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

eigenvalues-eigenvectorslinear algebramatricessymmetric matrices

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.

Best Answer

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 : $$X^{T}AX=(AX)^{T}X=\left<AX,X\right >=\left < \sum_{i=1}^n \alpha_i\lambda_i e_i,\sum_{i=1}^n\alpha_ie_i \right >=\sum_{i=1}^n \lambda_i\alpha_i^2 \ge 0.$$ $(\Leftarrow)$ 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 $1\le k\le n$ : $$\sum_{k=1}^n \lambda_i(A+B)\le\sum_{i=1}^k\left( \lambda_i(A)+\lambda_i(B) \right) .$$

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