Question

If a matrix can be written as \(A=uv^{T}\), where both \(u\) and \(v\) are \(n\)-dimensional real-valued non-zero column vectors, then the rank of the matrix \(A\) is _______ (answer in integer).

Hint:

Every column of \(A=uv^T\) is a scalar multiple of \(u\), so the column space is one-dimensional whenever \(u,v\) are both non-zero.

Answer 1

Correct Answer: 1

Step 1: Recall what an outer product looks like.
Here \(u\) and \(v\) are both \(n\times 1\) non-zero column vectors, so \(v^{T}\) is a \(1\times n\) row vector. The product \(A=uv^{T}\) is then an \(n\times n\) matrix, formed as the outer product of \(u\) and \(v\). Writing it out, the \((i,j)\) entry of \(A\) is \(A_{ij}=u_i v_j\), so every row of \(A\) is just the row vector \(v^{T}\) scaled by the corresponding entry of \(u\).

Step 2: Look at the column space of A.
Every column of \(A\) can be written as \(A_{:,j}=v_j\,u\), i.e. each column is a scalar multiple of the single vector \(u\). So all columns of \(A\) lie along the one direction \(u\); the column space of \(A\) is spanned by just the vector \(u\).

Step 3: Determine the rank.
The rank of a matrix equals the dimension of its column space. Since every column is a multiple of the single non-zero vector \(u\), the column space is one-dimensional (it cannot be zero-dimensional because \(u\) is non-zero, and \(v\) being non-zero guarantees at least one column has a non-zero multiple \(v_j\neq 0\), so \(A\) is not the zero matrix). Therefore the column space has dimension exactly 1.

Final Answer:
The rank of \(A=uv^{T}\), for non-zero vectors \(u\) and \(v\), is always 1. \[ \boxed{\text{rank}(A)=1} \]