# Tag Info

5

Hint: if $\lambda$ is an eigenvalue of $A$, what can you say about eigenvalues of $A^2$?

5

For any $m \times n$ and $n \times m$ matrices $A$ and $B$, the nonzero eigenvalues of $AB$ and $BA$ are the same. Namely, if $AB v = \lambda v$ with $\lambda \ne 0$ and $v \ne 0$, then $Bv \ne 0$ and $BA(Bv) = B(AB v) = \lambda B v$.

5

Go back to the definition: if $A$ has an eigenvalue $\lambda$ with corresponding eigenvector $x$, then $A x = \lambda x$. Now $A^2 x = \lambda^2 x$. This gives $\lambda^2 = -1$, contradiction!

4

The space of solutions of $Av=\lambda v$ is the eigenspace of the eigenvalue $\lambda$ so for example $W_{1}$ is the space of solutions for $Av=1\cdot v=v$ $$Av=v\iff(A-I)v=0$$ $$\begin{pmatrix}1-1 & 5 & 0\\ 0 & 1-1 & 0\\ 0 & 0 & 3-1 \end{pmatrix}\begin{pmatrix}a\\ b\\ c \end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0 \end{pmatrix}$$ $$... 3 We have that:$$\mathbf{A}^{2} = \mathbf{I} \implies \mathbf{A} = \mathbf{A}^{-1}$$But we know that if the eigenvalues of \mathbf{A} are \lambda_{1},\dots,\lambda_{n}, then the eigenvalues of \mathbf{A}^{-1} are \lambda_{1}^{-1},\dots,\lambda_{n}^{-1}, but if:$$\lambda_{i}=\lambda_{i}^{-1} \qquad \forall i \in \{1,\dots,n\}$$Then: ... 3 the nonzero eigenvalues of AB and BA are the same. the reason is tr(AB) = tr(BA), tr((AB)^2) = tr((BA)^2), \cdots implying that the the sum of the powers of eigenvalues of AB and BA are the same. through the newtons formula, this implies that the leading coefficients of characteristic polynomials are the same. 2 Say v\in \Bbb R^n is such that ABv = -v. Then I claim that u = Bv is a vector such that BAu = -u. In fact:$$ BAu = BABv = B(ABv) = B(-v) = -Bv = -u $$So if AB has eigenvalue -1, then so does BA. The opposite is of course true as well, by an analoguous argument. This argument also works for any other eigenvalue, so AB and BA share all ... 2 X^TX is a D\times D matrix, hence it has D eigenvalues (taken with multiplicity). Same goes for S^TS. Each element on the diagonal of S^TS is an eigenvalue of X^TX; moreover, the corresponding eigenvectors are columns of V; there are D such columns, so everything fits. 2 From where you left off,$$X = \frac{i \pm \sqrt{(-i)^{2}-4(2)}}{2} = \frac{i\pm\sqrt{-1-8}}{2} = \frac{i\pm \sqrt{-9}}{2} = \frac{i\pm i\sqrt{9}}{2} = \frac{i\pm 3i}{2} = 2i, -i$$2 An idea:$$C=S^{-1}DS\iff CS^{-1}=S^{-1}D\implies \color{red}{CS^{-1}x}=S^{-1}Dx=S^{-1}(\lambda x)=\color{red}{\lambda S^{-1}x}$$Thus, \;S^{-1}x\; is an eigenvalue of \;C\; corresponding to \;\lambda\; . Now, if the corresponding eigenspaces (for \;C, D\;) are one-dimensional, then we already get \;ky=S^{-1}x\;,\;\;k\in\Bbb F\;, so I'm not sure ... 2 The eigenvectors X^TX can be computed as follows: (1) Assume v is an eigenvector XX^T to eigenvalue \lambda\ne0. Then it holds XX^Tv=\lambda v, and$$ X^TX(X^Tv) = \lambda (X^Tv). $$Since \lambda \ne0, it follows X^Tv\ne0, hence X^Tv is an eigenvector of X^TX with eigenvalue \lambda. (2) The eigenvectors to eigenvalue zero are the ... 2 The multiplication operator (Mf)(x)=xf(x) on L^{2}[0,1] is a classical example of an operator with no eigenvalues, but its spectrum is [0,1]. M has no eigenvalue because Mf=\lambda f gives (x-\lambda)f=0, which forces f(x)=0 a.e.. To see that [0,1]\in\sigma(M), note that the constant function 1 is not in the range of (M-\lambda I) for ... 2 Spectral theory in infinite-dimensional spaces is quite a bit more complicated than in the finite-dimensional case. In particular, we have to distinguish between the spectrum \sigma(A) of an operator and its eigenvalues. Let A be a linear operator on a Banach space X over the scalar field C. We have$$ \sigma(A) = \{ \lambda \in C: (\lambda I - A) ...

