# Tag Info

1

The Harvard class page isn't actually using the trace method, as that computes each eigenvector from the other eigenvalue(s). It's just solving the equations directly. And since it took me way too long to realize that... Given an eigenvector $\lambda$ of the matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the associated eigenvector(s) ...

3

Hint: If $v$ is an eigenvector of $A$ for the eigenvalue $\lambda$ we have: $$Av=\lambda v \iff A^{-1}Av=\lambda A^{-1}v \iff A^{-1}v=\frac{1}{\lambda} v$$ so if the eigenvalues of $A$ are $\{\lambda_1 \cdots \lambda_n\}$, the eigenvalues of $A^{-1}$ are the inverses.

0

Define a matrix ${\bf{D}}=diag(d_1, \dots, d_n)$. Further define a matrix ${\bf{B}} = ({\bf{w}}_1,{\bf{w}}_2, \dots, {\bf{w}}_n)$. Note that ${\bf{B}}=\bf{VD}$. Hence, {\bf{B}} =\left ( {\bf{v}}_1, {\bf{v}}_2, \dots, {\bf{v}}_n \right ) \begin{pmatrix} d_1 &0 &\dots &0 \\ 0&d_2 &0 &\vdots \\ \vdots ...

0

Let $n = \dim V$. Choose $v \in V$ with $v \ne 0$. Then $$v, Tv,T^2 v, \dots , T^n v$$ is not linearly independent, because $V$ has dimension $n$ and we have $n+1$ vectors. Thus there exist complex numbers $a_0, \dots , a_n$, not all $0$, such that $$0 = a_0 v + a_1 Tv + \dots + a_n T^n v.$$ Note that $a_1, \dots, a_n$ cannot all be $0$, because ...

1

Your matrix $A-0I$ can be row-reduced to $$\pmatrix{0&a&b&c\cr0&0&1&f\cr0&0&0&2\cr0&0&0&0\cr}\ .$$ Now $A$ is diagonalisable if and only if this matrix has $2$ non-leading columns, and that occurs for one specific value of $a$. The values of $b,c,d,e,f$ are irrelevant. See if you can finish the problem from ...

0

A long-winded version of enzotib's answer: The $j$th row of $B^T$ is $w_j=d_jv_j$, so the $j$th element of $B^Tv_i$ is $d_j(v_i\cdot v_j)$, but the $v_i$ are orthonormal, so $B^Tv_i=d_i\mathbf e_i$, where $\mathbf e_i$ is the standard basis vector. The columns of $B$ are the images of the standard basis vectors, so B(d_i\mathbf ... 0 By definition $$B_{ij}=(w_j)_i,\qquad B^T_{ij}=(w_i)_j$$ Then \begin{align} (Av_k)_i&=A_{ij}(v_k)_j=B_{ih}B^T_{hj}(v_k)_j+d\delta_{ij}(v_k)_j\\ &=(w_h)_i(w_h)_j(v_k)_j+d(v_k)_i\\ &=(w_h)_id_h\delta_{hk}+d(v_k)_i\\ &=d_k^2(v_k)_i +d(v_k)_i\\ &=(d_k^2+d)(v_k)_i=c_k(v_k)_i \end{align} 0 Permuting rows two and three, as well as columns two and three yields the matrix $$A_2:= \begin{pmatrix} 1275 & 0 & -169 & -208 \\ 0 & 1275 &-208 & -256 \\ -169 & -208 & 1531& -208 \\ -208 & -256& -208 & 1444\\ \end{pmatrix}.$$ DenotingB:=\pmatrix{-169 ...

2

In a sense, those complex eigenvalues are the rotation. One way to think of a real eigenvalue is the amount by which a matrix stretches or shrinks things along a certain axis—the associated eigenvector. With a pair of complex eigenvalues (they always come in conjugate pairs for a real matrix), there’s no axis along which things are stretched, i.e., no real ...

3

Hint What can you say about the traces of the given matrices? (Alternatively, for three of the choices, one can find a suitable matrix $B$ for which the equation holds for all $A$.)

1

You can eliminate answers A and D by noting that the zero matrix B always satisfies that identity.

