$E$ 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!
$kI+E$ is positive definite where $E$ is an orthogonal projection and k>0.
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0Couldn't find anything on orthogonal projection which helped me solve this exercise. – 2012-10-26
2 Answers
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.
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.