# For Positive Matrices $A, B \in M_n(\mathbb R)$, prove that $(A^T)(A)$ and $A + B$ are positive?

Let $$A, B \in M_n(\mathbb R)$$ be positive matrices. Show that $$A^TA$$ and $$A + B$$ are positive.

I know that to show that $$A^TA$$ and $$A + B$$ are positive I need to show that:

1. The matrix is symmetric
2. The eigenvalues of the matrix are positive.

I am able to show that both $$A^TA$$ and $$A+B$$ are symmetric, but struggle to show that the eigenvalues are positive.

I know that the eigenvalues of $$A$$ and $$A^T$$ are equal, and therefore positive (since $$A$$ is positive) but do not understand what this says about the eigenvalues of $$A^TA$$.

Further, I have noticed through inspection that for positive $$A$$ and $$B$$ with eigenvalues $$\lambda_1 , \dots, \lambda_n$$ and $$\lambda '_1, \dots, \lambda '_n$$ respectively, the eigenvalues of $$A+B$$ are $$\lambda_1 + \lambda '_1, \dots, \lambda_n + \lambda '_n$$, but do not know how to prove this rigorously.

Any help is greatly appreciated.

• In linear algebra literature, a positive matrix usually means an element-wise positive matrix rather than a positive definite matrix. The terminology is different from operator theory. Please edit your question if you are talking about positive definite matrices. – user1551 Aug 7 at 6:58
• The eigenvalue criterion isn’t the easiest here. Work directly from the quadratic form definition: $M$ is positive if $x^tMx>0$ for all nonzero vectors $x$. You’ll see $A^tA$ is positive regardless of whether $A$ is positive (it merely needs to have trivial kernel, that is, be invertible), because $x^tA^tAx$ is the dot product of $Ax$ with itself, which is the squared magnitude of $Ax$ and thus must be positive for all nonzero $x$. – symplectomorphic Aug 7 at 7:01
• This is what I thought at first as well - would make the question incredibly simple. However, upon specifically inquiring about the distinction that you mention, I was informed that in the question "positive" means satisfying the above conditions, rather than being positive definite, which I have been led to believe (not sure why) is something different. – Marko Dakic Aug 7 at 7:03
• A real symmetric matrix satisfies the quadratic form definition if and only if all its eigenvalues are positive. So the only way the definitions could come apart is if your instructor is using one term to mean the quadratic form criterion plus symmetry, and the other term just the quadratic form criterion without necessarily being symmetric. That is highly nonstandard, though. – symplectomorphic Aug 7 at 7:09
• Would I be correct in saying that since we know A is positive, it is symmetric and so $A^T = A$ and by extension $A^TA = A^2$? From here, $\lambda^2$ is an eigenvalue of $A$ and so it will always be positive. – Marko Dakic Aug 7 at 17:15

Let $$S_n^{+}(\mathbb{R})$$ be the set of positive semidefinite matrices, i.e : $$S_n^{+}(\mathbb{R}):=\{A\in\mathcal{S}_n(\mathbb{R})|\text{Sp}(A)\subset \mathbb{R}_{\ge 0} \}.$$ First, we have the following equivalence :

Let $$A\in \mathcal{S}_n(\mathbb{R})$$. Then : $$A\in\mathcal{S}_n^{+}(\mathbb{R})\Leftrightarrow \forall X\in\mathbb{R}^n,\quad X^{T}AX\ge 0.$$

Proof -

$$(\Rightarrow)$$If $$A\in S_n^{+}(\mathbb{R})$$, by spectral theorem we can find an orthongonal basis of eigenvectors $$(e_1,...,e_n)$$ such that : $$\forall i\in\{ 1,...,n\},\quad Ae_i=\lambda_ie_i,$$ where $$\text{Sp}(A):=\{\lambda_1,...,\lambda_n\}$$.

Hence, if $$X=\displaystyle\sum_{i=1}^n \alpha_i e_i\in\mathbb{R}^n$$, then : $$X^{T}AX=(AX)^{T}X=\left=\left < \sum_{i=1}^n \alpha_i\lambda_i e_i,\sum_{i=1}^n\alpha_ie_i \right >=\sum_{i=1}^n \lambda_i\alpha_i^2 \ge 0.$$ $$(\Leftarrow)$$ With the same notations, for all $$i\in\{ 1,...,n\}$$ : $$0\le X_i^{T}AX_i=X_i^{T}\cdot\lambda_i X_i=\lambda_i ||X_i||^2=\lambda_i,$$ so : $$\text{Sp}(A)\subset \mathbb{R}_{\ge 0}. \blacksquare$$

Here, if $$A,B\in S_n^{+}(\mathbb{R})$$, then, for all $$X\in\mathbb{R}^n$$, $$A^{T}A$$ and $$A+B$$ are symmetric matrices, and for all $$X\in\mathbb{R}^n$$ : $$X^{T}A^{T}AX=(AX)^{T}(AX)=||AX||^2\ge 0,$$ and : $$X^{T}(A+B)X=X^{T}AX+X^{T}BX\ge 0,$$ so $$A^{T}A$$ and $$A+B$$ are in $$\mathcal{S}_n^{+}(\mathbb{R})$$.

If you want to get trained, here is another problem :

Problem - Let $$\mathcal{S}_n^{++}(\mathbb{R}):=\{A\in\mathcal{S}_n(\mathbb{R})|\text{Sp}(A)\subset \mathbb{R}_{> 0} \}$$ and let $$A\in\mathcal{S}_n^{++}(\mathbb{R})$$. Prove that : $$\text{Com}(A)\in\mathcal{S}_n^{++}(\mathbb{R}).$$

Moreover, your result about the eigenvalues of $$A+B$$ is false in general. For example, if you take : $$A=\begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} ,$$ and $$B=\begin{pmatrix}1&1\\1&1 \end{pmatrix},$$ then $$\text{Sp}(A)=\{0,1\}$$, $$\text{Sp}(B)=\{0,2\}$$ whereas $$\text{Sp}(A+B)=\left\{\frac{3\pm\sqrt 5}{2}\right\}$$.

However, the result can be true in particular cases. For example, if $$A$$ and $$B$$ commute, then you can prove that $$A$$ and $$B$$ are codiagonalisable, i.e you can find a singular matrix $$P$$ such that : $$A=P^{-1}D_1P$$ and $$B=P^{-1}D_2P,$$ where $$D_1=\text{Diag}(\lambda_1,...,\lambda_n)$$ and $$D_2=\text{Diag}(\mu_1,...,\mu_n)$$, and in this case : $$\text{Sp}(A+B)=\{\lambda_1+\mu_1,...,\lambda_n+\mu_n\}.$$ You also have the Ky-Fan inequalities :

Key-Fan inequlalities - Let $$A,B\in\mathcal{S}_n(\mathbb{R})$$, and $$\text{Sp}(A)=\{\lambda_1(A)\ge,...\ge\lambda_n(A)\}$$ and $$\text{Sp}(B)=\{\lambda_1(B)\ge ...\ge\lambda_n(B)\}$$. Then : $$\lambda_1(1+B)\le\lambda_1(A)+\lambda_1(B)$$ and for all $$1\le k\le n$$ : $$\sum_{k=1}^n \lambda_i(A+B)\le\sum_{i=1}^k\left( \lambda_i(A)+\lambda_i(B) \right) .$$