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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 $$\pmatrix{4 & 0 & 1 \\ -2 & 1 & 0 \\ -2 & 0 & 1} - \pmatrix{3 & ... 1 Hint: Let A be the 3\times 3 matrix. If \mathbf{v_1}=\begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} is the eigenvector that corresponds to the eigenvalue \lambda_1 =1, in order to find the eigenspace V_{\lambda_1} we may solve the system:$$A\cdot \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix} = 1\cdot \begin{bmatrix} x_1\\x_2\\x_3\end{bmatrix},$$with ... 0 This question is still unanswered. I'll write the outline of a solution. Rewrite T_n as a sum of weighted graphs T_{i,j}, i < j, where (i,j) \in K_n and T_{i,j} contains the (weighted) edges of the path from i to j in T_n. In each of the summands, the weight of a given edge (i,j) is \frac{1}{paths(i,j)} where paths_{n}(i,j) denotes ... 1 A is diagonalizable if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that P^{−1}AP=B is a diagonal matrix. Or A=PBP^{−1}, where B is is a diagonal matrix. We have A^3=A, then A^3= PBP^{−1}PBP^{−1}PBP^{−1}=PB^3P^{−1}= A =PBP^{−1}. Thus B^3=B, note that the eigenvalues of A are the diagonal entries of ... 2 Tr(A^2)=\sum_{i}\lambda_i^2. Let the number of 1's be k, no. of -1's be l, then, Tr(A^2)=k+l and rank(A)= no. of nonzero eigenvalues of A, which is again k+l. 3 You are correct that eigenvalue \lambda = 2 has multiplicity n, but the dimension of the \lambda = 1 eigenspace is equal to n^2 - n. This linear transformation is diagonalizable. Think about a set of eigenvectors with eigenvalue 1: It is given by the set of matrices of all 0's except for a 1 in the i, j entry with i\neq 2. The determinant ... 1 For k^nA^n to converge to the zero matrix, k^n, k^n5^n and k^n7^n must all converge to zero. For k^n, k^n5^n and k^n7^n to converge to zero when n \rightarrow\infty , the value of k, 5k and 7k must be between -1 and 1. Since 7 is the greatest number in the diagonal matrix, the maximum value of k will correspond to that number. Therefore,$$-1 ...

1

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.

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 $B'^2=A'$, ...

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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 ...

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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.

0

Hint Note that \left(\begin{align}1 \\ 0 \\ 0\\ 0\\ 1\end{align}\right) and \left(\begin{align}1 \\ 0 \\ 0\\ 0\\ -1\end{align}\right) are two linearly independent eigenvectors. Also note that \left(\begin{align}0 \\ 1 \\ 1\\ 1\\ 0\end{align}\right), \left(\begin{align}0 \\ -2 \\ 1\\ 1\\ 0\end{align}\right) and \left(\begin{align}0 \\ 1 \\ 1\\ ... 0 LetA$and$B$be repectively the following matrices: Then$A= B +I_5$where$I_5$is the identity matrix. Since B and I commutes then the$eigenvalues(A)=eigenvalues(B) +eigenvalues(I_5)$(with the same order), that is$eigenvalue(A)=eigenvalue(B) +1$(for every eigenvalue of B). Now commuting the eigenvalues of$B$is easy since in the first row ... 2 Hint: Note that$(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$3$. 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 ... 0 Consider$B=A-\lambda_1 \frac{v_1 v_1^T}{\| v_1 \|^2}$. This will have the same eigenvectors but$v_1$will have eigenvalue$0$. Note that we need the symmetry to ensure that this does not alter the other eigenvectors or eigenvalues, because the eigenvectors for different eigenvalues are orthogonal and therefore$\left (A-\lambda_1 \frac{v_1 v_1^T}{\| v_1 ...

