Finding the difference in dimension for Null Spaces We have $N(A + vv^T) = N(A) ∩ N(vv^T)$ where $N$ represents the null space. A is a symmetric positive semidefinite matrix. We know that $\mathrm{dim}(R(vv^T)) = 1$, and therefore $\mathrm{dim}(N(vv^T)) = n-1$. However, does anyone know why $\mathrm{dim}(N (A + vv^T
)) ≥ \mathrm{dim}(N (A)) − 1$ is true based on the facts given?
Given , $N(A+vv^{T})=N(A)∩N(vv^{T})$
And, $dim(N(vv^{T}))=n-1$
$dim(N(A+vv^{T}))=dim(N(A)∩N(vv^{T}))=dim(N(A))+dim(N(vv^{T}))-dim(N(A)+N(vv^{T}))$
$\implies dim(N(A+vv^{T}))= dim(N(A))+(n-1)-dim(N(A)+N(vv^{T}))$
$\implies dim(N(A+vv^{T}))= dim(N(A))-1+(n-dim(N(A)+N(vv^{T})))$
Now, since , $N(A) $ and $N(vv^{T})$ are subspaces of some space $\mathbb{F}^{n}$ of dimension $n$.
So, $N(A)+N(vv^{T})$ is also subspace of $\mathbb{F}^{n}$.
Hence, $dim(N(A)+N(vv^{T}))\le n $
So, $dim(N(A+vv^{T}))\ge dim(N(A))-1$
Update
Actually the statement is true even without $N(A + vv^T) = N(A)$ $\cap$ $N(vv^T)$ or symmetric positive semidefiniteness as the following proof demonstrates. In fact this easily follows form A learner's answer because while they prove $dim(N(A)$ $\cap$ $N(vv^T))\geq ndim(N(A))-1$, we know that $$N(A) \cap N(vv^T) \subseteq N(A + vv^T)$$

Example
As an example suppose $v=e_5=\begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}$, and $A=\begin{bmatrix}1&amp;0&amp;0&amp;0&amp;0\\1&amp;0&amp;0&amp;0&amp;0\\1&amp;0&amp;0&amp;0&amp;0\\0&amp;1&amp;0&amp;0&amp;0\\0&amp;0&amp;1&amp;0&amp;0\end{bmatrix}$
then $N(vv^T):=span(e_1, e_2, e_3, e_4)$, $N(A):=span(e_4, e_5)$ and $N(A+vv^T):=span\left(e_4, \begin{bmatrix}0\\0\\1\\0\\-1\end{bmatrix}\right)$
which is not the intersection of the first two and yet $dim(N(A+vv^T))=2\geq 1=dim(N(A))-1$.

This is a prescriptive proof. We will find an orthonormal basis $B$ of $N(A + \mathbf{v}\mathbf{v}^T)$ with $|B|\geq dim(N(A))-1$
Let $k=dim(N (A))$. Let an orthonormal basis of the $N(A)$ be denoted by the set $\{a_1, a_2, a_3...a_{k}\}$. Firstly, if $\mathbf{v}\cdot a_i=0$ for all $a_i$, $B:=\{a_1, a_2, ..., a_k\}$ and $|B|=k$. Let us assume this is not so. 
Then the following vector is non-zero: $$\mathbf{v_{\parallel}}=\sum\limits_{i=1}^{k}(\mathbf{v}\cdot a_i)\,a_i$$
Notice that $\mathbf{v_{\perp}} = \mathbf{v}-\mathbf{v_{\parallel}}$ is such that for all $a_i$, we have $\mathbf{v_{\perp}}\cdot a_i=0$. 
Now the set $\{a_1, a_2, a_3...a_{k}, \mathbf{v_{\parallel}}\}$ is clearly linearly dependant but spans a subspace of dimension only $k$. We now drop any one of the $a_i$s, where $\mathbf{v}\cdot a_i\neq 0$, say $a_1$ was one such vector. Now we do a Gram Schmidt orthogonalization on $\{a_2, a_3...a_{k}, \mathbf{v_{\parallel}}\}$ to get a basis with $\mathbf{v_{\parallel}}$ in it. Consider the vectors in this new basis other than $\mathbf{v_{\parallel}}$, let's say $\{a'_1, a'_2, a'_3...a'_{k-1}\}$. Now each $a'_j$ is in $N(A)$ because $a'_j$ must be in the span of $\{a_2, a_3..., a_{k}, \mathbf{v_{\parallel}}\}$ and thus in the span of $\{a_1, a_2, a_3...a_{k}\}$. But since $a'_j\cdot\mathbf{v_{\parallel}}=0$, notice that $a'_j\cdot\mathbf{v} = a'_j\cdot(\mathbf{v_{\parallel}}+\mathbf{v_{\perp}}) = a'_j\cdot\mathbf{v_{\perp}}=0$. The last equality is true because $a'_j$ is in the span of $\{a_1, a_2, a_3...a_{k}\}$ and as we noted, $\mathbf{v_{\perp}}\cdot a_i=0$ for each $a_i$. These thus constitute $B$ and $|B|=k-1$