Spectral Decomposition Range Let  $A_k=\lambda_1 u_1u_1^T+\cdots +\lambda_k u_ku_k^T$. 
$A$ is a symmetric matrix, $k$ is an integer, and $u$ are unit eigenvectors of $A$.
$A_k$ is a partial reconstructed matrix of the original symmetric matrix A up to k parts. 
My textbook says that the range of the corresponding linear transformation has dimension $k$. Why is that the set of all possible images for the partial decomposition have $k$ possible vectors?
Also why does each 'part'of the decomposition have rank 1?
 A: Suppose $A$ is an $n \times n$ symmetric matrix, and suppose that $\{ u_1, \dots, u_n \}$ is an orthonormal basis of eigenvectors for $A$, whose corresponding eigenvalues are $\lambda_1, \dots, \lambda_n$.
Let's think about how the matrix $A_k$ acts on vectors. Since $\{u_1, \dots, u_n\}$ is a basis, we know that any vector $v \in \mathbb R^n$ can be written in the form
$$ v = c_1 u_1 + \dots + c_k u_k + c_{k+1}u_{k+1} + \dots + c_n u_n.$$
Using the fact that $u_i^T u_j$ is equal to $1$ if $i = j$ and is equal to $0$ otherwise, we can show that
$$ A_k v = \lambda_1 c_1 u_1 + \dots \lambda_k c_k u_n.$$
So the image of $A_k$ is precisely the span of $\{ u_1, \dots, u_k \}$ (assuming that all the $\lambda_k$'s are non-zero). Hence $A_k$ has rank $k$.
To address your final question, an individual contribution $\lambda_j u_ju_j^T$ (for fixed $j$) maps $v = \sum_{i=1}^n c_i u_i$ to the vector $\lambda_j c_j u_j$. So the image of $\lambda_j u_ju_j^T$ is precisely the one-dimensional vector space spanned by $u_j$ (assuming that $\lambda_j$ is non-zero), hence this linear transformation has rank one.
