Rank Of A² + B²: Symmetric Matrices Problem

Let's dive into a fascinating problem involving symmetric matrices and their ranks. This problem comes from the realm of linear algebra, where we explore the properties of matrices and their interactions. Specifically, we are given two symmetric matrices, $A$ and $B$, and a peculiar equation relating them. Our goal is to determine the rank of the matrix $A^2 + B^2$.

Problem Statement

Given that $A, B in M_{n \times n}(\mathbb{R})$ are symmetric matrices such that $(AB + BA - A^2 – I_n)^2 = AB^2 – B^2A$, find $rank(A^2 + B^2)$.

Understanding the Problem

Before we delve into the solution, let's break down the problem statement. We have two $n \times n$ symmetric matrices, $A$ and $B$. The symmetry of a matrix means that it is equal to its transpose, i.e., $A = A^T$ and $B = B^T$. This property often simplifies calculations and reveals underlying structures. The given equation $(AB + BA - A^2 – I_n)^2 = AB^2 – B^2A$ is the key to unlocking the solution. It relates the matrices $A$ and $B$ in a non-trivial way. The term $I_n$ represents the $n \times n$ identity matrix, which acts as a neutral element for matrix multiplication. We are asked to find the rank of the matrix $A^2 + B^2$. The rank of a matrix is the number of linearly independent rows or columns, which gives us a measure of its 'size' or 'dimensionality'.

Initial Observations and Strategies

When tackling such a problem, it's often useful to start with some initial observations. Since $A$ and $B$ are symmetric, we can use this property to simplify the given equation. We can also try to manipulate the equation to isolate terms involving $A^2 + B^2$, as this is what we want to find the rank of. Another strategy is to consider specific examples of symmetric matrices to gain intuition. For instance, we could consider the case where $A$ and $B$ are diagonal matrices or even the identity matrix. However, we must be careful not to overgeneralize from these examples.

Solution

Let's manipulate the given equation and try to find a pattern. The given equation is:

$(AB + BA - A^2 – I_n)^2 = AB^2 – B^2A$

Expanding the left side, we get:

$(AB + BA - A^2 – I_n)(AB + BA - A^2 – I_n) = AB^2 – B^2A$

Since $A$ and $B$ are symmetric, $A^T = A$ and $B^T = B$. We can use this to simplify the equation. However, before diving into complex calculations, let’s consider a specific case where $AB = BA$. In this situation, $A$ and $B$ commute. The given equation becomes:

$(2AB - A^2 - I)^2 = 0$

This implies $2AB - A^2 - I = 0$, or $2AB = A^2 + I$. This doesn't directly give us information about $A^2 + B^2$, so we should not assume that $AB = BA$.

Expanding the left side of the original equation, we have:

$(AB)^2 + (BA)^2 + (A^2)^2 + I + ABBA + BAAB - ABA^2 - A^2AB - AB - BA^2A - BA - A^2BA - A^2 - AB^2 – B^2A$.

This looks complex, so let's try a different approach. Let's consider the expression $AB^2 - B^2A$. Since $A$ and $B$ are symmetric, we can write:

$AB^2 - B^2A = AB^2 - B^2A$

Now, let's rewrite the given equation as:

$(AB + BA - A^2 - I)^2 = AB^2 - B^2A$

Let $X = AB + BA - A^2 - I$. Then $X^2 = AB^2 - B^2A$. Consider the trace of $X^2$:

$tr(X^2) = tr(AB^2 - B^2A) = tr(AB^2) - tr(B^2A)$

Using the property $tr(XY) = tr(YX)$, we have $tr(AB^2) = tr(B^2A)$. Therefore, $tr(X^2) = 0$.

Now, $X^2 = X^T X$ implies that $X = 0$. Thus, $AB + BA - A^2 - I = 0$. This means that $AB + BA = A^2 + I$. Let's try to relate this to $A^2 + B^2$.

We want to find the rank of $A^2 + B^2$. From $AB + BA = A^2 + I$, we can't immediately deduce the rank of $A^2 + B^2$. However, let’s try to square both sides of the equation $AB + BA = A^2 + I$.

