Decomposition of $4 \times 4$ matrix of rank $3$ I have to show that any $4 \times 4$ matrix of rank $3$ is decomposable as the sum of three $4 \times 4$ matrices of rank $1$, but I have no clue. Can you help me, please?
 A: One way is the SVD (singular value decomposition): any matrix $A$ can be decomposed as $A = \sum_i \sigma_i \mathbf{u}_i \mathbf{v}_i^T$, where $\{\mathbf{u}_i\}$ is an orthonormal collection of vectors, $\{\mathbf{v}_i\}$ is an orthonormal collection of vectors and $\sigma_i>0$. Note that each term in the sum is obviously rank 1 (since it is the outer product of 2 vectors, you can write each row/column as a scaled multiple of another). 
The rank is the number of $\sigma_i$'s. Thus, take the SVD, and one choice of matrices will be $\sigma_1\mathbf{u}_1 \mathbf{v}_1^T, \sigma_2\mathbf{u}_2 \mathbf{v}_2^T, \sigma_3\mathbf{u}_3 \mathbf{v}_3^T$. 
A: Let $T : \mathbb R^4 \to \mathbb R^4$ be the corresponding linear transformation. Since the matrix has rank $3$, the image of $T$ is a 3 dimensional subspace of $\mathbb R^4$.
Let $v_1,v_2,v_3$ be an orthonormal basis in $Im(T)$. Define 
$$P_j : \mathbb R^4 \to \mathbb R^4$$ by
$$P_j(u) ={\rm proj}_{v_j}(u)$$
Since $v_1,v_2,v_3$ is an orthonormal basis in $Im(T)$ we have
$$P_1(u)+P_2(u)+P_3(u)=u ; \forall u \in Im(T)$$
Therefore 
$$T= P_1 \circ T + P_2 \circ T + P_3 \circ T$$
and 
$$Im(P_j \circ T)= \mathbb R v_j$$
therefore $P_j \circ T$ has rank 1.
A: If you are familiar with the SVD decomposition, this can be obtained with its dyadic form. Otherwise, if $V\in \mathbb{R}^{4 \times 4}$ is a matrix of rank 3, then let us write as
$$ V = \begin{bmatrix}
    \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \mathbf{v}_4
  \end{bmatrix}$$
where $\mathbf{v}_j \in \mathbb{R}^{4 \times 1}$ are the column vectors for $j =1,2,3,4$. Since rank$(V) = 3$, $\{ \mathbf{v}_1, \mathbf{v}_2 ,\mathbf{v}_3, \mathbf{v}_4 \}$ is a linear dependent set. Hence, at least one of the $\mathbf{v}_j$ can be expressed as a linear combination of the others. WLG, let us suppose that
$$\mathbf{v}_4 = a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + a_3 \mathbf{v}_3.   $$ Therefore, the matrix V can be expressed as
$$ \begin{array}  VV &=& \begin{bmatrix}
    \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & \mathbf{v}_4
  \end{bmatrix} \\ &=& \begin{bmatrix}
    \mathbf{v}_1 & \mathbf{v}_2 & \mathbf{v}_3 & a_1\mathbf{v}_1 + a_2 \mathbf{v}_2 + a_3 \mathbf{v}_3  
  \end{bmatrix}\\ 
 &=& \underbrace{\begin{bmatrix}
   \mathbf{v}_1 & 0 & 0 & a_1\mathbf{v}_1 
  \end{bmatrix}}_{V_1}+ \underbrace{\begin{bmatrix} 0
     & \mathbf{v}_2 & 0  &   a_2 \mathbf{v}_2 
  \end{bmatrix}}_{V_2} + \underbrace{\begin{bmatrix}
    0 & 0  & \mathbf{v}_3 & a_3 \mathbf{v}_3  
  \end{bmatrix}}_{V_3} \\
&=& V_1 + V_2+V_3 \end{array} $$ 
where the matrices $V_1, V_2$ and $V_3$ have rank 1. 
