MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 deleted 78 characters in body

A stronger fact holds: If $\lambda_1\le\lambda_2\le\dots\le\lambda_m$ are the eigenvalues of $A$, and $\mu_1\le\mu_2\le\dots\le\mu_p$ are eigenvalues of $Z^TAZ$, then $\mu_k\ge\lambda_k$ for all $k=1,\dots,p$. Update: This well-known fact is similar to well-known as Cauchy interlacing theorem. I remember that it also bears some name but don't remember which one.

Proof. Let $Q$ be the quadratic form on $\mathbb R^m$ defined by $A$ (that is, $Q(x)=x^TAx$ for all $x\in\mathbb R^m$), $L:\mathbb R^p\to\mathbb R^m$ the (isometric) linear map defined by $Z$. Then $Z^TAZ$ is the matrix of the quadratic form $Q'$ on $\mathbb R^p$ given by $Q'(x)=Q(L(x))$. I suggest you think of $Q'$ as the restriction of $Q$ to the subspace $L(\mathbb R^p)$ of $\mathbb R^m$.

Suppose that $\mu_k<\lambda_k$ for some $k$. Let $V$ be the $k$-dimensional subspace of $\mathbb R^p$ spanned by the first $k$ eigenvectors of $Q'$. Then $Q'(x)\le \mu_k |x|^2$ for all $x\in V$. Let $W$ be the $(m-k+1)$-dimensional subspace of $\mathbb R^m$ spanned by the eigenvectors corresponding to $\lambda_k,\lambda_{k+1},\dots,\lambda_m$. Then $Q(x)\ge\lambda_k|x|^2$ for all $x\in W$. The subspaces $W$ and $L(V)$ have nonzero intersection since the sum of their dimensions is greater than $m$. Hence there exists a nonzro vector $x\in V$ such that $L(x)\in W$. For this vector, we have $Q'(x)\le \mu_k |x|^2 <\lambda_k|x|^2$ but $Q'(x)=Q(L(x))\ge \lambda_k|L(x)|^2=\lambda_k|x|^2$, a contradiction. Q.E.D.

Similarly (inverting all inequalities in the argument) one sees that the $k$-th largest eigenvalue of $Q'$ is no greater than the $k$-th largest eigenvalue of $Q$, answering your second question.

A stronger (and well-known) fact holds: If $\lambda_1\le\lambda_2\le\dots\le\lambda_n$ \lambda_1\le\lambda_2\le\dots\le\lambda_m$are the eigenvalues of$A$, and$\mu_1\le\mu_2\le\dots\le\mu_p$are eigenvalues of$Z^TAZ$, then$\mu_k\ge\lambda_k$for all$k=1,\dots,p$. Update: This well-known fact is similar to Cauchy interlacing theorem. I remember that it also bears some name but don't remember which one. Proof. Let$Q$be the quadratic form on$\mathbb R^n$R^m$ defined by $A$ (that is, $Q(x)=x^TAx$ for all $x\in\mathbb R^n$)R^m$),$L:\mathbb R^p\to\mathbb R^n$R^m$ the (isometric) linear map defined by $Z$. Then $Z^TAZ$ is the matrix of the quadratic form $Q'$ on $\mathbb R^p$ given by $Q'(x)=Q(L(x))$. I suggest you think of $Q'$ as the restriction of $Q$ to the subspace $L(\mathbb R^p)$ of $\mathbb R^n$R^m$. Suppose that$\mu_k<\lambda_k$for some$k$. Let$V$be the$k$-dimensional subspace of$\mathbb R^p$spanned by the first$k$eigenvectors of$Q'$. Then$Q'(x)\le \mu_k |x|^2$for all$x\in V$. Let$W$be the$(n-k+1)$-dimensional (m-k+1)$-dimensional subspace of $\mathbb R^n$ R^m$spanned by the eigenvectors corresponding to$\lambda_k,\lambda_{k+1},\dots,\lambda_n$. \lambda_k,\lambda_{k+1},\dots,\lambda_m$. Then $Q(x)\ge\lambda_k|x|^2$ for all $x\in W$. The subspaces $W$ and $L(V)$ have nonzero intersection since the sum of their dimensions is greater than $n$. m$. Hence there exists a nonzro vector$x\in V$such that$L(x)\in W$. For this vector, we have$Q'(x)\le \mu_k |x|^2 <\lambda_k|x|^2$but$Q'(x)=Q(L(x))\ge \lambda_k|L(x)|^2=\lambda_k|x|^2$, a contradiction. Q.E.D. Similarly (inverting all inequalities in the argument) one sees that the$k$-th largest eigenvalue of$Q'$is no greater than the$k$-th largest eigenvalue of$Q$, answering your second question. 1 A stronger (and well-known) fact holds: If$\lambda_1\le\lambda_2\le\dots\le\lambda_n$are the eigenvalues of$A$, and$\mu_1\le\mu_2\le\dots\le\mu_p$are eigenvalues of$Z^TAZ$, then$\mu_k\ge\lambda_k$for all$k=1,\dots,p$. Proof. Let$Q$be the quadratic form on$\mathbb R^n$defined by$A$(that is,$Q(x)=x^TAx$for all$x\in\mathbb R^n$),$L:\mathbb R^p\to\mathbb R^n$the (isometric) linear map defined by$Z$. Then$Z^TAZ$is the matrix of the quadratic form$Q'$on$\mathbb R^p$given by$Q'(x)=Q(L(x))$. I suggest you think of$Q'$as the restriction of$Q$to the subspace$L(\mathbb R^p)$of$\mathbb R^n$. Suppose that$\mu_k<\lambda_k$for some$k$. Let$V$be the$k$-dimensional subspace of$\mathbb R^p$spanned by the first$k$eigenvectors of$Q'$. Then$Q'(x)\le \mu_k |x|^2$for all$x\in V$. Let$W$be the$(n-k+1)$-dimensional subspace of$\mathbb R^n$spanned by the eigenvectors corresponding to$\lambda_k,\lambda_{k+1},\dots,\lambda_n$. Then$Q(x)\ge\lambda_k|x|^2$for all$x\in W$. The subspaces$W$and$L(V)$have nonzero intersection since the sum of their dimensions is greater than$n$. Hence there exists a nonzro vector$x\in V$such that$L(x)\in W$. For this vector, we have$Q'(x)\le \mu_k |x|^2 <\lambda_k|x|^2$but$Q'(x)=Q(L(x))\ge \lambda_k|L(x)|^2=\lambda_k|x|^2$, a contradiction. Q.E.D. Similarly (inverting all inequalities in the argument) one sees that the$k$-th largest eigenvalue of$Q'$is no greater than the$k$-th largest eigenvalue of$Q\$, answering your second question.