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While solving Linear algebra and Its application by Gilbert Strang, I am not getting any idea how to solve the problem 6.4.14, which says

From the zero submatrix decide the signs of the $n$ eigenvalues: $$\pmatrix{0&.&0&1 \\ .&.&0&2 \\0&0&0&.\\1&2&.&n}$$

Since the matrix has rank $2$, $n-2$ eigenvalues will be zero. Only two remains to be decided. How to determine the sign of other two.

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  • $\begingroup$ do you know how to compute determinants by cofactor expansion? $\endgroup$ – Jon Warneke Mar 17 '16 at 18:10
  • $\begingroup$ @PaulSinclair: That's consistent with rank 2. The top $n-1$ rows only give rank $1$ on their own. $\endgroup$ – Greg Martin Mar 17 '16 at 18:25
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What I am going to write is not fully rigorous, but this site is also a place where people practising mathematics show sometimes how they have an intuitive grasp on certain situations.

I have thought a certain time asking myself what do they mean when they say "from the zero submatrix decide the signs of the eigenvalues", and the interlacing property came me back.

The important result we are going to apply (that you may have not seen) is the Cauchy interlacing theorem for real symmetrical (or hermitian) matrices ; see the theorem 2.3 in http://www.caam.rice.edu/~caam440/chapter2.pdf): when one adds a south-east border to a $(n-1) \times (n-1)$ matrix having distinct eigenvalues $\lambda_i$ i.e., adds it a n-th row and and a n-th column, the new "bordered" $n \times n$ matrix has eigenvalues $\mu_j$ such that:

$$\mu_1<\lambda_1<\mu_2 <\cdots<\lambda_{n-1}<\mu_n \ \ \ (1)$$

Distinct eigenvalues, you said; but here we have bordered a zero matrix, thus all its eigenvalues $\lambda_1=\lambda_2= \cdots =\lambda_{n-1}$ are zero? All right, let us do a classical trick: we perturbate this zero matrix (for example by adding it a diagonal matrix $diag(\epsilon, -\epsilon^2, \cdots(-1)^{n-1}\epsilon^{n-1}$) with distinct values which besides are its eigenvalues; in this way, using (1), $\mu_1<0$ and $\mu_n>0$. And now we do $\epsilon \rightarrow 0$. Using a continuity argument of the eigenvalues with respect to the matrix coefficients, that we don't want to discuss here, we will still have $\mu_1<0$ and $\mu_n>0$ at the limit (it is impossible that they tend to zero because the rank of the matrix is 2).

Once again, all this has to be made more rigorous...

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Find the eigenvalues using the definition: $$\pmatrix{0&.&0&1 \\ .&.&0&2 \\0&0&0&.\\1&2&.&n}\pmatrix{v_1\\ v_2\\.\\v_n} = \lambda \pmatrix{v_1\\ v_2\\.\\v_n} \Rightarrow \\ \begin{cases} v_n & = \lambda v_1 \\ 2v_n & = \lambda v_2 \\ \ldots \\ (n-1)v_n & = \lambda v_{n-1} \\ \sum_{i=1}^n iv_i & = \lambda v_n \end{cases}.$$

From the first $n-1$ equations, we get that:

$$v_i = \frac{iv_n}{\lambda},$$

with $\lambda \neq 0$. Using the last equation of the system, we have:

$$\sum_{i=1}^n iv_i = \lambda v_n \Rightarrow \sum_{i=1}^{n-1} i\frac{iv_n}{\lambda} + nv_n = \lambda v_n \Rightarrow \\ \frac{v_n}{\lambda}\sum_{i=1}^{n-1} i^2 + nv_n = \lambda v_n \Rightarrow \sum_{i=1}^n i^2 = \lambda(\lambda -n) \Rightarrow\\ \lambda^2 - \lambda n - \sum_{i=1}^n i^2 = 0.$$

Using Descartes rule, we get that the previous equation has one positive and one negative solution ($n$ and $\sum_{i=1}^ni^2$ are both positive).

Concluding: you have $n-2$ null eigenvalues, 1 positive eigenvalue and 1 negative eigenvalue.

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