A question about derivation of kalman gain

Question

I read the derivation of Kalman gain. In the derivation, we have: $$(\mathbf{H}\mathbf{P_{n,n-1}})^T = \mathbf{K_n}(\mathbf{H}\mathbf{P_{n,n-1}}\mathbf{H}^T+\mathbf{R_n})$$ where $\mathbf{H}$ is observation matrix, $\mathbf{P_{n,n-1}}$ is the predicted estimate uncertainty, $\mathbf{R_n}$ is the measurement uncertainty, and $\mathbf{K_n}$ is our desired Kalman gain. Then, $$\mathbf{K_n} = (\mathbf{H}\mathbf{P_{n,n-1}})^T(\mathbf{H}\mathbf{P_{n,n-1}}\mathbf{H}^T+\mathbf{R_n})^{-1}$$ My question is why $\mathbf{H}\mathbf{P_{n,n-1}}\mathbf{H}^T+\mathbf{R_n}$ must have inverse?

Answer 1

Claim: $\mathbf{H}\mathbf{P_{n,n-1}}\mathbf{H}^T+\mathbf{R_n}$ is the addition of two P.D. matrices, and is therefore P.D.

Now you need to figure out yourself why $\mathbf{P_{n,n-1}}$ and $\mathbf{R_n}$ are P.D.?

Answer 2

Since $\mathbf{P_{n,n-1}}$ and $\mathbf{R_n}$ are covariance matrices, they are positive (semi)definite. (We believe in this case, they are positve definite). And one can show that $\mathbf{H}\mathbf{P_{n,n-1}}\mathbf{H^T}$ preserves positive (semi)definite, and addition of two positive definite matrix is still postive definite. Thus $\mathbf{H}\mathbf{P_{n,n-1}}\mathbf{H^T}+\mathbf{R_n}$ is positive definite. And positive definiteness implies all positive eigenvalue(s).