Eigenvectors Practice Q.2

Question

Let $A$ be an $n \times n$ real matrix. Consider the following two statements about the matrix $B = A^T A$:

     1.  Every eigenvector of $B$ is also an eigenvector of $A$.
     2. Every eigenvector of $A$ is also an eigenvector of $B$.

Which of the following is necessarily true?

     Both (i) and (ii) are true.
     (i) is true but (ii) is false.
     (ii) is true but (i) is false.
     Neither (i) nor (ii) is true.

Answer 1

To share all eigenvectors, matrices $A$ and $B = A^T A$ must commute:
\[
MN = NM.
\]

Thus, for $A$ and $B = A^T A$ to share all eigenvectors, we require:
\[
A(A^T A) = (A^T A)A.
\]

Compute both sides:

\begin{align*}
A(A^T A) &= AA^T A, \\
(A^T A)A &= A^T A^2.
\end{align*}

Equating them:
\[
AA^T A = A^T A^2.
\]

Rearranging:
\[
AA^T A - A^T A^2 = 0 \quad \Rightarrow \quad (AA^T - A^T A)A = 0.
\]

This implies that the matrix $(AA^T - A^T A)$, when applied to any vector in the image of $A$, yields zero. However, for this to hold for all vectors (i.e., for $A$ and $B$ to commute), we must have:
\[
AA^T - A^T A = 0,
\]
or equivalently,
\[
\boxed{AA^T = A^T A}.
\]

This condition defines a normal matrix.

However, the problem states that $A$ is a general real matrix — not necessarily normal.

Therefore, in general, $A$ and $A^T A$ do not commute.

Hence Both statements are Wrong.

Option D = Correct