Doubts about a question I asked a long time ago (eigenvalues)

Rank $1$: For any vector $a$, note that $v^Ta$ is a scalar, so $vv^Ta$ is a scalar multiple of $v$. So the range of $vv^T$ is within the span of $v$, which is one-dimensional, so $vv^T$ has rank $1$.

All but $1$ eigenvalue is $0$: Since the rank of $vv^T$ is $1$, the nullspace has dimension $n-1$, by the rank-nullity theorem. That means we can find $n-1$ linearly independent vectors in the nullspace. Since every vector in the nullspace has eigenvalue $0$, $0$ is an eigenvalue with multiplicity $n-1$. That leaves room for only one more eigenvalue, which we have already shown is $\sum x_i^2$.

Diagonalizability: As we have shown above, we can find $n-1$ linearly independent eigenvectors of $0$. We can also find one eigenvector of $\sum x_i^2$, to make a basis in which $vv^T$ is diagonal.

A is diagonizable because it is real valued and symmetric $A^T=A$ (This is the spectral theorem)

$A=vv^T$ so $A^T=(vv^T)^T=(v^T)^Tv^T=vv^T=A$

https://en.wikipedia.org/wiki/Spectral_theorem

To see that $v v^T$ has rank at most one, we see that for any arbitrary vector $u$ ,we have: $$ (vv^T)x = v(v^Tx) $$ The right hand side is a scalar times $v$. So every vector $x$ is mapped to a scalar multiple of $v$, with the scalar determined by $x$, meaning that the linear transformation represented by $vv^T$ has rank $1$ and so $vv^T$ has rank $1$.