# Eigenvalues of a Real Symmetric Bilinear Form Independent of Basis

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.

• after reading about sylvester's law I suspect such matrices only have the same number of positive, negative and zero eigenvalues Jan 3, 2012 at 2:06
• If a matrix is symmetric, it's Hermitian. Jan 3, 2012 at 2:07
• I just wrote up a counter-example in question 96904. Compare the starting and the last matrices' eigenvalues.
– user13838
Jan 14, 2012 at 5:03
• If the question does not mean a similarity transformation by change of basis that is by the way.
– user13838
Jan 14, 2012 at 5:13

Let $f$ be a symmetric bilinear form on a finite dimensional vector space over a field $K$ whose characteristic in not two and which has at least four elements, and let $A$ be the matrix of $f$ with respect to some basis $B$ of $V$. Prove or disprove: The eigenvalues of $A$ are independent of the choice of $B$.
Putting $$V:=K,\quad f(x,y):=xy,\quad B:=\{b\},$$ we get $A=(b^2)$. In view of the assumptions on $K$, we can choose nonzero $b$ and $b'$ in $K$ so that $b^2\neq b'^2$.
If you regard those $$Q$$ as change of basis, then only rigid transformations (isometries/preserving the inner product) preserve eigenvalue. Otherwise if you scale the principal axes like Gaillard does in his answer - $$b\neq b’$$ - the bilinear form applies a quadratic effect which alters the eigenvalue. Pratically you need $$Q$$ to be unitary.