Suppose the matrix $A\in\mathbb{R}^{m \times n}$ has full rank, $x\neq0\in\mathbb{R}^n, b\in\mathbb{R}^m$ and $ Ax \neq b$. Denote $I \in \mathbb{R}^{n\times n}$ the identity matrix.

Does it hold for all $\alpha > 0 \in \mathbb{R}$ that the following matrix is nonsingular:

$ J = \begin{pmatrix}A^T A + \alpha I & x \\ (Ax-b)^T A & 0 \end{pmatrix} ? $

The matrix $A^T A + \alpha I$ is obviously positive definite and hence invertible. Since $\det(J) = -(Ax-b)^T A(A^T A + \alpha I )^{-1} x\hspace{0.3cm} \underbrace{\det(A^T A + \alpha I)}_{> 0}$, it would be sufficient to prove that the Schur complement $S = -(Ax-b)^T A(A^T A + \alpha I )^{-1} x$ is nonzero.


1 Answer 1


This is not true. Consider, e.g. $$ m=n=2,\ A=I_2,\ x=\pmatrix{1\\ 1},\ b=\pmatrix{-1\\ 3},\ \alpha=1, \ J=\pmatrix{2&0&1\\ 0&2&1\\ 2&-2&0}. $$ In fact, when $A$ and $\alpha$ are given and $(m,n)\ne(1,1)$, we can always pick some vectors $x$ and $b$ such that $Ax-b\ne0$ but $-(Ax-b)^TA(A^TA+\alpha I)^{-1}x=0$:

  • When $m>n$, pick any $x\in\mathbb R^n$, any nonzero vector $z\in\ker(A^T)$ and set $b=Ax-z$.
  • When $n>m$, pick any $x\in\ker\left(A(A^TA+\alpha I)^{-1}\right)$ and any $b\ne Ax$.
  • When $m=n>1$, pick any nonzero vector $y\in\mathbb R^n$ such that $y^Tx=0$ and set $z^T=y^T(A^TA+\alpha I)A^{-1}$ and $b=Ax-z$.

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .