# Is this a kind of “sketching”?

Say one is given a matrix (assume real and symmetric if necessary) and its $n-$dimensional columns be say $v_1,v_2,..,v_n$. Now is it possible to find a set of $d<n$ lower dimensional vectors ($w_1,..,w_d$) such that they form a $d \times d$ matrix $B$ (assume real and symmetric) with either of these properties,

• That the $d$ eigenvalues of $B$ are "close" to the top$-d$ eigenvalues of $A$? (and hence the correspomding eigenvectors are also "close" in some sense?)

• That the spectral norm of $B$ is "close" to the spectral norm of $A$.

Will this be called "sketching"? Does Johnson-Lindenenstrauss lemma help in doing these?

No, matching the eigenvalues doesn't imply that the eigenvectors of $B$ are somehow related in any way to the eigenvectors of $A$.
Given a value $n$ you can easily construct a matrix $B$ whose spectral norm is $n$, so that too can be exactly matched.