1

Note that $-x^3+6x^2+9x-14=-(x-1)(x+2)(x-7)$, so we may assume that $M$ is the diagonal matrix with entries $1,-2,7$ on the diagonal. Then it's easy to see that the characteristic polynomial of $M^{-3}$ is given by $(x-1)(x+(1/2)^3)(x-(1/7)^3)$.

0

A nice trick for finding eigenvectors for $2$ and $3$ dimensional matrices is to exploit the Cayley-Hamilton theorem (it works for higher dimensions too, but the cost of multiplying the matrices makes it less attractive). The Cayley Hamilton theorem says that if $p(x)$ is the characteristic polynomial of a square matrix $A$, then $p(A) = 0$ (scalars in this ...

1

Note that $A = \operatorname{Re}A +i\operatorname{Im}A$. Also $$z(x) = \begin{bmatrix}\operatorname{Re}x\\\operatorname{Im}x\end{bmatrix}\qquad z^{-1}\left(\begin{bmatrix}v_1\\v_2\end{bmatrix}\right) = v_1 + iv_2$$ So, since $K = z\circ A\circ z^{-1}$, \begin{align}K\begin{bmatrix}v_1\\v_2\end{bmatrix} &= zAz^{-1}\begin{bmatrix}v_1\\v_2\end{bmatrix}\\ ... 1 Up to a minor bug, your ideas are correct. You eigenvector is v_1=\begin{pmatrix}1\\0\\0\end{pmatrix}. You want v_2 such that (A-2I)v_2=v_1, or, as you said, \begin{pmatrix}a\\1\\0\end{pmatrix} for arbitrary a (because first generalisd eigenvector is defined up to an eigenvector). For the v_3 you want (A-2I)v_3=v_2 ... 1 By diagonalization, you can find that any diagonal matrix A can be represented asA = P D P^{-1},$$where D is a diagonal matrix in which each element is an eigenvalue, and then P is a nonsingular matrix (i.e. its columns are linearly independent). Now, the power of a diagonalization is that you can easily figure out the power of a matrix (pun ... 5 Hint: a rotation that is neither 180^\circ nor 0^\circ has no real eigenvalues. 0 The matrix A_iP_i, is a matrix containing a lot of zeros, and its i-th row is equal to the i-th row of P_i. That is, the resulting matrix has rows all with Euclidean norm equal to 1. This implies \|C\|_2 \le \sqrt n:$$ \|Cx\|_2^2 =\sum_i (c_i^Tx)^2\le \sum_i \|c_i\|^2 \|x\|^2 = n \|x\|^2, $$where c_i is the i-th row of C. This bound is ... 0 (a) The unit normal vector to the plane is (3, -2, 1)/\sqrt{14}. We could find a vector within the plane by choosing an arbitrary vector and subtracting its projection onto the unit normal vector, but it is easier to just eyeball it:$$(1,1,-1)\cdot(3,-2,1) = 0To get a 2nd vector, an easy method (since we are in \Bbb R^3) is to just take the cross ... 0 It's an easy thing to show that \sigma(T) \subseteq \{ \lambda : |\lambda| \le \|T\| \} because the following inverse series converges in operator norm for \|T\| < |\lambda| \begin{align} (T-\lambda I)^{-1} &= \frac{1}{\lambda}(\frac{1}{\lambda}T-I)^{-1} \\ & = -\frac{1}{\lambda}(I-\frac{1}{\lambda}T)^{-1} \\ ... 0 It fails if the columns of E are linearly dependent. This would include all a_i = b_i, for example. This would also include all b_i = (1/2) a_i. In general, the existence of a triple of real constants \gamma, \alpha, \beta such that \gamma + \alpha a_i + \beta b_i = 0 for all i would mean your E^T E has rank two or one, is positive ... 3 Suppose T is a bounded operator on a Banach space X. \lambda\in\rho(T) iff T-\lambda I is a linear bijection. In that case, the inverse (T-\lambda I)^{-1} is automatically continuous by the closed graph theorem. There are three basic things that can stand in the way of T-\lambda I being invertible. T-\lambda I is not injective. Equivalently, ... 1 For T to have \lambda as an eigenvalue, T-\lambda I must be non-injective. For \lambda to be in the spectrum of T, it must only be non-invertible. These are equivalent when T is an operator on a finite-dimensional space, but not in general! For example, let T be the shift operator (x_0,x_1,\dots) \mapsto (0,x_0,x_1,\dots) on your favorite ... 0 Hint: \lambda is an eigenvalue of A if \det(A - \lambda I) = 0. On the other hand, \lambda is an eigenvalue of A - \alpha I if \det(A - (\lambda + \alpha)I) = 0. If Ax = \lambda x, try computing (A - \alpha I)x 1 That's quite essential in the study of Linear Algebra and in the study of Linear Transformations so it' written in more or less every book of the subject. Anyway the main thing is this: You're interested in finding the vector that are invariant with the transformation i.e.Av=\lambda v$$Thus you rewrite$$Av - \lambda v =0$$which is$$(A - \lambda I) ...

