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8

Let $v$ be an eigenvector of $A$ with eigenvalue $a$, and of $B$ with eigenvalue $b$. Then $$(AB-BA)v = A(bv)-B(av) = b(Av)-a(Bv) = bav-abv=(ba-ab)v = 0.$$ This is true of any joint eigenvectors. Further, $(AB-BA)$ is a linear operator, so it is also zero on any linear combination of these eigenvectors. Since you have $n$ linearly independent ...

5

From $p$ alone it is not clear that $A$ is diagonalizable. Fortunately, we easily find that the kernel is two-dimensional so that indeed $A$ is similar to a diagonal matrix, namely $$A'=\begin{pmatrix}0&0&0&0\\0&0&0&0\\0&0&4&0\\0&0&0&1\end{pmatrix}.$$ We can immediately name four matrices $B'$ with $B'^2=A'$, ...

4

$P$ will diagonalize $A$ if the columns of $P$ are eigenvectors of $A$. It looks like the 3rd column of $P$ is an eigenvector for the eigenvalue 7, so then you just need to see if the other three columns span the eigenspace for the eigenvalue 1.

3

If the eigenvalues are linearly independent, they form a base of $\Bbb R^n$. Their image also completely fixes the actions of $A$ and $B$, so you just need to check that each of the eigenvectors is mapped to the same image by $AB$ and $BA$, but this is trivial.

3

You are correct that eigenvalue $\lambda = 2$ has multiplicity $n$, but the dimension of the $\lambda = 1$ eigenspace is equal to $n^2 - n$. This linear transformation is diagonalizable. Think about a set of eigenvectors with eigenvalue 1: It is given by the set of matrices of all $0$'s except for a $1$ in the $i, j$ entry with i$\neq 2$. The determinant ...

2

Generally, it means not much in paricular, just that it is composed of Jordan block corresponding to the same eigenvalue. From the Jordan decomposition theorem, we see that $A = V^{-1} J V$, with $J$ having constant diagonal entries, which are the eigenvalues of $A$. However, if the matrix has all the eigenvalues the same, and is in addition normal, you ...

2

$Tr(A^2)=\sum_{i}\lambda_i^2$. Let the number of $1$'s be $k$, no. of $-1$'s be $l$, then, $Tr(A^2)=k+l$ and $rank(A)=$ no. of nonzero eigenvalues of $A$, which is again $k+l$.

2

I think you got confused. If $A$ is invertible, then $Ax = b \implies x = A^{-1}b$, and so $x$ is unique and determined by the expression $A^{-1}b$. Now consider: $$\begin{cases} x-y = 0 \\ 2x-2y = 0\end{cases}$$Your matrix $A = \begin{bmatrix} 1 & -1 \\ 2 & -2\end{bmatrix}$ is not invertible and the system has the solution set $\{ (x,x) \in \Bbb R^2 ... 2 What you've written isn't actually quite right. If$A$is invertible, then$A^{-1}$exists, and thus $$Ax = b \iff A^{-1}b = A^{-1}Ax = x.$$ Thus this is the only solution, and it always exists. (Note that$b = 0$implies$x=0$.) If$A$is not invertible, then$Ax = b$may have no solutions, an exact solution ($x^*$) or an uncountably large family of ... 2 This is actually fairly straightforward. Let the symmetric matrix be$A$. Let's start with the classic result: Let$\lambda$be an eigenvalue of$A$with eigenvector$v$. Then$\lambda \in \mathbb{R}$. Proof:$v^{\dagger} v$is real for any complex vector. By definition, $$\lambda v^{\dagger}v = v^{\dagger} A v.$$ Taking conjugate transpose on both ... 2 Hint: Note that$(1,0,0,0,1)^T$and$(0,1,1,1,0)^T$are eigenvectors by inspection. What are their corresponding eigenvalues? (You just need to compute$Av$for each of the above.) Now observe that the matrix has only$2$linearly independent columns, so its kernel has dimension$3$. Thus$0$is also an eigenvalue with multiplicity$3$. 2 Yes, eigenvectors belonging to distinct eigenvalues of symmetric positive matrix are orthogonal, and your solution is correct. More generally, this follows from the fact, that symmetric positive matrices are hermitian, and therefore normal (which is the most general class of matrices having this property). 1 Here is a shorter solution: You need only the symmetry of the matrix. Let$\lambda_1\neq\lambda_2$and$Ax=\lambda_1x$,$Ay=\lambda_2y$. Then $$(Ax,y)-(x,Ay)=(\lambda_1x,y)-(x,\lambda_2y)=(\lambda_1-\lambda_2)(x,y)$$ But the left side is$0$, because of the symmetry :$(Ax,y)=Ax\cdot y=x\cdot A^Ty=x\cdot Ay=(x,Ay)$. Since$\lambda_1-\lambda_2\neq ...

