# Tag Info

### Closed form formula for $(\mathbb I+R)^{-1}$ where $R$ is an orthogonal matrix?

If $-1$ is an eigenvalue of $R$, $\mathbb I + R$ is not invertible. Otherwise, let $P(x)= \det(x\mathbb I - R)$ be the characteristic polynomial of $R$. Note that $P(-1) = \det(-\mathbb I - R) \ne 0$....
• 453k
Accepted

### Is $\text{Id}-T$ always invertible when $\lim_{n\to\infty} T^n = 0$?

Injective: Let us first prove that $I-T$ is injective. For this we check that the kernel is trivial. Let $v$ be in the kernel of $I-T$, i.e. $(I-T)v=0$, then $v=Iv=Tv$. However, this cannot be as ...
• 20.4k

• 32.7k
1 vote

### Prove that the given function is invertible

Let \begin{align} x+y+z^2&=v &&\\ x-y+z&=u&&\\ 2x+y-z&=w \end{align} In this system, $v,u,w$ are expressed in terms of $x,y,z$. Can you express $x,y,z$ in terms of $v,u,w$?
• 1,442
1 vote
Accepted

### If $X\sim\mathcal N(x,\Sigma)$, what is $\operatorname E\left[\left\|\Sigma^{-1}(X-x)\right\|^2\right]$?

The law of $X \sim \mathcal{N}(x, \Sigma)$ can be realized via $$X = x + \Sigma^{1/2} Z$$ where $Z \sim \mathcal{N}(0, I_d)$ and $\Sigma^{1/2}$ is the unique PSD square root of $\Sigma$. Hence, ...
• 170k
1 vote
Accepted

### Is this special BLOCK upper triangular matrix diagonalizable?

Since the two diagonal sub-blocks are diagonalisable (because they are Hermitian), $A$ is diagonalisable if and only if it is similar to $B=\pmatrix{A_{1,1}\\ &A_{2,2}}$. Consequently, by Rothâ€™s ...
• 141k
1 vote

### Is this special BLOCK upper triangular matrix diagonalizable?

If $A_{1,1}$ and $A_{2,2}$ have distinct eigenvalues the matrix is straightforwardly diagonalizable since all eigenvalues of $A$ are distinct. If $A_{1,1}$ and $A_{2,2}$ share eigenvalues, it may ...
• 70
1 vote

### The inverse of a lower triangular matrix is lower triangular

Let $\mathbb{K} \in \{\mathbb{R},\mathbb{C}\}$ be either the field of real numbers or the field of complex numbers. Definition (Lower triangular matrix). A (symmetric) lower triangular matrix is a ...

Only top scored, non community-wiki answers of a minimum length are eligible