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Question is from Artin's Algebra, p. 263.

If $A$ is the matrix of a symmetric bilinear form, prove or disprove: The eigenvalues of $A$ are independent of the choice of basis.

I suspect the result is true.

Real & Symmetric $\Rightarrow$ Hermitian

Then by Corollary (4.12) the matrices which represent the same hermitian form are $= QAQ^*$, where $Q\in \operatorname{GL}_n(\mathbb{C})$.

Does this mean that all of these matrices are similar? If so I would be done since similar matrices have the same eigenvalues.

  • 3
    after reading about sylvester's law I suspect such matrices only have the same number of positive, negative and zero eigenvalues2012-01-03
  • 1
    If a matrix is symmetric, it's Hermitian.2012-01-03
  • 0
    I just wrote up a counter-example in [question 96904](http://math.stackexchange.com/questions/96904). Compare the starting and the last matrices' eigenvalues.2012-01-14
  • 1
    If the question does not mean a similarity transformation by *change of basis* that is by the way.2012-01-14

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