explicit formula for the inverse of matrix $A_{i,j}=a_ib_j$

Is there an explicit formula for the inverse of matrix ? $$A_{i,j}=a_ib_j$$

What is this kind of matrix called?

EDIT: It turns out that $\det A=0$ because all the columns are propertional to each other so $A$ has no inverse.

EDIT: This problem is related to another question approximate a function by linear combination of its asympototics with fractional argument.

• You mean $A$ is the product of a column vector $a$ and a row vector $b$ (of the same length)? Such a matrix will not be invertible (except in the trivial case where $a$ and $b$ and therefore $A$ are $1\times 1$, with $a_i, b_j \ne 0$). Commented Jun 5, 2014 at 3:44
• Indeed @ EDIT :P Commented Jun 5, 2014 at 3:47
• @M.Vinay Thanks for the answer. I just used mathematica to invert a 3-by-3 $A$ matrix and found out the $\det A=0$. Commented Jun 5, 2014 at 3:47
• What you can invert explicitely is $\alpha I + \beta A$ for $\alpha\neq 0$. Commented Jun 5, 2014 at 4:39

The matrix can be written as a standard product in vector algebra $\mathbf{a} \mathbf{b}^T$. It is called direct vector product, dyadic product, or rank-1 matrix (unless, of course, $a$ or $b$ are the zero vector). It is also a special case of a Kronecker product.
In case $\mathbf{a} = \mathbf{b}$ and $|\mathbf{a}|=1$ it is also a projection matrix (on a 1d subspace).
I'm not sure that this kind of matrix is called anything, but it must be of the form $$\left[ \begin{array}{c} a_1 \\ \vdots \\ a_n \end{array} \right] [b_1 \ \cdots \ b_n].$$ For such a matrix, notice that any two rows $i$ and $j$ with $a_i,a_j \neq 0$ will be equal to $a_i[b_1 \ \cdots \ b_n]$ and $a_j[b_1 \ \cdots \ b_n]$, and hence multiples of each other. A matrix with one row equal to a multiple of another row cannot be invertible (the rows would not be linearly independent). On the other hand if $a_i = 0$ for some $i$ then there is a zero row in the matrix and again the matrix is not invertible.