# Positive semi-definiteness of the adjoint matrix

I am studying the conditions of positive semi-definiteness of a $$(n+1)\times(n+1)$$ symmetric matrix $$\mathbf{M}$$ built in the following way: $$\mathbf{M}=\begin{pmatrix} \mathbf{A} & \mathbf{b} \\ \mathbf{b}^T & c \end{pmatrix}$$ where $$\mathbf{A}$$ is a symmetrix $$n\times n$$ matrix, $$\mathbf{b}$$ is a $$n$$-dimensional column vector and $$c$$ is a real number.
The first $$n$$ leading principal minors of $$\mathbf{M}$$ are the leading principal minors of $$\mathbf{A}$$, so $$\mathbf{A}$$ should be positive semi-definite.
The last condition is $$\det\mathbf{M}=|\mathbf{M}|\geq0$$. By a simple calculation, I obtained $$|\mathbf{M}|=c|\mathbf{A}|-\mathbf{b}^T\mathbf{A}^*\mathbf{b}\geq0$$ where $$\mathbf{A}^*$$ is the adjoint matrix of $$\mathbf{A}$$, i.e. the transpose of the matrix of cofactors.
This condition can be written $$c|\mathbf{A}|-\mathbf{b}^T\mathbf{A}^*\mathbf{b}= \begin{cases} |\mathbf{A}|\left(c-\mathbf{b}^T\mathbf{A}^{-1}\mathbf{b}\right), & \text{if }|\mathbf{A}|>0 \\ -\mathbf{b}^T\mathbf{A}^*\mathbf{b}, & \text{if }|\mathbf{A}|=0 \end{cases}$$ So, when $$|\mathbf{A}|>0$$ the condition simply becomes $$c\geq\mathbf{b}^T\mathbf{A}^{-1}\mathbf{b}\geq0,$$ given that $$\mathbf{A}^{-1}$$ is positive definite.
When $$|\mathbf{A}|=0$$ the condition becomes $$\mathbf{b}^T\mathbf{A}^*\mathbf{b}\leq0,$$ so I am interested to know if $$\mathbf{A}^*$$ is positive semi-definite when $$\mathbf{A}$$ is positive semi-definite.
In the case $$|\mathbf{A}|>0$$, using spectral decomposition $$\mathbf{A}=\sum_{i=1}^n\lambda_i\mathbf{e}_i\otimes\mathbf{e}_i,$$ where $$\lambda_i$$ are the eigenvalues and $$\mathbf{e}_i$$ the unit eigenvectors, so we have $$\mathbf{A}^*=|\mathbf{A}|\mathbf{A}^{-1}=\left(\prod_{k=1}^n{\lambda}_k\right)\sum_{i=1}^n\frac{1}{\lambda_i}\mathbf{e}_i\otimes\mathbf{e}_i = \sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k\right)\mathbf{e}_i\otimes\mathbf{e}_i,$$ so $$\mathbf{A}^*$$ is positive definite when $$\mathbf{A}$$ is, given that its eigenvalues are expressed as the product of eigenvalues of $$\mathbf{A}$$, excluded one in turn.
I suspect that this last expression represents $$\mathbf{A}^*$$ also when $$|\mathbf{A}|=0$$, probably by considering a positive semi-definite matrix with vanishing determinant as the limit of a positive definite matrix when one or more eigenvalues tends to zero.

So my questions:

1. are my calculation correct?
2. the last expression of $$\mathbf{A}^*$$ is valid also when $$|\mathbf{A}|=0$$?
3. how can this be proved?
• Note that what you are referring to as the "adjoint" is more typically referred to as the "adjugate" or "classical adjoint". Commented Jul 12, 2020 at 12:47

Yes, your equations are correct. Yes, the last expression you wrote is valid when $$|A| = 0$$. Note in particular that $$\mathbf A^* = 0$$ whenever the kernel of $$\mathbf A$$ has dimension at least $$2$$.

For a quick proof, we could simply note that both sides of the equation $$\mathbf{A}^* = \sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k\right)\mathbf{e}_i\otimes\mathbf{e}_i$$ are continuous functions of the entries of $$\mathbf A$$. If the equation holds for all strictly positive definite $$\mathbf A$$, then it must hold for positive semidefinite $$\mathbf A$$ "by continuity". In particular, if we define $$\mathbf A_{\epsilon} = \mathbf A + \epsilon \mathbf I$$ and $$\lambda_{k}^{\epsilon}$$ to be the $$k$$th eigenvalue of $$\mathbf A_{\epsilon}$$, then we can say that for a positive semidefinite $$\mathbf A$$ we have $$\mathbf{A}^* = \lim_{\epsilon \to 0^+}\mathbf{A}_{\epsilon}^* = \lim_{\epsilon \to 0^+}\sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k^{\epsilon}\right)\mathbf{e}_i\otimes\mathbf{e}_i = \sum_{i=1}^n\left(\prod_{k=1,k\neq i}^n{\lambda}_k\right)\mathbf{e}_i\otimes\mathbf{e}_i.$$

For a direct proof: we note that $$\dim\ker \mathbf A \geq 2$$ implies that $$\mathbf A^* = 0$$, which is positive semidefinite. For the case where $$\dim\ker \mathbf A = 1$$, we see that $$\mathbf A$$ is symmetric and $$\mathbf A \mathbf A^* = 0$$ implies that $$\mathbf A^*$$ has rank at most $$1$$, which means that $$\mathbf A^*$$ can be written in the form $$\mathbf A^* = k \mathbf {xx}^T$$ for some unit vector $$\mathbf x$$ and some $$k \in \Bbb R$$. We note that $$k$$ satisfies $$\operatorname{tr}(\mathbf A^*) = k$$.

With that, it suffices to note that $$\operatorname{tr}(\mathbf A^*) = -\frac{d}{dt}|_{t = 0} \det(t\mathbf I - \mathbf A) = -\frac{d}{dt}|_{t = 0} (t - \lambda_1) \cdots (t - \lambda_n).$$

• You mean "$\mathbf{A}^*$ has rank at most $1$"? Commented Jul 12, 2020 at 13:13
• @enzotib Yes.${}{}$ Commented Jul 12, 2020 at 13:14
• So, when $|\mathbf{A}|=0$ the condition $\mathbf{b}^T\mathbf{A}^*\mathbf{b}\leq0$ is verified by $\mathbf{b}\in\ker\mathbf{A}^*$, thas is the whole space or a subspace of dimension $n-1$, right? Commented Jul 12, 2020 at 13:24
• @enzotib I don't understand your use of the word "verified." To answer the question I believe you're asking: when $A$ has rank $n-1$, $b^T A^* b \leq 0$ occurs if and only if $b \in \ker(A^*) = \operatorname{im}(A)$ (which is a subspace of dimension $n-1$). When $A$ has smaller rank, $A^* = 0$ which means that $b^TA^*b \leq 0$ always holds. Commented Jul 12, 2020 at 13:33
• @enzotib A very important point, however: if $M$ is not invertible, then it is not necessarily true that it is positive definite if its principal minors are invertible. In fact, it is generally necessary to check all minors of $M$ to verify that it is positive semidefinite. For instance, even if $A$ is positive semidefinite with rank below $n-2$, there always exists a vector $b$ for which your matrix fails to be positive semidefinite (given a fixed constant $c$). Commented Jul 12, 2020 at 13:40