Answer: Can A² + B² Be Invertible?

Key Idea / Intuition

The two conditions together are screaming "factor something." Notice that $A^3 - B^3$ and $A^2B - B^2A$ are both zero — if you combine them cleverly, you can show that $(A^2 + B^2)(A - B) = 0$. Since $A \neq B$, the matrix $A - B$ is nonzero, so $A^2 + B^2$ must have a nontrivial kernel — meaning it cannot be invertible.

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Formal Proof / Solution

Step 1: Compute $A^3 - B^3$ using the conditions.

We are given $A^3 = B^3$, so: $A^3 - B^3 = 0.$

Now factor this expression cleverly. We write: $A^3 - B^3 = A^3 - A^2B + A^2B - AB^2 + AB^2 - B^3.$

Group as: $= A^2(A - B) + AB(A - B) + B^2(A - B)... \quad \text{(careful: matrices don't commute!)}$

Let's be more careful. Use the given condition $A^2B = B^2A$ to control cross terms.

Step 2: Use both conditions together.

Compute $(A^2 + B^2)(A - B)$: $(A^2 + B^2)(A - B) = A^3 - A^2B + B^2A - B^3.$

Now apply both given conditions:

• $A^3 = B^3$, so $A^3 - B^3 = 0$.
• $A^2B = B^2A$, so $-A^2B + B^2A = 0$.

Therefore: $(A^2 + B^2)(A - B) = (A^3 - B^3) + (B^2A - A^2B) = 0 + 0 = 0.$

Step 3: Conclude non-invertibility.

Since $A \neq B$, the matrix $A - B \neq 0$. But $(A^2 + B^2)(A - B) = 0$ with $A - B \neq 0$ means $A^2 + B^2$ has a nontrivial right null vector (any nonzero column of $A - B$ works).

Therefore $A^2 + B^2$ is not invertible.

Conclusion: No, $A^2 + B^2$ cannot be invertible under these conditions.

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Remark on elegance: The entire proof collapses to one line once you see the right factorization. The conditions $A^3 = B^3$ and $A^2B = B^2A$ are precisely the two pieces needed to make $(A^2+B^2)(A-B)$ telescope to zero. This is the kind of algebraic identity that feels magical but is perfectly natural in hindsight.