Traces and determinant identity

Question

$A$ is a $2\times2$ matrix with $\det(A)=0$ and tr(A) not equal to -1.

How to prove that:

$\displaystyle (I+A)^{-1}=I-\frac{1}{1+\operatorname{tr}(A)}A$.

Answer 1

Let $A$ be a $2\times 2$ matrix with $\text{det}A=0$, then $A$ has one eigenvalue which is zero and another eigenvalue which is real, let's call is $\lambda$. Then $\text{tr}A = 0+\lambda = \lambda$ and the characteristic polynomial is $$c_A(x)=x(x-\lambda)=x^2-\lambda x$$ so by the Cayley-Hamilton theorem: $$A^2=\lambda A.$$ By definition of the inverse, $(I+A)^{-1}(I+A)=I$, and $$\left( I-\frac{1}{1+\lambda }A \right)(I+A)=I^2-\frac{1}{1+\lambda }A+A-\frac{1}{1+\lambda}A^2$$ Then plugging in $A^2=\lambda A $, the expression above becomes $$I+ \left(1-\frac{1}{1+\lambda}-\frac{\lambda}{1+\lambda} \right)A=I+\left(1-1 \right)A=I$$ therefore $(I+A)^{-1}=\left(I-\frac{1}{1+\lambda}A \right)$ provided that $\lambda \neq -1$.

Answer 2

What you have described is a rank-1 matrix.
The defining property of a rank-1 matrix is $$A^2={\rm tr}(A)\,A = t A$$ The related matrix $P=\frac{A}{t}$ is idempotent, i.e. $P^2=P$.
The eigenvalues of an idempotent matrix are either $0$ or $1$.

Using Cayley-Hamilton, we can write any function of a 2x2 matrix as a quadratic polynomial $$f(X) = \alpha_2X^2 + \alpha_1X + \alpha_0I$$ In the case of an idempotent matrix, this reduces to a linear polynomial $$f(P) = \alpha_1P + \alpha_0I$$ whose $\alpha$-coefficients can be determined by substituting the eigenvalues into the scalar version of the polynomial. This is known as Hermite-Sylvester interpolation.

Using the function $f(\lambda) = (t\lambda+1)^{-1}$ and the eigenvalues $\lambda=\{0,1\}$ leads to $$\eqalign{ f(0) &= 1 = \alpha_0 \cr f(1) &= \frac{1}{t+1} = \alpha_1 + \alpha_0 \cr }$$ yielding the coefficients $$\eqalign{\alpha_0&=1 ,\,\,\, \alpha_1&=-\frac{t}{t+1}}$$ So the matrix function is $$\eqalign{ (tP+I)^{-1} &= I - \frac{tP}{t+1} \cr }$$ or in terms of the original matrix $$\eqalign{ (A+I)^{-1} &= I - \frac{A}{t+1} \cr\cr }$$ This may seem like a lot of trouble, but you can calculate any function of $A$ the same way.
For example, let $$f(\lambda)=\sin(t\lambda)$$ then $$\eqalign{ f(0) &= \sin(0) = \alpha_0 \cr f(1) &= \sin(t) = \alpha_1 + \alpha_0 \cr &\implies \alpha_0=0 ,\,\,\, \alpha_1=\sin(t) \cr }$$ and $$\eqalign{ \sin(tP) &= \alpha_1P + \alpha_0I = \sin(t) P \cr \sin(A) &= \frac{\sin(t)}{t} A \cr }$$