# Tag Info

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Hint: a rotation that is neither $180^\circ$ nor $0^\circ$ has no real eigenvalues.

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 $A$.)

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You have to reconstruct the (right) eigenbasis of $A$ first. Since two eigenvectors are given, there is enough information to find the second right eigenvector. Let $v$ denote this eigenvector, then $Ap_{r2}=-p_{r2}$. Multiplying this from the left by $p_{l1}^T$, we get $$-p_{l1}^Tp_{r2} = p_{l1}^TAp_{r2} = 2p_{l1}^Tp_{r2},$$ which proves ...

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, ...

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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 ...

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A question you might wish to answer first: What is the relationship between the eigenvalues of $A$ and the eigenvalues of $A^3$?

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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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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 ...

2

$P_{\lambda}(A^{T})=\det(\lambda I_{n}-A^{T})= \det((\lambda I_{n}-A)^{T})= P_{\lambda}(A)$

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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 ...

2

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

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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: $x(t) = \sum\limits_{i=1}^n ... 1 For$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)$on your favorite ... 1 By diagonalization, you can find that any diagonal matrix$A$can be represented as $$A = P D P^{-1},$$ 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 ... 1 You can eliminate answers A and D by noting that the zero matrix B always satisfies that identity. 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 ... 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 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 ... 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. $$\left( \begin{array}{rrrrrrrrrr} 1 & ... 1 Suppose A(x)=cx, B(A(x))=cB(x)=A(B(x)), this implies that B(x) is an eigeinvector of A associated to c, since A has n distinct eigenvalues, the eigenspace associated to c has dimension 1, thus B(x)=dx. 1 Hint: the determinant is equal to: \operatorname{det}(A-\lambda I) = (1-\lambda)\operatorname{det}\begin{pmatrix} 2-\lambda & c\\ c & 3-\lambda \end{pmatrix} + (-1) \cdot a \cdot \operatorname{det} \begin{pmatrix} a & c \\ b & 3- \lambda \end{pmatrix} + \dots Can you finish it yourself? 1 This inequality does not hold if M is a positive scalar multiple of the identity matrix. More generally, this inequality does not hold (strictly) if all the column sums of M are equal, i.e., if \mathbf{1}^TM = \alpha\mathbf{1}^T, for some positive scalar \alpha < s. To see this, observe that$$ \mathbf{1}^T(sI - M) = (s - \alpha)\mathbf{1}^T ... 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$(x-1)(x+(1/2)^3)(x-(1/7)^3)$. 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 Here’s a hint, then: What’s the relationship between the determinants of$A$and$A^T$? 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 ... 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 $$(A - \lambda I) ... 1 Up to a minor bug, your ideas are correct. You eigenvector is v_1=\begin{pmatrix}1\\0\\0\end{pmatrix}. You want v_2 such that (A-2I)v_2=v_1, or, as you said, \begin{pmatrix}a\\1\\0\end{pmatrix} for arbitrary a (because first generalisd eigenvector is defined up to an eigenvector). For the v_3 you want (A-2I)v_3=v_2 ... 1 Your matrix A-0I can be row-reduced to$$\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 ...