If $\mathbf M=\left[\begin{matrix}\bf A & \bf b\\\bf b^\top & \bf d \\\end{matrix}\right]$ such that $\bf A$ is positive definite, under what conditions is $\bf M$ positive definite, positive semidefinite and indefinite?

It is readily seen that $\det(\mathbf M)=\alpha\det(\mathbf A)$, where $\alpha=\mathbf d-\bf b^\top A^{-1}b$


  • $\alpha>0\Rightarrow \det(\mathbf M)>0$ $\quad(\det(\mathbf A)>0$ by hypothesis$)$

This is not enough for $\bf M$ to be p.d. as I need all its leading principal minors to be positive as well. But the answers tell me that $\alpha>0$ ensures all of that.

  • $\alpha=0\Rightarrow \bf M$ is singular

As it turns out, this condition ensures that $\bf M$ is p.s.d, but as far as I know this is not the case in general.

  • $\alpha<0\Rightarrow \det(\mathbf M)<0$

This condition is apparently sufficient in this case for $\bf M$ to be indefinite. But if $\bf M$ is indefinite, it may be a singular matrix as well (i.e., $\alpha$ might as well have vanished in this case). Moreover, as I note in the above two cases, this isn't a sufficient condition either for a matrix to be indefinite.

I would like to clarify my doubts and a possible proof sketch would be helpful too.


Extending the problem from a bordered matrix to a partitioned matrix:

Suppose $\mathbf M=\left[\begin{matrix}\bf A & \bf B\\\bf B^\top & \bf D \\\end{matrix}\right]$ is symmetric and $\bf A$ is square.

Then show that $\bf M$ is p.d. iff $\bf A$ and $\mathbf D-\bf B^\top A^{-1}B$ are p.d.

What should be the correct approach now? I can calculate the determinant like before but the same problem still bothers me. Thinking in terms of quadratic forms and using the definitions of definiteness do not seem to be the way to go about it.

  • $\begingroup$ I have edited my answer such that it covers the general case. $\endgroup$ – Reinhard Meier May 15 '17 at 16:52

You can use the following definitions for a matrix $M\in\mathbb{R}^{n\times n}$: $$ \begin{eqnarray} M\;\mbox{positive definite}\; & \Longleftrightarrow & x^TMx>0\;\forall x\in\mathbb{R}^{n}\setminus\{0\} \\ M\;\mbox{positive semidefinite}\; & \Longleftrightarrow & x^TMx\geq 0\;\forall x\in\mathbb{R}^{n} \\ M\;\mbox{indefinite}\; & \Longleftrightarrow & \exists x_1,x_2\in\mathbb{R}^{n}:x_1^TMx_1>0>x_2^TMx_2 \end{eqnarray} $$ Let $A\in\mathbb{R}^{(n-m)\times (n-m)}$, $A$ positive definite, $B\in\mathbb{R}^{(n-m)\times m},\;D\in\mathbb{R}^{m\times m}$. Let $u\in\mathbb{R}^{n-m}$, $c\in\mathbb{R}^{m},\; x=\binom{u}{c}.$

With $v=u+A^{-1}Bc$, we find the following: $$ x^T M x = \begin{pmatrix}u \\ c\end{pmatrix}^T \begin{pmatrix} A & B \\ B^T & D \end{pmatrix} \begin{pmatrix}u \\ c\end{pmatrix} = v^T\!Av+c^T\alpha c $$ with $\alpha = D-B^TA^{-1}B \in\mathbb{R}^{m\times m}$.

Now it is easy to check if $M$ is positive definite, positive semidefinite or indefinite.

If $\alpha$ positive definite and $c\neq 0$, then $x^TMx = v^T\!Av+c^T\alpha c \geq c^T\alpha c > 0$.

If $\alpha$ positive definite and $c=0$ and $u\neq 0$, then $x^TMx = u^T\!Au > 0$.

If $\alpha$ positive semidefinite, then $v^T\!Av+c^T\alpha c \geq 0$. We can see that $0$ is actually attained by choosing $c$ such that $c^T\alpha c = 0$ and then by setting $u=-A^{-1}Bc$.

If $\alpha$ indefinite, negative definite or negative semidefinite, we choose a vector $c_2\in\mathbb{R}^{m}$ such that $c_2^T\alpha c_2 <0.$ Then we set $x_1=\binom{u_1}{0}$ with $u_1\in\mathbb{R}^{n-m}\setminus\{0\}$ and $x_2 = \binom{u_2}{c_2} = \binom{-A^{-1}Bc_2}{c_2}.$ Then $v_2=u_2+A^{-1}Bc_2 = 0$ and $$ x_1^TMx_1 = u_1^T\!Au_1 > 0 > c_2^T\alpha c_2 = v_2^T\!Av_2 + c_2^T\alpha c_2 = x_2^TMx_2 $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.