1

The $2 \times 2$ case is quite simple: such an $A$ causes a shear transformation: one vector is fixed, the one perpendicular to it maps to one with the same perpendicular component, but the parallel component is changed (think of the effect on the basis consisting of $(1,0)$ and $(0,1)$: $(1,0) \mapsto (1,0)$ and so is unchanged, whereas $(0,1) \mapsto ... 1 Think about (simple) eigenspaces. The eigenvalue tells you the scaling factor for how that matrix acts on the corresponding eigenspace. (There are some details about generalized eigenvalues that aren't super relevant here) So in the eigenvalue 0 case, the eigenvector tells you what subspace is getting collapsed to the origin. 1 For$k^nA^n$to converge to the zero matrix,$k^n, k^n5^n$and$k^n7^n$must all converge to zero. For$k^n, k^n5^n$and$k^n7^n$to converge to zero when n$\rightarrow\infty$, the value of k, 5k and 7k must be between -1 and 1. Since 7 is the greatest number in the diagonal matrix, the maximum value of k will correspond to that number. Therefore, $$-1 ... 1 The elements of your numerical eigenvector all have the same sign, so there is a representative of the eigenspace which has all nonnegative entries (e.g. the one taken by multiplying all the entries of your numerical eigenvector by -1). Of course the representative of probabilistic interest is the one with all nonnegative entries and which has a sum of ... 1 At the 4th row of your working, It is supposed to be (-7-\lambda)(-6+\lambda-6\lambda+\lambda^2-8) not (-7-\lambda)[-6+\lambda-6\lambda+\lambda^2+8). Hope you can continue to work it from here. 1 Hint: Let A be the 3\times 3 matrix. If \mathbf{v_1}=\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} is the eigenvector that corresponds to the eigenvalue \lambda_1 =1, in order to find the eigenspace V_{\lambda_1} we may solve the system:$$A\cdot \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = 1\cdot \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix},$$with ... 1 The eigenspace associated with the eigenvalue \lambda will be the set of all solutions to the equation$$(A-\lambda I)x=0$$One of your eigenvalues is 3, let's look at that one. What we need to do here is solve$$(A-3I)x=0$$So first off, what's A-3I? It's$$\pmatrix{4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1} - \pmatrix{3 & ... 1 Since$|\lambda I-A|$is of degree$n$, it has$n$roots and so$A$has$n$eigenvalues. Let$\lambda_i$be any eigenvalue of$A$and$x\ne 0$be eigenvector for$\lambda_i$. Then$Ax=\lambda_i x$. So$\overline{x}^TAx=\lambda_i \overline{x}^Tx$. Take conjugate transpose on it, we have $$... 1 Correct. The Hamiltonian is (at least, should be) self-adjoint since it is an observable. The spectral theorem says that the eigenfunctions \psi_n form an orthonormal basis, and so H^2 can have no more eigenfunctions that are linearly independent of those of H. H\psi_n=E_n\psi_n, and we have f(H) \psi_n = f(E_n) \psi_n, certainly for any polynomial ... 1 If you introduce three new variables, x_1, x_2 and y_2 and set them equal to \frac{d}{dt}w_1, \frac{d}{dt}w_2 and \frac{d^2}{dt^2}w_2 respectively, then you can write it as a system of five coupled linear differential equations. In order to do this you have to take the derivative of the second equation, rewrite it as an expression for ... 1 Let's say that we want to find an eigenvector that corresponds to the eigenvalue \lambda_1=1. We can solve the equation:$$\begin{array}{l}A\cdot \mathbf{x} = 1\cdot \mathbf {x}\\[2ex] \begin{bmatrix} 0.8&0\\2&1\end{bmatrix}\cdot \begin{bmatrix}x_1\\x_2\end{bmatrix}= \begin{bmatrix}x_1\\x_2\end{bmatrix}\\[3ex] \left\{\begin{array}{l} ... 1 for$\lambda = 0.8$you need to solve $$\begin{bmatrix}0.8 & 0\\ 2 & 1\end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = 0.8\begin{bmatrix}x \\ y \end{bmatrix}$$ so $$\begin{bmatrix}0.8 x \\ 2x+y \end{bmatrix} =\begin{bmatrix}0.8x \\0.8 y \end{bmatrix}$$$0.8x=0.8x$doesn't tell you anything but$2x+y=0.8y$gives you$y=-10x$so the ... 1 The eigenvalues of$A^2$are the squares of the eigenvalues of$A$. The sum of the eigenvalues of any matrix (with algebraic multiplicity) is the trace. So the sum of the squares of the eigenvalues of$A$(with algebraic multiplicity) is the trace of$A^2$. 1 By the given condition, we see that$A=\frac12I+K$for some skew-Hermitian matrix$K$. Since skew-Hermitian matrices have purely imaginary eigenvalues, it follows that the real part of every eigenvalue of$A$is$\frac12$. The eigenvalues can be non-real. E.g. consider$A=\frac12\pmatrix{1&-1\\ 1&1}$. Its eigenvalues are$\frac1{\sqrt{2}}e^{\pm ...

1

$A$ is diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that $P^{−1}AP=B$ is a diagonal matrix. Or $A=PBP^{−1}$, where $B$ is is a diagonal matrix. We have $A^3=A$, then $A^3= PBP^{−1}PBP^{−1}PBP^{−1}=PB^3P^{−1}= A =PBP^{−1}$. Thus $B^3=B$, note that the eigenvalues of $A$ are the diagonal entries of ...

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