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

Let's say that we want to find an eigenvector that corresponds to the eigenvalue $\lambda_1=1$. We can solve the equation: $$\begin{array}{l}A\cdot \mathbf{x} = 1\cdot \mathbf {x}\\[2ex] \begin{bmatrix} 0.8&0\\2&1\end{bmatrix}\cdot \begin{bmatrix}x_1\\x_2\end{bmatrix}= \begin{bmatrix}x_1\\x_2\end{bmatrix}\\[3ex] \left\{\begin{array}{l} ... 0 Your first system of equations is: -0.2x + 0y = 0\\2x + 0y = 0. This tells us x = 0, but does not restrict y in any way. Thus ANY non-zero value of y lead to an eigenvector, (0,y) = y(0,1). Since any non-zero scalar multiple of an eigenvector is ALSO an eigenvector, we can choose ANY basis vector generating the subspace \{(0,y):y \in \Bbb R\}. ... 1 By the given condition, we see that A=\frac12I+K for some skew-Hermitian matrix K. Since skew-Hermitian matrices have purely imaginary eigenvalues, it follows that the real part of every eigenvalue of A is \frac12. The eigenvalues can be non-real. E.g. consider A=\frac12\pmatrix{1&-1\\ 1&1}. Its eigenvalues are \frac1{\sqrt{2}}e^{\pm ... 0 Let (,) denote the inner product. Let v be an eigenvector of AB with eigenvector k. WLOG assume we have it so that v is unit length: i.e. (v,v)=1. Then (ABv,v)=k(v,v)=k. Moreover, (v,ABv)=(v,kv)=\bar{k}(v,v)=\bar{k}. Now, k=(ABv,v)=(Bv,Bv)=(v,ABv)=\bar{k} so k=\bar{k} implies k\in\mathbb{R}. The other direction is similar. If you know ... 1 The eigenvalues of A^2 are the squares of the eigenvalues of A. The sum of the eigenvalues of any matrix (with algebraic multiplicity) is the trace. So the sum of the squares of the eigenvalues of A (with algebraic multiplicity) is the trace of A^2. 0 Trace of a Matrix is the sum of its Eigenvalues Proof that the Trace of a Matrix is the sum of its Eigenvalues 2 What you've written isn't actually quite right. If A is invertible, then A^{-1} exists, and thus$$ Ax = b \iff A^{-1}b = A^{-1}Ax = x.$$Thus this is the only solution, and it always exists. (Note that b = 0 implies x=0.) If A is not invertible, then Ax = b may have no solutions, an exact solution (x^*) or an uncountably large family of ... 2 I think you got confused. If A is invertible, then Ax = b \implies x = A^{-1}b, and so x is unique and determined by the expression A^{-1}b. Now consider:$$\begin{cases} x-y = 0 \\ 2x-2y = 0\end{cases}$$Your matrix A = \begin{bmatrix} 1 & -1 \\ 2 & -2\end{bmatrix} is not invertible and the system has the solution set \{ (x,x) \in \Bbb R^2 ... 1 The 2 \times 2 case is quite simple: such an A causes a shear transformation: one vector is fixed, the one perpendicular to it maps to one with the same perpendicular component, but the parallel component is changed (think of the effect on the basis consisting of (1,0) and (0,1): (1,0) \mapsto (1,0) and so is unchanged, whereas (0,1) \mapsto ... 2 Generally, it means not much in paricular, just that it is composed of Jordan block corresponding to the same eigenvalue. From the Jordan decomposition theorem, we see that A = V^{-1} J V, with J having constant diagonal entries, which are the eigenvalues of A. However, if the matrix has all the eigenvalues the same, and is in addition normal, you ... 1 Here is a shorter solution: You need only the symmetry of the matrix. Let \lambda_1\neq\lambda_2 and Ax=\lambda_1x, Ay=\lambda_2y. Then$$(Ax,y)-(x,Ay)=(\lambda_1x,y)-(x,\lambda_2y)=(\lambda_1-\lambda_2)(x,y)$$But the left side is 0, because of the symmetry :(Ax,y)=Ax\cdot y=x\cdot A^Ty=x\cdot Ay=(x,Ay). Since \lambda_1-\lambda_2\neq ... 2 Yes, eigenvectors belonging to distinct eigenvalues of symmetric positive matrix are orthogonal, and your solution is correct. More generally, this follows from the fact, that symmetric positive matrices are hermitian, and therefore normal (which is the most general class of matrices having this property). 1 If you introduce three new variables, x_1, x_2 and y_2 and set them equal to \frac{d}{dt}w_1, \frac{d}{dt}w_2 and \frac{d^2}{dt^2}w_2 respectively, then you can write it as a system of five coupled linear differential equations. In order to do this you have to take the derivative of the second equation, rewrite it as an expression for ... 0 for \lambda=2, a=(1,0,0) for \lambda=-3, b=(0,i,1) for \lambda=3, c=(0,-i,1) Note that \color{red}{\langle U,V\rangle=U\bar V^T} so \langle c,b\rangle=c\bar b^T=0.0+i.i+1.1=-1+1=0... (I think your pro was in definition of inner product) 1 Since |\lambda I-A| is of degree n, it has n roots and so A has n eigenvalues. Let \lambda_i be any eigenvalue of A and x\ne 0 be eigenvector for \lambda_i. Then Ax=\lambda_i x. So \overline{x}^TAx=\lambda_i \overline{x}^Tx. Take conjugate transpose on it, we have$$ ...

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 ...

1

It is not true. For example, let A be the identity, and any two vectors will satisfy the equation, since for the identity, every vector is an eigenvector with eigenvalue 1.

