Is it necessary that $A$ has $m-1$ eigenvectors corresponding to $\lambda$, which are orthogonal to vector of all ones If $A$ is real symmetric matrix and has an eigenvalue $\lambda$ with multiplicity $m$, Is it necessary that $A$ has $m-1$ eigenvectors corresponding to $\lambda$, which are orthogonal to vector of all ones.
Proof outline : Say $\{x_1,x_2....x_m\}$ are those vectors. Then
$x_i=\alpha_i1+\beta_is_i$ where $1$ is vector of all ones and $s$ is in orthogonal complement of $1$.
Now taking suitable linear combination of $x_1$ with each $x_i ; i\not=1$ we can produce required vectors.
Is this approach a correct one or am I missing some details?
where real symmetric property is used?
 A: Yes, one can formulate it in a more abstract way. If $\mathcal{S}$ is a nonzero subspace of a finite-dimensional inner-product space $\mathcal{V}$ and $x\in\mathcal{V}$ is nonzero, then since $\dim x^\perp=\dim\mathcal{V}-1$, either we have $\dim(\mathcal{S}\cap x^\perp)=\dim\mathcal{S}$ (if $\mathcal{S}\subseteq x^\perp$) or, if $\mathcal{S}$ is not a subspace of $x^\perp$, $\dim(\mathcal{S}\cap x^\perp)=\dim\mathcal{S}-1$.
The first case is obvious (if $\mathcal{S}\subseteq x^\perp$, then $\mathcal{S}\cap x^\perp=\mathcal{S}$). For the other case, we can use the identity
$$
\dim(\mathcal{S}\cap x^\perp)=\dim\mathcal{S}+\dim x^\perp-\dim(\mathcal{S}+x^\perp)=\dim\mathcal{S}+\dim\mathcal{V}-1-\dim(\mathcal{S}+x^\perp)
$$
Obviously, $\dim(\mathcal{S}+x^\perp)\leq\dim\mathcal{V}$. On the other hand, since $\mathcal{S}\not\subseteq x^\perp$, there is a nonzero $y\in\mathcal{S}$ such that $y\not\in x^\perp$. Any vector of $\mathcal{V}$ can be expressed as a linear combination of $y$ and a vector from $x^\perp$, so $\dim\mathcal{V}\leq\dim(\mathcal{S}+y)$. Putting this to the identity above gives
$$
\dim(\mathcal{S}\cap x^\perp)=\dim\mathcal{S}-1.
$$
Now set $x$ to be a vector of all ones and $\mathcal{S}$ to be the eigenspace associated with $\lambda$.
A: The eigenvectors of a real symmetric matrix can be chosen to form an orthonomal basis of $\mathbb{R}^n$.  In particular, the geometric multiplicity of an eigenvalue equals the algebraic multiplicity: if $\lambda$ has multiplicity $m$, then the eigenspace corresponding to $\lambda$ has dimension $m$.  Intersecting this eigenspace with the codimension $1$ subspace of vectors orthogonal to the vector of all ones is a subspace of dimension $\geq m-1$ of vectors which are both eigenvectors with eigenvalue $\lambda$ and orthogonal the vectors of all ones, as you conjectured.  
