Is the following matrix diagonalizable over $\mathbb{R}$?

Question

I am getting conflicting answers depending on how I approach the problem and I don't understand why. Let $A=\begin{pmatrix}2&-5&7\\1&-2&0\\0&0&3 \end{pmatrix}$. If I go directly to the characteristic polynomial I get $\det(A-I\lambda)=\det\begin{pmatrix}2-\lambda&-5&7\\1&-2-\lambda&0\\0&0&3-\lambda \end{pmatrix}$ and then calculating the determinant I get that $\lambda_1=3, \lambda_{2,3}=\pm i$.

I then conclude that $A$ is not diagonalizable over $\mathbb{R}$ as we have non-real eigenvalues.

However if I put $A$ into an upper diagonal matrix I have that $A=\begin{pmatrix}2&-5&7\\0&0.5&-3.5\\0&0&3 \end{pmatrix}$

I then see the eigenvalues as the entries of our upper diagonal matrix, i.e. distinct and real. Then isn't this representation of $A$ diagonalizable? But $A$ hasn't changed. What am I misunderstanding?

Thanks

Answer 1

Your new matrix is equivalent to $A$, but not similar. The latter condition is stronger and required to preserve the Eigenvalues.

Answer 2

The operation you are performing on the matrix $A$ is equivalent to pre-multiplying by another elementary matrix $E$. Just because $EA$ is diagonalizable, does not mean that $A$ is. In particular, as you just demonstrated, the eigenvalues of $A$ and $EA$ are not necessarily the same.

Perhaps a more elementary example of such "innocent" alterations leading to changing eigenvalues are flipping the rows of $$M = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},$$ which results in the identity matrix. The original matrix $M$ has eigenvalues $-1,1$ and the resulting identity has only $1$ instead (with multiplicity $2$)...

Answer 3

That means every matrix of order $n\times n$ with rank $n$ is diagonalizable. But this is certainly not true. The matrix you got after you row reduced your $A$ is not simlar to $A$ since there does not exist any invertible $P$ such that $A=PBP^{-1}$ where $B$ is the row reduced version of $A$