Let $A$ be a real symmetric matrix of order $n$ with eigenvalues $\mu_1\ge \mu_2,\ldots, \mu_{n}$. $B=\begin{bmatrix} x&x&x&x&x \end{bmatrix}$, where $x$ is non zero column vector in $R^n$. Construct a new matrix: $$C=\begin{bmatrix} A&B\\B^T &0 \end{bmatrix}$$ Removing four zeros of $C$ (due to identical columns) arrange the remaining eigenvalues as $\lambda_1\ge \lambda_2,\ldots, \lambda_{n+1}$.

How to show that $\lambda_i\ge\mu_i\ge\lambda_{i+1}$ for $i=1,2,\ldots,n$.

I know that removing $4$ identical columns, we are left with matrix of order $n+1$ and there this result hold due to interlacing. But eigenvalues of that matrix and of matrix $C$ are different. So , how can I conclude here?


1 Answer 1


You almost solved your problem. The only thing that you forgot to do was a change a basis.

Let $e_1,e_2,\ldots,e_{n+5}$ denote the canonical basis of ${\mathbb R}^{n+5}$. The identical columns in $C$ force the subspace $Z$ spanned by $e_{k}-e_{n+1} (n+2\leq k \leq n+5)$ to be included in the kernel of $C$. We construct an orthonormal basis in which $Z$ will be spanned by the last, rightmost vectors (so we can concentrate on the interesting leftmost part).

Formally, define a new basis $F=(f_1,f_2,\ldots,f_{n+5})$ by $f_k=e_k$ for $k\leq n$, and

$$ \begin{array}{lcl} f_{n+1} &=& \frac{e_{n+1}+e_{n+2}+e_{n+3}+e_{n+4}+e_{n+5}}{\sqrt{5}} \\ f_{n+2} &=& \frac{e_{n+1}+e_{n+2}+e_{n+3}+e_{n+4}-4e_{n+5}}{\sqrt{20}} \\ f_{n+3} &=& \frac{e_{n+1}+e_{n+2}+e_{n+3}-3e_{n+4}}{\sqrt{12}} \\ f_{n+4} &=& \frac{e_{n+1}+e_{n+2}-2e_{n+3}}{\sqrt{6}} \\ f_{n+5} &=& \frac{e_{n+1}-e_{n+2}}{\sqrt{2}} \\ \end{array} $$

Then $F$ is orthonormal. In this new basis, $A$ has not changed while $C$ has become

$$ C'=\begin{bmatrix} D&0\\0 &0 \end{bmatrix} $$

where $D$ is the $(n+1)\times(n+1)$ matrix

$$ D=\begin{bmatrix} A&\sqrt{5}x\\\sqrt{5}x &0 \end{bmatrix} $$

The eigenvalues of $D$ are exactly the $\lambda_1\ge \lambda_2,\ldots \geq \lambda_{n+1}$. Then the usual interlacing property you mention in your post applies.

  • $\begingroup$ Further, can we say something about strict interlacing? Can we restrict $A$ such that there is strict interlacing? $\endgroup$ May 2, 2016 at 6:49
  • $\begingroup$ It is likely that the interlacing will be strict in $C$ if it is already strict in $A$ and $0$ is not an eigenvalue for $A$. $\endgroup$ May 2, 2016 at 7:27
  • $\begingroup$ Do you mean that if all eigenvalues of $A$ i.e $\mu_i$'s are simple and none of them is zero, then $\lambda_i>\mu_i>\lambda_{i+1}$? $\endgroup$ May 2, 2016 at 7:36
  • $\begingroup$ We can say Interlace of eigenvalues of $A$ with $C$. But what does interlacing in $A$ itself mean? $\endgroup$ May 2, 2016 at 7:39
  • $\begingroup$ You’re right, I shouldn't have sad "interlacing in $A$" but rather simple eigenvalues for $A$. And yes, I presume that $\lambda_i > \mu_i > \lambda_{i+1}$ in this case (though I don't know how to prove it). $\endgroup$ May 2, 2016 at 8:17

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