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No, it's famously false. The usual counterexample is $A=\begin{pmatrix}0&1&0&0\\ 0&0&0&0\\ 0&0&0&1\\ 0&0&0&0\end{pmatrix}, B=\begin{pmatrix}0&1&0&0\\ 0&0&1&0\\ 0&0&0&0\\ 0&0&0&0\end{pmatrix}$ for both of which $0$ has algebraic multiplicity $4$ and geometric ...

3

First, if $A$ is a matrix $m\times n$, then $B$ has to be a $n\times m$ matrix (otherwise it doesn't make sense to talk about $tr(AB)$.) Now, you can see $B\mapsto tr(AB)$ as a function from $f:\mathbb{R}^{n\times m}\to \mathbb{R}$ and $\frac{d}{dB}[tr(AB)]$ will be the usual gradient of $f$. This gradient is expected to be some "vector" in $\mathbb{R}^{n\... 3 Hint: What does$T(T+2\operatorname{Id}_V)$equal to? 3 You have $$(A\otimes B)^n = A^n \otimes B^n$$ so $$\exp(A\otimes I) = \sum_{n=0}^\infty \frac{1}{n!}(A\otimes I)^n = \sum_{n=0}^\infty \frac{1}{n!} A^n \otimes I = \exp(A) \otimes I$$ 2 It's tricky to speak definitively here, since we are missing some context, but I think there are some issues here. Specifically, you seem to be too reliant on citing results that say, more or less directly, what it is you're supposed to be proving. For example,$iii) \implies iv)$Because column rank = row rank for any matrix, the row rank must be n ... 2 I think the key point is to understand the meaning of the derivative in this context where$f:\mathbb R^{m \times n} \to \mathbb R$. My favorite way to think about the gradient of a function$f:\mathbb R^{m \times n} \to \mathbb R$is $$\tag{1} f(B + \Delta B) \approx f(B) + \langle \nabla f(B), \Delta B\rangle.$$ In this equation,$\Delta B$is a matrix ... 1 As I understand it, you want to know the origins of the equation $$(A-\lambda I)v = 0$$ where$A$is a matrix,$v$is an eigenvector of$A$and$\lambda$is an eigenvalue. The fact of the matter is that this is a quite opaque expression. It has lost a lot of resemblance to the original intent because it has been heavily manipulated algebraically. The ... 1 We have$U= span \{(1,1,0,0), (0,0,1,1)\}.$Try$W= span \{(1,0,0,0), (0,0,0,1)\}.$1 If you have a system like this you can use forward substitution,$Ax=b$then $$x_{1} = \frac{b_{1}}{a_{1,1}}$$ and $$x_{i} = \frac{b_{i} - \sum_{j=1}^{i-1} a_{i,j} x_{j} }{ a_{i,i} }$$ 1 counter example: $$\begin{bmatrix} -1 & 1 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & -1 \end{bmatrix}$$ $$\begin{bmatrix} -1 & 1 & 0 & 0\\ 0 & -1 & 1 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ \end{bmatrix}$$ these two matrices have the same eigen ... 1 [Begin Edit] My initial answer was incorrect, but I believe it is interesting to explain how and why it is incorrect and provide some comments elaborating upon the correct answers posted here (I have left my initial incorrect answer unedited below). The Jordan normal form classifies all matrices up to similarity transformations. It shows that matrices have ... 1 From$T^2 + 2T + I = 0 \tag 1$we can of course write$T(T + 2I) = -I, \tag 2$or$T(-(T + 2I)) = I, \tag 3$which of course shows that$T^{-1} = -(T + 2I), \tag 4$in accord with the answer given by user10354138; indeed, any time an operator such as$T$satisfies a polynomial with non-vanishing constant term,$\displaystyle \sum_0^n \beta_i T^i = 0,...

1

Take the equation $$AX-XA=C$$ Pick $v_i, v_j$ and calculate the following: $$\langle v_i,(AX-XA)v_j\rangle=\langle v_i,AXv_j\rangle-\langle v_i,XAv_j\rangle$$ Since $A$ is symmetric, we can move it to $v_i$ in the first term. So we get: $$\langle v_i,(AX-XA)v_j\rangle=(\lambda_i-\lambda_j)\langle v_i,Xv_j\rangle$$ Motivated by this, we define $X$ by  ...

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