1

Let's forget about the basis of eigenvectors and just concentrate on change of basis. Say we're given the matrix $T_{\mathcal B}$ which represents some linear transformation $T: \Bbb R^n \to \Bbb R^n$ relative to some basis $\mathcal B = \{b_1, \dots, b_n\}$. But our vectors (column matrices) are given with respect to another basis $\mathcal C = \{c_1, ... 3 Consider the matrix $$\begin{pmatrix}0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0\end{pmatrix}$$ Its off-diagonal entries (in fact, all its entries) are nonpositive, but it has$i$as an eigenvalue. 0 The first assumption you applied assumes that$\theta\approx 0$, which allows you to linearize the differential equation and turns it into a linear autonomous and homogeneous differential equation, however such differential equation will always have an equilibrium solution at zero. The correct way to solve this is to first find the equilibrium solution of ... 0 You find the "true" or first order eigenspace by solving$\det({\bf A}-\lambda{\bf I}) = 0$, then second order eigenvector$\det(({\bf A}-\lambda{\bf I})^2)=0$which is true for the full space but not in the first space. This leaves the only one left$(0,1,0)^T$.${\bf A}(0,1,0)^T = (1,2,-1)^T$which gives us a hint that it is actually$(1,0,-1)^T$it ... 1 You know that if you have a vector $$u = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$$ then you have $$(A- 2I)u = \begin{pmatrix} b \\ 0 \\ -b \end{pmatrix} = v$$ You also know that$v$is an eigenvector of$A$, with eigenvalue 2, since it is in your calculated eigenspace (it can be written as$v = bv_1 - bw_1$). Hence$u$will be a generalized eigenvector ... 0 Important Observation That two eigenvalues are the same could mean we have either one or two eigenvectors. For instance the matrix $${\bf A} = \left[\begin{array}{cc} 1&1\\ 0&1 \end{array}\right]$$ has one eigenvector$[1,0]^T$, but two eigenvalues "1". The space corresponding to this eigenvalue is a generalized eigenspace, which has one true ... 0 An$n \times n$matrix is diagonalizable iff the dimensions of its eigenspaces sums to$n$. The eigenspace of the eigenvector$\lambda=2$is spanned by vectors$x$such that$Ax=2x$and this means that$(A-2I)x=0$, so the space of the eigenvectors has the same dimension as the kernel of$A-2I$, and since this dimension is$2$, and the dimension of the ... 3 An$n \times n$matrix is diagonalizable if(f) we can find$n$linearly independent eigenvectors. In particular, this is equivalent to saying that for each eigenvector$\lambda$, we have a number of eigenvalues equal to the algebraic multiplicity of that eigenvalue (i.e. the associated exponent in the characteristic polynomial). Because the rank of$A - ...

0

Since the rank is 1, you are able to have a free family of 2 $\lambda$-eigenvectors, which is a good start to find an eigenvector base (length 3). A in such a base would be a diagonal matrix

3

$\det(\lambda I-C)=\det\pmatrix{\lambda I&-A\\ -B&\lambda I}$. Since all square subblocks have the same sizes and the two subblocks at bottom commute, the determinant is equal to $\det(\lambda^2 I - AB)$. Therefore, the eigenvalues of $C$ are the square roots of eigenvalues of $AB$. That is, for each eigenvalue $t$ of $AB$, the two roots of ...

1

@A.G. idea is a good start: If we assume that $A,B$ commute, then we get an easy result. $C^2$ is symmetric, so it is diagonalisable. Let $P$ be a matrix which is the change to the diagonal basis. $(PC^2 P^{-1}) = (PCP^{-1})^2 = D$, where $D$ some diagonal matrix, so $PCP^{-1} = D^{1/2}$, which you can compute by taking the square roots of the entries of ...

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$0=\det(A)=\det(A-0\cdot I)\iff \text{$0$is an eigenvalue of$A$}$, iff there's a vector $x\neq0$ such that $Ax=0x$.

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$8$ will not be an eigenvalue of $A$ and $1$ may or may not be (depending upon whether $A$ has complex-valued entries or not). This is because: Let $M$ be an $n\times n$ matrix with eigenvalue $\lambda$ and corresponding eigenvector $\vec{x}$. Then we have: $A^n\vec{x}=A^{n-1}A\vec{x}=\lambda A^{n-1}\vec{x}=\lambda A^{n-2}A\vec{x}=\lambda^2 ... 1 The idea for the reverse is as follows: Since$\det A=0$, then it means any row of$A$can be written as a linear combination of the other rows. If that is true then (suppose$A$is$n\times n$)$\mathrm{rank} A<n$. In other words the linear transformation (Suppose the vector space is$V$,$n$-dimensional)$T:V\to V$given by$x\mapsto Ax$, is not onto. ... 2 Expansion (if you want to learn more) It can be any nilpotent matrix. Nilpotency of the matrix$\bf A$means that${\bf A}^k = {\bf 0}$for some$k \in \mathbb{N_+}$. All of it's eigenvalues needs to be$0$, it can also be written as${\bf A= VUV}^{-1}$, where$\bf U$is a strictly upper-triangular matrix. Such a matrix gets one more superdiagonal of 0 for ... 4 You need to use the Cayley Hamilton theorem.$A$satisfies its own characteristic polynomial which has degree 4. If$A^5=0$, then the minimal polynomial$p(A)$divides$A^5$. It is a polynomial of degree at most 4 since it has to divide the degree 4 characteristic polynomial. So the minimal polynomial$p(A)$is either$A$,$A^2$,$A^3$, or$A^4\$. In any ...

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