If $A$ is a symmetric, prove that, $dvv^T$ is of rank one.

Question

If $A$ is a symmetric matrix of size $n$, and $v_{n\times 1}$ is eigenvector corresponding to the eigen value $d$ with norm $1$.

Prove that, $$dvv^T$$ is of rank one.

Answer 1

Let $v=(v_1,v_2,\dots,v_n)^T$.

Then all columns of $dvv^T$ are proportional to $v$, the coefficient for column $j$ being $dv_j$. Thus the rank of $dvv^T$ is $1$, whenever $v$ and $d$ are nonzero.

Here, you assume $v$ has norm $1$ so it's nonzero, but the eigenvalue $d$ may be $0$, then $dvv^T$ would have null rank.

Answer 2

By the spectral theorem you can find an orthonormal basis of eigenvectors for $A$. Without loss of generality assume $v$ is normalized. Notice that $v$ is an eigenvector of $dvv^T$ and any vector in the orthogonal complement of $span\{v\}$ is in the kernel.