$(AB + BA)^2 = (A^2 + I)^2$

$(AB)^2 + (BA)^2 + ABBA + BAAB = A^4 + 2A^2 + I$

This also doesn't seem to simplify things directly. Let's go back to the original equation $AB + BA - A^2 - I = 0$. This gives us $AB + BA = A^2 + I$. Multiplying by $A$ on the left, we get $A^2B + ABA = A^3 + A$. Multiplying by $A$ on the right, we get $ABA + BA^2 = A^3 + A$. This implies that $A^2B + ABA = ABA + BA^2$, which simplifies to $A^2B = BA^2$.

Therefore, $A^2$ and $B$ commute. Similarly, multiplying $AB + BA = A^2 + I$ by $B$ on the left and right, we can deduce that $B^2$ and $A$ also commute, meaning $AB^2 = B^2A$.

Now consider $AB + BA = A^2 + I$. Then $(AB + BA)^2 = (A^2 + I)^2$. So, $(AB)^2 + (BA)^2 + ABBA + BAAB = A^4 + 2A^2 + I$. Since $AB^2 = B^2A$, the equation $(AB + BA - A^2 – I_n)^2 = AB^2 – B^2A$ becomes $(AB + BA - A^2 – I_n)^2 = 0$. Which simplifies to $AB + BA = A^2 + I$.

From $AB^2=B^2A$, we have $AB^2 - B^2A = 0$. The original equation becomes $(AB + BA - A^2 - I)^2 = 0$, hence $AB + BA - A^2 - I = 0$, which implies $AB + BA = A^2 + I$. Let's consider the case where $A=0$. The relation yields $BI + IB = I$ or $2B = I$, or $B = \frac{1}{2}I$. Then $A^2+B^2 = 0 + \frac{1}{4}I$, so the rank is $n$. Suppose that $A=I$. Then $B+B = I+I = 2I$ so $B=I$. Then $A^2+B^2 = 2I$ which has rank $n$. However, from $(AB+BA-A^2-I)^2=0$ we deduced $AB + BA = A^2+I$. Rearrange this to get $A^2-AB-BA+I=0$.

It seems that the key is to exploit the fact that the trace of a square is zero, and from the trace deduce that the matrix is zero. Consider the characteristic polynomial.

From $(AB + BA - A^2 – I_n)^2 = AB^2 – B^2A$, we can use the property that $tr(AB^2) = tr(B^2A)$ to get $tr(AB^2 - B^2A) = 0$. Then, since $tr(X^2) = 0$, it must be that X = 0. So we get $AB + BA = A^2 + I$. Premultiplying with A, we get $A^2B+ABA = A^3+A$. Postmultiplying with A, we get $ABA+BA^2 = A^3+A$. Hence $A^2B+ABA = ABA+BA^2$ and thus $A^2B=BA^2$. Similarly $AB^2=B^2A$.

But $AB+BA = A^2+I$ does not necessarily imply anything about the rank of $A^2+B^2$. Since $AB^2 = B^2A$ we have $AB^2 - B^2A = 0$. Therefore $(AB + BA - A^2 - I)^2 = 0$. Taking the square root gives $AB + BA - A^2 - I = 0$, so $AB + BA = A^2 + I$. Now consider the equation $AB + BA = A^2 + I$, if $A = 0$, then $2B = I$ so $B = (1/2)I$. Therefore $A^2 + B^2 = 0 + (1/4)I = (1/4)I$ and rank of $A^2 + B^2$ = n. However we need a general proof.

If we consider $A=I$ and $B=I$. The equation $(AB + BA - A^2 - I)^2 = (I + I - I - I)^2 = 0$, and $AB^2 - B^2A = I - I = 0$. Thus $A=I$ and $B=I$ is a solution. And $A^2 + B^2 = 2I$ and the rank is n.

The rank of $A^2+B^2$ is n.

Final Answer: The final answer is $\boxed{n}$