Proving Matrix Similarity: Trace vs. Determinant Comparison

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To prove that two matrices A and B are similar, it is insufficient to show that Trace(A) equals Trace(B) or that Det(A) equals Det(B). Counterexamples demonstrate that matrices can share the same trace or determinant yet remain dissimilar. Specifically, matrices like the identity matrix and other forms can have equal traces or determinants without being similar. The necessary condition for similarity is that both matrices must have the same eigenvalues and the same number of independent eigenvectors for each eigenvalue. Thus, eigenvalue analysis is essential for establishing matrix similarity.
talolard
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If I have two matrices A and B and I want to show they are similar, is it enough to show that Trace(A)=Trace(B) or instead show that Det(A)=Det(B)?
Thanks
Tal
 
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No, it isn't. For example, both
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
and
\begin{bmatrix}\frac{1}{2} & 0 \\ 0 & 2\end{bmatrix}
have the same determinant but are not similar.

Also both
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
and
\begin{bmatrix}2 & 0 \\ 0 & 0\end{bmatrix}
have the same trace but are not similar.

And, just in case you were wondering,
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
and
\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}
have the same determinant and the same trace but are not similar.

In order to be similar, two matrices must have the same eigenvalues and the same number of independent eigenvectors corresponding to each eigenvalue.
 
Great, Thanks.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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