Prove that the Moore-Penrose inverse of a symmetric matrix have the same nullspace as the original.

Question

Prove that the Moore-Penrose inverse of a symmetric matrix have the same nullspace as the original.

The definition in my textbook is as follows:

Corresponding to any $m \times n$ matrix A, there is a unique $n \times m$ matrix G such that

(1) AGA=A (i.e., G is a generalized inverse of A)

(2) GAG=G (i.e., A is a generalized inverse of G)

(3) (AG)'=AG (i.e., AG is symmetric); and

(4) (GA)'=GA (i.e., GA is symmetric).

Then G is called the Moore-Penrose inverse of A.

Answer 1

I suspect that your textbook establishes the following relationship between the Moore Penrose pseudoinverse and the SVD.

Suppose that $A = U\Sigma V^T$ is a singular value decomposition (with $U \in \Bbb R^{m \times m}$ and $V \in \Bbb R^{n \times n}$ orthogonal and $\Sigma \in \Bbb R^{m \times n}$ diagonal with non-negative, non-increasing diagonal entries). Then the psuedoinverse of $A$ may be computed as $$ A^\dagger = V\Sigma^\dagger U^T, $$ where $\Sigma^\dagger$ is obtained by transposing $\Sigma$ and replacing any non-zero (diagonal) entries in $\Sigma^T$ with their reciprocal.

Now, to prove the result at hand. Of course, if $A$ is invertible, then both $A$ and $A^{\dagger} = A^{-1}$ have a trivial nullspace so that the result holds. On the other hand, if $A$ is square and not invertible, then its singular value decomposition can be written as $$ A = \pmatrix{U_1 & U_2} \pmatrix{\Sigma & 0\\0 & 0}\pmatrix{V_1 & V_2}^T, $$ where $\Sigma$ has size $r\times r$ (with $r$ equal to the rank of $A$) and $U_1,V_1$ each have $r$ columns. It can be shown that because of this decomposition, the columns of $V_2$ form a basis for the nullspace of $A$ and the columns of $U_2$ form a basis of the nullspace of $A^T$.

On the other hand, $A^\dagger$ has a similar looking decomposition: $$ A^\dagger = \pmatrix{V_1 & V_2}\pmatrix{\Sigma^{-1} & 0\\0 & 0}\pmatrix{U_1 & U_2}^T. $$ With a similar line of reasoning, we can deduce that the columns of $U_2$ form a basis for the nullspace of $A^\dagger$ and the columns of $V_2$ form a basis for the nullspace of $(A^\dagger)^T$.

Thus, it is generally true that for a square matrix $A$, $A^T$ will have the same nullspace as $A^\dagger$. If $A$ is symmetric, then this means that $A$ (which is equal to $A^T$) has the same nullspace as $A^\dagger$, which is what we wanted.