Question

If \[ f(x)=\frac{1+x}{1-x} \] and \(A\) is a matrix such that \(A^{3}=0\), then \[ f(A)= \]

• A. $I+2A+2A^{2}$
• B. $I+2A+A^{2}$
• C. $I-2A+A^{2}$
• D. $I+A+A^{2}$

Hint:

An alternative algebraic trick to avoid infinite geometric expansions is to set up a direct matrix equation: let $f(A) = X \implies (I - A)X = I + A$. Since $A^3 = 0$, multiply both sides by $(I + A + A^2)$ to use the factorization rule $(I - A)(I + A + A^2) = I - A^3 = I$. This instantly yields $X = (I + A + A^2)(I + A) = I + 2A + 2A^2$.

Answer 1

The Correct Option is A

Concept: When evaluating a matrix function $f(A)$ defined by a rational scalar function $f(x) = \frac{1+x}{1-x}$, we convert the fractional structure into a matrix polynomial expansion using a geometric series power expansion: $$(1 - x)^{-1} = 1 + x + x^2 + x^3 + \dots$$ For a nilpotent matrix of order $k$ (where $A^k = 0$), the infinite matrix series terminates into a finite matrix polynomial, since all higher-power matrix terms collapse to zero. Step 1: Rewrite the scalar function as a series function.
The given function can be factored into a product of a linear numerator and an inverse linear denominator: $$f(x) = (1 + x)(1 - x)^{-1}$$ Using the standard infinite geometric series power expansion for $|x| < 1$: $$f(x) = (1 + x)\left(1 + x + x^2 + x^3 + x^4 + \dots\right)$$ Distribute the $(1 + x)$ terms across the infinite sum: $$f(x) = \left(1 + x + x^2 + x^3 + \dots\right) + \left(x + x^2 + x^3 + x^4 + \dots\right)$$ Group the matching power coefficients together: $$f(x) = 1 + 2x + 2x^2 + 2x^3 + 2x^4 + \dots$$

Step 2: Map the scalar series directly to the matrix variable $A$.
Replacing the scalar variable $x$ with the matrix variable $A$ transforms the scalar constant $1$ into the identity matrix $I$: $$f(A) = I + 2A + 2A^2 + 2A^3 + 2A^4 + \dots$$

Step 3: Apply the nilpotent boundary constraint condition.
We are given that the matrix $A$ is nilpotent with an index of 3, meaning: $$A^3 = 0$$ This structural condition implies that any matrix power greater than or equal to 3 is also identically zero: $$A^4 = A^3 \cdot A = 0 \cdot A = 0$$ $$A^5 = 0, \quad A^6 = 0, \quad \dots$$

Step 4: Collapse the matrix series to find the final polynomial expression.
Substitute these vanishing terms ($A^3 = A^4 = \dots = 0$) into our expanded matrix series for $f(A)$: $$f(A) = I + 2A + 2A^2 + 2(0) + 2(0) + \dots$$ $$f(A) = I + 2A + 2A^2$$ This matches the exact expression given in option (A).