For an $x \in \mathbb{R}^n$, and $n$-by-$n$ identity matrix $I_n$, we are given that $$ \begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix} \succeq 0.$$

What is the easiest way to show that $$ \begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix} \succeq 0$$ holds?

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    $\begingroup$ Symmetric permutation? $\endgroup$ – user251257 Mar 22 at 18:37
  • $\begingroup$ seems so. is there such a theory which concludes? $\endgroup$ – independentvariable Mar 22 at 18:40
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    $\begingroup$ Schur complement? $\endgroup$ – user251257 Mar 22 at 19:02
  • $\begingroup$ They dont reduce to the same condition, do they? $\endgroup$ – independentvariable Mar 22 at 19:22

This is due to the identity

$$ \underbrace{\begin{pmatrix} 0_{1,n} & 1 \\ I_n & 0_{n,1} \end{pmatrix}}_{J^T} \underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A} \underbrace{\begin{pmatrix} 0_{n,1} & I_n \\ 1 & 0_{1,n} \end{pmatrix}}_{J} = \underbrace{\begin{pmatrix} 1 & x^T \\ x & I_n \end{pmatrix}}_{B} $$

(notation $0_{m,n}$ is for a zero block with $m$ lines and $n$ columns).

Indeed, $J$ being a permutation matrix, it is an orthogonal matrix, with $J^T=J^{-1}$. We can conclude that $A$ and $B$ are similar, thus have the same spectrum (Similar matrices have the same eigenvalues with the same geometric multiplicity) with positive eigenvalues, thus are both semi-definite positive.

Besides, $A$ being symmetric, one can conclude from $J^TAJ=B$ that $B$ is symmetric as well.

Appendix : There is a pending question : is there a criteria on $x$ for positive semi-definiteness of $A$. ? The answer is yes :

$A$ is semi-definite positive iff $\|x\| \leq 1$.

This is due, as we are going to see it, to an analysis of the rather particular spectrum of $A$. Let us obtain it explicitly.

First of all, let us establish that $A$ (which is a $(n+1) \times (n+1)$ matrix) has eigenvalue $1$ with order of multiplicity at least $n-1$.

Consider hyperplane $x^{\perp}$ of $\mathbb{R}^n$ defined as the set of vectors $y$ that are orthogonal to $x$. Let $(y_1,y_2,\cdots y_{n-1})$ be a basis of $x^{\perp}$ ; then,

$$\underbrace{\begin{pmatrix} I_n & x \\ x^T & 1 \end{pmatrix}}_{A}\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k}=1\underbrace{\begin{pmatrix} y_k\\ 0 \end{pmatrix}}_{V_k} \ \ \ \text{for} \ \ k=1,2, \cdots (n-1),$$

proving that $V_k$ is an eigenvector associated with eigenvalue $1$.

Due to the fact that trace$(A)=n+1$, the two remaining eigenvalues are of the form $\alpha$ and $\beta:=2-\alpha$. We can assume, WLOG that $\alpha \leq 1 \leq \beta$.

Besides, using the so-called Schur determinant identity (Eigenvalues of a Block Matrix from Schur Determinant Identity) for the computation of the determinant of a $2 \times 2$ block matrix, we obtain :


As the determinant is also the product of eigenvalues, we get the following identity :


Thus, one can compute explicitly the two remaining eigenvalues by solving quadratic equation (1), with the following explicit solutions (if we assume that $\alpha$ is the smallest eigenvalue)

$$\alpha=1 - \|x\| \ \ \ \implies \ \ \ \beta:=2-\alpha=1 + \|x\|\tag{2}$$

As the criteria for a symmetric matrix do be semi-definite positive is that must have all eigenvalues $\geq 0$, this criteria becomes $\alpha \geq 0$, i.e., $\|x\| \leq 0$. $\square$

Remark : eigenvalues $\alpha$ and $\beta$ can be associated with eigenvectors $\begin{pmatrix} x\\ -\|x\| \end{pmatrix}$ and $\begin{pmatrix} x\\ \|x\| \end{pmatrix}$ resp.

Let us take an example in the case $n=4$ ; let $m=1/n$ ; consider matrix :

$$A:=\left(\begin{array}{rrrr|r} 1 & & & & m \\ & 1 & & & m \\ & & 1 & & m\\ & & & 1 & m\\ \hline m & m & m & m & 1 \\ \end{array}\right)$$

One can check, using (2), that the spectrum of $A$ is

$$(\tfrac12, 1 , 1, 1, 1, \tfrac32).$$

Just now, I "googled" with keywords "bordered identity matrix" : I found in (Eigenvalues of a certain bordered identity matrix) a somewhat similar computation that I did in the Appendix.

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    $\begingroup$ Thank you for your answer! My main purpose was showing $||x||_2 \leq 1$ holds iff the matrices I gave are Psd. I proved one by Schur complement theory on PSD matrics, but for the equivalence I just followed the definition of PSD and by contradiction showed that if one holds, the other one should hold etc.. $\endgroup$ – independentvariable Mar 23 at 19:28
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    $\begingroup$ But yours seem the better way, not the 'dirty' $a^T X a \geq 0 $ for all $a$ approach... I don't like it, it seems too manual $\endgroup$ – independentvariable Mar 23 at 19:35

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