if P is invertible, which was not stated in the given information. Since this was not stated I thought the statement should be false, but I'm not 100% certain. Any insight?

As for 2, I'm pretty sure that it is true because all invertible matrices are similar to the identity matrix, which is a diagonal matrix. Is this sufficient to prove that statement, or is this not accurate?

Re: Eigenvalues and Diagonalization

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**Chaobunny** The first is "Let A be the matrix of the linear transformation T. Without writing A, find an eigenvalue of A and describe the eigenspace. T is the transformation on R3 that rotates points about some line through the origin."

If $\displaystyle V$ is the line passing through the origin then, $\displaystyle T(x)=x=1x$ for all $\displaystyle x\in V$ and there exists at least one $\displaystyle 0\neq x\in V$ that is, $\displaystyle \lambda=1$ is an eigenvalue of $\displaystyle T$ and the corresponding eigenspace is $\displaystyle V$ .

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My original assumption was that A would not have any eigenvalues because it is a rotation. But someone told me that because A is in R3 and not R2, any point on the line of rotation would be an eigenvalue because it would be the same after the transformation. Could someone confirm this or perhaps clarify it? Do rotation transforms only not contain eigenvalues in R2?

Rotations around the origin in $\displaystyle \mathbb{R}^2$ and angle $\displaystyle \theta$ have no eigenvalues if $\displaystyle \theta\neq 0$ and $\displaystyle \theta\neq \pi$ . If $\displaystyle \theta=0$ we get the identity map $\displaystyle I$ (eigenvalue $\displaystyle \lambda=1$ double). If $\displaystyle \theta=\pi$ we get $\displaystyle -I$ (eigenvalue $\displaystyle \lambda=-1$ double).

Re: Eigenvalues and Diagonalization

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Originally Posted by

**Chaobunny** 1. If AP = PD, with D diagonal, then the nonzero columns of P must be eigenvectors of A.

False. It is easy to find a counterexample.

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2. If A is invertible, then A is diagonalizable.

Also false. Choose for example $\displaystyle A=\begin{bmatrix}{1}&{1}\\{0}&{1}\end{bmatrix}$

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As for 2, I'm pretty sure that it is true because all invertible matrices are similar to the identity matrix, which is a diagonal matrix. Is this sufficient to prove that statement, or is this not accurate?

All invertible $\displaystyle n\times n$ matrices are **equivalent** (not necessarily similar) to the identity matrix.