2

the eigenvalue will change but not the eigenvector. that is if $Au = \lambda u,$ then $(A + 3I)u=(\lambda + 3)u.$ in general if $p(\lambda)$ is a polynomial, then $p(A)$ ahs an eigenvalue $p(\lambda).$ for eaxmple, an eigenvalue of $A^2 + 2A + 3I$ is $\lambda^2 + 2\lambda + 3.$

2

More precisely (allowing for infinite-dimensional vector spaces): there is no finite-dimensional invariant subspace with odd dimension. Hint: consider the characteristic polynomial of the restriction of $T$ to an odd-dimensional invariant subspace. What do you know about real roots of polynomials of odd degree?

2

Don't use the Cayley-Hamilton theorem; it is less elementary than what you need. And in any case in Axler's book it (8.20) follows results to the effect you are asking about (8.10, 8.18). In fact Axler defines the (algebraic) multiplicity of $\lambda$ as $\dim(G_\lambda)$, and then goes on to define the characteristic polynomial to be the product over ...

1

Items $\textbf{(a)}, \textbf{(b)}$ and $\textbf{(c)}$ are just meant to see if you can write the system in matrix form. That is, if you can recognize that: $$\begin{cases} x_1' = x_1 + 2x_2 \\ x_2' = 3x_1 + 2x_2 \end{cases} \iff \begin{bmatrix} x_1' \\ x_2' \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 ... 1 I think the definition of flag basis you are looking for is: If T is a linear transformation T : V \to V, then a flag basis for T is a basis \mathcal B = \{v_1, v_2, \dots, v_n\} of V such that$$Tv_j = \sum_{i = 1}^j b_{ji} v_inote that the summation is only taken up to i = j. This means that T is triangular in the basis \mathcal B. So, ... 1 Linear stability analysis for a discrete dynamical system only make sense close to an equilibrium whose e.v. are all strictly inside the unit ball (have negative real parts) as pointed out CTNT. In general usually for highly nonlinear systems like yours stability analysis via linearisation does not work. If you have eigenvalues with modulus very close to ... 1 The eigenvalues are the same, only X^TX has 0 as an additional eigenvalue. Eigenvectors of the same eigenvalue in both spaces are mapped to one another by X and X^T. 1 Substituting C with D you get  D S x = \lambda S x, so S x is an eigenvector of D. 1 Suppose v is an eigenvector of A with eigenvalue \lambda. What is A^2 v? 1 Hints: 1) The matrix's characteristic polynomial is \;x^4\; (i.e., the matrix is nilpotent) 2) A matrix is diagonalizable iff its minimal polynomial is the product of different linear factors 3) Thus, a nilpotent matrix is diagonalizable iff ... 1 Definition: The definition of the algebraic multiplicity k of a root \lambda_0 of a function f_A additionally contains that k is maximal. So, \begin{align*} f_A(\lambda)=(\lambda-\lambda_0)^kg(\lambda)\quad\text{and}\quad g(\lambda_0)\ne 0\tag{1} \end{align*} Roots: The roots \lambda_0 are elements of the domain of f_A. ... 1 If \lambda is the largest eigenvalue, then for "generic" initial vector v_0, we have (AA')^nv_0\approx \lambda^nv_0. Now use that (AA')^{n+1}=A(A'A)^nA' 1 Av = \lambda v \Rightarrow A^2v = \lambda Av = v $$So$$ \lambda Av = v \Rightarrow \lambda \lambda v = \lambda ^2v = v $$This shows \lambda = 1 or \lambda = -1. 1 Quadratic equation has real roots iff D \ge 0. In your case  D = (tr(A))^2 - 4det(A) \ge 0 . Hence matrix has real eigenvalue iff  (tr(A))^2 \ge 4det(A) , equivalent  det(A) \le (tr(A)/2)^2  1 It is easy to obtain the eigenvalues. The matrix B is at most rank-two (as a sum of two (at most) rank-one matrices) and can be expressed as$$ B=CD^T, \quad C:=[u,w], \quad D:=[v,z]. $$Since CD^T has the same nonzero eigenvalues as D^TC (note that a few answers prove exactly this in contrast to the original question), the (at most two) nonzero ... 1 As noticed in the comments, your optimization problem can be formulated as$${\rm max}_{\mathbf{x},\mathbf{y}} \mathbf{x}^T S(\mathbf{y}) \mathbf{x} ~~~{\rm s.t.} \|\mathbf{y}\|_{2} = 1,\|\mathbf{x}\|_{2} = 1,$$where$$ S(\mathbf{y})= S = \left [ \begin{array}{ccc} \mathbf{y}^{H}S_{11}\mathbf{y} & \ldots & \mathbf{y}^{H}S_{1n}\mathbf{y}\\ \vdots ...

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