# Relation between the maximum eigenvalues of symmetric positive definite matrix $A$ and $B A B^\dagger$

Suppose $A$ is a symmetric positive definite matrix of $n \times n$ dimensions. Matrix $B \in \mathbb{R}^{m \times n}$ is a full-ranked real-valued matrix with $m$ strictly smaller than $n$, i.e., $m < n.$

Through Monte-Carlo experiments, I noticed that

$$\lambda_{\max} (A) \geq \lambda_{\max}(BA B^\dagger),$$ where $\lambda_{\max}(\cdot)$ denotes the maximum eigenvalue of the argument. Moreover, matrix $B^\dagger$ stands for the pseudo-inverse of $B$, i.e., $B^\dagger =B^T(BB^T)^{-1}$.

I am wondering where this inequality comes from.

For more illustration, one can run the following short code in Matlab.

n=30;
m=29;
c=0;

for i=1:1000;

Q = randn(n,n);
eigen_mean = 0.1; %can be made anything,
A = Q' * diag(abs(eigen_mean+randn(n,1))) * Q;  %A random symmetric positive definite
B=randn*randn(m,n);

if max(eig(A))< max(eig(B * A * pinv(B)))
c = c +1 ;
end
end

c


Everytime $c$ is returned zeor, since $\lambda_{\max} (A)$ is apparently never smaller than $\lambda_{\max}(BA B^\dagger).$

• What is $B^\dagger$? Also, is symmetric implied by positive definite? Jun 8, 2018 at 17:53
• As mentioned above, $B^{\dagger}$ is the pseudo-inverse of matrix $B$. Besides being positive definite, matrix $A$ is also symmetric. I edited the question. Thank you. Jun 8, 2018 at 17:58
• Notice that this formula for the pseudo-inverse applies only if $B$ is full rank. Jun 8, 2018 at 19:05
• Thanks again. Matrix $B$ is guaranteed to be full-ranked in my application which I forgot to mention. Edited! Jun 8, 2018 at 19:12

Let $C=BAB^+=BAB^T(BB^T)^{-1}$. Note that $BAB^T$ and $(BB^T)^{-1}$ are $m\times m$ symmetric $>0$ matrices and, therefore, their product $C$ is diagonalizable and has only $>0$ eigenvalues.

More precisely, $C$ is similar to the following $>0$ symmetric mtrix

$S=(BB^T)^{-1/2}BAB^T(BB^T)^{-1/2}=[(BB^T)^{-1/2}B]A[(BB^T)^{-1/2}B]^T$.

For every vector $x\in\mathbb{R}^m$, $x^TSx=y^TAy$ where $y=[(BB^T)^{-1/2}B]^Tx$.

Then $x^TSx\leq \rho(A)||y||^2$ where $||y||^2=x^T(BB^T)^{-1/2}BB^T(BB^T)^{-1/2}x=||x||^2$ and we are done.

• Just to make it more readable, $S=X^{-1}\left[BAB^T{\left(BB^T\right)}^{-1}\right]X$ with $X={\left(BB^T\right)}^{1/2}$. Great answer! +1 Jun 9, 2018 at 19:47
• What a nice solution! Thanks a lot Jun 10, 2018 at 0:17
• Thanks for the above compliments.
– user91684
Jun 11, 2018 at 16:41