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I require for a proof that if a n x n real symmetric matrix (of rank n) has n distinct non-zero eigenvalues, a principal submatrix, in particular the one obtained by removing the first row and first column, has (n-1) distinct non-zero eigenvalues.

The following is addendum:

I have managed to find however, that if a real symmetric matrix is of full rank n, the principal submatrix obtained by removing the first row and first column, is not necessarily of rank n-1, even if it must be of at least rank n-2. This would mean that some principal submatrices of this form would not have n-1 distinct non-zero eigenvalues.

I do so by constructing a counterexample since by performing Gaussian elimination on

$A_1=\begin{pmatrix} a & b & c \\ b & d & d \\ c & d & d\end{pmatrix}$

We obtain

$A_2=\begin{pmatrix} 0 & 0 & c-b \\ b-c & 0 & 0 \\ 0 & d & 0\end{pmatrix}$

Thus the matrix is of full rank even though in A1 the bottom right principal submatrix is degenerate. I have empirical evidence, using wolfram alpha to calculate the eigenvalues, that 2 of the eigenvalues of this matrix are identical. This supports my idea that if a real symmetric matrix has a principal submatrix not of full rank, it would not have distinct non-zero eigenvalues.

However, a matrix of full rank may not have distinct non-zero eigenvalues, so I was hoping that given the condition it does and if it is symmetric, somehow this would force the principal submatrix, which is also symmetric, to be of full rank and also have to full spectrum of distinct eigenvalues.

So quite obviously, my question is, how does one go about proving it?

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    So what is the question? – 2017-01-26
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    *"if a real symmetric matrix is of full rank n, the principal submatrix obtained by removing the first row and first column, is not necessarily of rank n-1, even if it must be of at least rank n-2. This would mean that some principal submatrices of this form would not have n-1 distinct eigenvalues."* How exactly do you think rank is related to eigenvalues? – 2017-01-26
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    The bottom right submatrix has the distinct eigenvalues $2d$ and $0$ – 2017-01-26
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    @Omnomnomnom The rank is a bound on the maximum number of distinct eigenvalues a matrix can have. Thus, if a matrix is of rank n-2, it cannot have n-1 distinct eigenvalues. That much is clear. – 2017-01-26
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    @Rascalniikov That much is not clear; in fact, it's wrong. All we know is that the matrix with size $n$ and rank $r$ has the eigenvalue $0$ with geometric multiplicity $n-r$. For example, the matrix $$ \pmatrix{1&1\\1&1} $$ has the distinct eigenvalues $0,2$. – 2017-01-26
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    @Omnomnomnom Indeed, What I meant was distinct, non-zero eigenvalues. I will revise the question. – 2017-01-26

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The answer: the Cauchy interlacing theorem which is presented here for hermitian matrices but, that, of course, can be "downgraded" to real symmetric matrices.

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    I see... as a matter of fact, the proof I was working on, which required the crucial step presented in the question, was basically, as I have just found out, the Cauchy interlacing theorem. This means that I wanted to prove the interlacing theorem using one of its corollaries. Sadly, this means there was another way to tackle to proof, and I simply did not see it. – 2017-01-26
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    I see... Is [Gauss-Lucas theorem](https://en.wikipedia.org/wiki/Gauss%E2%80%93Lucas_theorem) of any help ? – 2017-01-26