1
$\begingroup$

$E$ is an orthogonal projection on a subspace $W$ of $V$ (please explain what that means), and $k>0$. Prove that $kI+E$ is positive definite. Thanks!

  • 0
    Couldn't find anything on orthogonal projection which helped me solve this exercise.2012-10-26

2 Answers 2

2

Let $(v,\lambda)$ be a pair of eigenvector and eigenvalue of $kI+E$, then $(kI+E)v=\lambda v,$ or $Ev=(\lambda-k)v.$ So $v$ is also a eigenvector of $E$, which is an orthogonal projection matrix, so $E$ has eigenvalue $0$ and/or 1.

Since $k>0$, we can see $\lambda>0$, so $kI+E$ is PD.

0

Another way to look at it would be that for any matrix $A$, the eigen values of $A+kI$ for a constant $k$ would be the eigen values of $A$ incremented by $k$. This follows directly from schur Triangularization. $A=UTU^{H}$ where $T$ is a triangular matrix with its diagonal entries as eigen values of $A$ and $U$ is a unitary matrix. Now \begin{align} A+kI=UTU^{H}+kI= UTU^{H}+U(kI)U^{H} =U(T+kI)U^{H} \end{align} Apply this to your case and observe that your projection matrix has only $0/1$ as their eigen values.