Inverse of a matrix!

Question

Let $I_n$ be the $n$ by $n$ identity matrix and $b$ and $c$ be two vectors in ${\mathbb R}^n$ such that $b^Tc\ne 0$. Then one can easily see that the $n+1$ by $n+1$ matrix $X$ defined as following $$ X = \begin{bmatrix} I_n&b\\c^T& 0 \end{bmatrix} $$ has an inverse matrix $X^{-1}$. Is the any close and simple formula for $X^{-1}$?

Answer 1

Following the link in the comment, we have $$ X^{-1} = \frac{1}{c^Tb}\pmatrix{ (c^Tb)I - bc^T & b\\ c^T & -1} $$

Answer 2

Wiki does not show how to get the solution. Here I give a simple way to prove it using the idea of row operations to get the inverse $$ (X,I_{n+1})\longrightarrow(I_{n+1}, X^{-1}) $$ First multiply a left matrix whose role is similar to elementary row operation of adding the 1st row multiplied by $-c^T$ to the 2nd row. So we have $$ \pmatrix{I_n&0\\-c^T&1}\pmatrix{I_n&b&I_n&0\\c^T&0&0&1}=\pmatrix{I_n&b&I_n&0\\0&-c^Tb&-c^T&1} $$ Then multiply a left matrix whose role is similar to elementary row operation of adding the 2nd row multiplied by $b/(c^Tb)$ to 2nd row. So we get $$ \pmatrix{I_n&b/(c^Tb)\\0&1}\pmatrix{I_n&b&I_n&0\\0&-c^Tb&-c^T&1}=\pmatrix{I_n&0&I_n-bc^T/(c^Tb)&b/(c^Tb)\\0&-c^Tb&-c^T&1} $$ Finally multiply a left matrix whose role is similar to elementary row operation of multiplying $-1/(c^Tb)$ to the 2nd row. So we have $$ \pmatrix{I_n&0\\0&-1/(c^Tb)}\pmatrix{I_n&0&I_n-bc^T/(c^Tb)&b/(c^Tb)\\0&-c^Tb&-c^T&1}=\pmatrix{I_n&0&I_n-bc^T/(c^Tb)&b/(c^Tb)\\0&1&c^T/(c^Tb)&-1/(c^Tb)} $$ So we have $$ X^{-1}=\frac1{(c^Tb)}\pmatrix{(c^Tb)I_n-bc^T&b\\c^T&-1} $$

Answer 3

Let's look for an inverse of the form $$ \begin{bmatrix} A & x \\ y^T & z \end{bmatrix} $$ and do block multiplication: $$ \begin{bmatrix} I_n & b \\ c^T & 0 \end{bmatrix} \begin{bmatrix} A & x \\ y^T & z \end{bmatrix} = \begin{bmatrix} I_nA+by^T & I_nx+bz \\ c^TA & c^Tx \end{bmatrix} $$ Then we get $$ \begin{cases} A+by^T = I_n\\[3px] x+bz = 0 \\[3px] c^TA = 0 \\[3px] c^Tx = 1 \end{cases} $$ The first gives $A=I_n-by^T$, so $c^T-c^Tby^T=0$ and therefore $y^T=d^{-1}c^T$ (where $d=c^Tb=b^Tc\ne0$, by hypothesis).

This also determines $A=I-b(d^{-1})c^T=I-d^{-1}bc^T$.

Now multiply the second equation by $c^T$, so $c^Tx+c^Tbz=0$, or $1+dz=0$, so $z=-d^{-1}$ and $x=-d^{-1}b$. Thus $$ X^{-1}= \begin{bmatrix} I-d^{-1}bc^T & -d^{-1}b \\ d^{-1}c^T & -d^{-1} \end{bmatrix} $$

Note also that the condition $c^Tb\ne0$ is necessary for invertibility, because $$ \begin{bmatrix} I_n & b \\ c^T & 0 \end{bmatrix} \begin{bmatrix} b\\ -1 \end{bmatrix} = \begin{bmatrix} 0\\ c^Tb \end{bmatrix} $$