2

Your lecturer is correct: there is a lot wrong with the proof above. The result (that every eigenvalue of a real symmetric matrix is real) has nothing to do with the Fundamental Theorem of Algebra---if the characteristic polynomial had no complex roots, then there would be no eigenvalues and the claimed result would be trivially true. The quantifier in the ...

1

Since $A$ is positive semi-definite, $A^{1/2}$ exists and is Hermitian too. Since $tr(AB)=tr(BA)$ $$tr(AB)=tr(A^{1/2}A^{1/2}B)=tr(A^{1/2}BA^{1/2})$$ Since $$(A^{1/2}BA^{1/2})^*=(A^{1/2})^*B^*(A^{1/2})^*=A^{1/2}BA^{1/2}$$ it is Hermitian and thus all its eigenvalues are real. So $tr(A^{1/2}BA^{1/2})$ is real.

1

Regarding the fishy solutions, when doing the discretization it can be important to regularize the solution or really strange things can happen when you try and solve the matrix-vector equation. As I have not investigated very many real problems in physics but mostly mimicked these types of equations for applications to other fields, I can't really say ...

0

Start finding the Jordan Canonical form of A, which in this case is just a diagonal matrix since eigen values are $\frac{1}{2}(9\pm i\sqrt{71})$, i.e. $$A=S^{-1}DS$$ Then you have $$A^{n}=S^{-1}DSS^{-1}DS...S^{-1}DS=S^{-1}D^{n}S$$ In this case since D is diagonal $D^{n}$ is just the diagonal matrix ...

0

If Bv=0, then as v is non-zero, so B cannot be invertible. So, det(B)=0 Hence det(AB)=det(BA)=0 So AB and BA both have eigenvalue 0. Not if c is any other non-zero eigenvalue of AB,then AB(v)=cv Proceeding as you did, if Bv=0, then we get c=0. So Bv can't be zero and hence c is an eigenvalue of BA corresponding to eigenvector Bv.

0

No. Counterexample: consider the scalar case, where $A=2$, every $I-B_{ii}$ (with $i>1$) is equal to $\frac14$ and $$M = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0&\frac12&0&\frac32\\ 0&0&\frac12&\frac32\\ 0&0&\frac32&\frac12 \end{bmatrix}.$$ The eigenvalues of $M$ are $0,\frac12,-1$ and $2$. By continuity, if you ...

1

Suppose $A(x)=cx, B(A(x))=cB(x)=A(B(x))$, this implies that $B(x)$ is an eigeinvector of $A$ associated to $c$, since $A$ has $n$ distinct eigenvalues, the eigenspace associated to $c$ has dimension 1, thus $B(x)=dx$.

0

Step 1) Ax=λx Step 2) y'Ax=λy'x Step 3) y'Ax-λy'x=0 Step 4) (y'A-λy')x=0 Step 5) (µy'-λy')x=0 Step 6) (µ-λ)y'x=0 How: µ $\neq$ λ -> µ-λ $\neq$ 0 This way: y'x=0

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Hint: Calculate $y'Ax$ two different ways, and relate the answer to $y'x$.

0

Matrix cookbook says, $det(A) = \prod _{i} \lambda _{i}$ where, $\lambda _{i} = eig(A)$

1

The geometric multiplicity of an eigenvalue cannot (necessarily) be detected from the characteristic polynomial $\det(\lambda I - A)$ so it is not clear how calculating a determinant will help. Instead, if $\lambda = 4$ is an eigenvalue of $A$ with geometric multiplicity $2$ then you must have $\mathrm{rank} (A - 4I) = 4 - 2 = 2$. Thus, you need to find ...

1

This seems to address your problem, I found a very nice basis for the eigenvectors of a matrix with all entries $1.$ The reason we know the columns are independent is that they are perpendicular to each other, ordinary dot product of columns is zero. I am encouraging you to do something along these lines. $$\left( \begin{array}{rrrrrrrrrr} 1 & ... 0 All eigenvectors sharing a common eigenvalue \lambda  form a subspace: the kernel of the operator A-\lambda I. Using the Gram-Schmidt process we can construct an orthonormal basis for this subspace. The vectors of this basis have the needed properties. 3 The typical approach to this problem is not to show directly that an eigenvalue with multiplicity m for a symmetric matrix has an m-dimensional space of corresponding eigenvectors but to use an inductive argument which shows it indirectly. In order to do that, it is more comfortable to talk about self-adjoint maps instead of real symmetric matrices ... 0 Eigenvalues of a matrix A are those values,say \lambda which satisfy the equation \det(\lambda I_{n}-A)=0. This equation is called the characterstic equation.But we can see that the characterstic equations for A and A^{T} are the same and hence,their eigenvalues. Just try seeing how their characterstic equations are same. 0 hint Let |M| denote the determinant of matrix M. Then$$|A-xI|=|(A-xI)^T|= |A^T-xI|.$$2  P_{\lambda}(A^{T})=\det(\lambda I_{n}-A^{T})= \det((\lambda I_{n}-A)^{T})= P_{\lambda}(A)  0 Theorem. If A is an n \times n matrix, then A is similar to A^{\top}. (Try to find out what meaning we assign to "similar"). Theorem. The determinant function is multiplicative. The statement under consideration follows from these two theorems: If A is an n \times n matrix, then there is some invertible Q such that A = Q^{-1}A^{\top}Q, ... 1 Here’s a hint, then: What’s the relationship between the determinants of A and A^T? 2 Every square matrix is similar to a matrix in what's called Jordan Canonical Form. This has various properties, but most important here is that it is upper triangular, and the eigenvalues (of both the new and original matrix) are on the diagonal of the resulting matrix. The way to think about this process is that we change bases, and in that new basis ... 1 Indeed in 2-D, the only nontrivial nilpotent matrix is \left(\begin{array}{cc}0 & \epsilon\\ 0 & 0 \end{array}\right), \epsilon\neq0, (or it's transpose), and the equality can hold iff q=\frac{d_{ii}}{d_{jj}}. So this argument suggests it cannot be done. Next, why not suppose that N is upper triangular. Since N has all zeros on the ... 1 In control theory and dynamical systems you have modal decomposition, which is a very useful tool to quickly create the dynamic equation for a given (real life) system Given a system of differential equation: \dot x(t) = Ax(t), x(0) = x_o, A has distinct eigenvalues Then the solution to this equation is given as: x(t) = \sum\limits_{i=1}^n ... -1 If your problem comes from a more strongest formulation like, for example, this ODE -u''(x)+\mu u(x)=f(x) with some boundary conditions try to use the condition of eigenvalue there. Continuing with the example, you know that f \neq 0 will be an eigenvector with eigenvalue \lambda if and only if -\lambda f''(x)+\mu \lambda f(x)=\lambda f(x). So you ... 1 This is fairly standard stuff. If A v_k = \lambda_k v_k then let V = \begin{bmatrix} v_1 & \cdots & v_n\end{bmatrix} and \Lambda = \operatorname{diag}(\lambda_1, \cdots , \lambda_n ). Note that V is invertible. Then the above equations can be written as AV = V \Lambda which gives V^{-1} A V = \Lambda. If A is similar to a diagonal ... 2 Well, one way to look at it is considering the identity$$e^{i\theta}=\cos(\theta)+i\sin(\theta) which tells us that the function $e^{i\theta}$ traces out the unit circle of the complex plane. One might note that if we have an eigenvalue of the form $e^{i\theta}$ then powers of it are of the form $e^{ni\theta}$ - so they are just rotating around the unit ...

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