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0

If $\lambda=1$ is an eigenvalue, then there is a vector $r$ such that $Ar=r$ Let $r=xi+yj+zk$ $Ar=r$ would mean that $3x-2y+3z=x$ or $2x-2y+3z=0$ $-y+3z=y$ or $3z=2y$ $-x+2y-2z=z$ or $-x+2y-3z=0$ Are these equations consistent?


2

$$A-I=\begin{bmatrix} 3 & -2 & 3\\ 0 & -2 & 3\\ -1 & 2 & -3 \end{bmatrix}$$ And you can see the two last columns are proportionate so the matrix is not invertible and $\lambda=1$ is an eigenvalue My computation of $\det(A-\lambda I)=-\lambda^3+\lambda^2+13\lambda-13$


2

no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$ for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$ pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = ...


1

Assume $v$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then $v$ is eigenvector of $M^2$ wih eigenvalue $\lambda^2\ge0$. Since $M^2$ has only negative eigenvalues $-1$ and $-1-\epsilon$, $M$ has no eigenvectors. Thus the matrix $A$ that maps $e_1\mapsto e_1$ and $e_2\mapsto Me_1$ is invertible and since $M$ maps $Me_1$ to $-e_1$ we find that ...


2

Let $M=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. Then we want $M^2=\left(\begin{matrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{matrix}\right)=\left(\begin{matrix}-1&0\\0&-1-\epsilon\end{matrix}\right)$. Now we can't have $b=0$ or $c=0$ (why?). Therefore $a=-d$ and ...


4

(I assume $M$ is supposed to be real; if it is allowed to be complex, then the exercise is trivial.) Hint What are the possible eigenvalues of $M$? What can one say about the eigenvalues of real $2 \times 2$ matrices...?


0

If $b \ne 1$, the matrix $A = \begin{bmatrix} 1 & a \\ 0 & b \end{bmatrix} \tag{1}$ is always diagonalizable, since it has distinct eigenvalues $1$ and $b$; thus as is well-known eigenvectors $v_1$, $v_b$ corresponding to $1$ and $b$ respectively are linearly independent, so the matrix $S$, the columns of which may be take to be such $v_1$ and ...


1

If $b$ is anything other than 1 then the matrix is diagonalizable. If $b=1$ then the matrix is diagonalizable only if $a=0$ (its already in diagonal form when this is true, in fact), since if $a \neq 0$ then \begin{equation} A-I=\begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix},\end{equation} the space of solutions of the homogenous system associated with ...


0

By plotting $p(\lambda)$ for the case where $k = 0$, we obtain the following curve: By varying $k$, we are free to move the curve up or down. Now recall that having three distinct real eigenvalues corresponds to making $p(\lambda)$ cross the horizontal axis exactly three times. By computing the local extrema, notice that this can happen by vertically ...


0

the characteristic equation is $$3\lambda - \lambda^3 = k.$$ the graph has a local max $(1,2)$ and local min at $(-1-2).$ so if $$ -2 < k < 2, \text{the char equation has three distinct roots}\\ k = \pm 2, \text{the char equation has a repeated real root}\\ |k| >2 \text{char equation has one real root and two complex conjugate roots.}$$


1

You are practically done - If $B=\{v_{1},v_{2},v_{3}\}$ is a basis of eigenvectors then $v_{1},v_{2},v_{3}$ are all eigenvectors that correspond to the eigenvalue $1$ (since this is the only eigenvalue) But the eigenspace of the eigenvalue $1$ have a dimension $1$ hence cannot contain $3$ lineally independent vectors .


1

I suppose $T_n, T$ are all endomorphisms of $E$, otherwise there is no meaning on speaking about eigenvalues. Let $n= \dim E$. Denote by $m(S)= \min \{ |\lambda| : \lambda \mbox{ is an eigenvalue of } S\}$ And recall that $|\det S| = \prod_{\lambda \in \sigma(S)} |\lambda| \ge (m(S))^n$ for all endomorphisms $S$. Since the determinant is continuous, you ...


1

Note that your matrix can be written as product of a column matrix and a row matrix: $$ A=\pmatrix{w_1\\w_2\\\vdots\\w_n}\pmatrix{1/w_1&1/w_2&\ldots&1/w_n}, $$ which explains why it has rank$~1$ (provided that $n>0$; the case $n=0$ is trivial and I will henceforth not consider it). Now as you can for instance read in this answer, such a ...


0

A linear vector space can be represented by a set of linear equations; I give you a couple of examples: $W$ is the set of vectors for which $x+y-z=0$ If we know by other means that $W$ is a subspace of $\mathbb{R^3}$, its dimension would be 2 (a plane) because you put an arbitrary value to two variables and you can get the the third: $x=0$ $y=1$ then ...


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 $Av=1\cdot v=v$ $$ Av=v\iff(A-I)v=0 $$ $$ \begin{pmatrix}1-1 & 5 & 0\\ 0 & 1-1 & 0\\ 0 & 0 & 3-1 \end{pmatrix}\begin{pmatrix}a\\ b\\ c \end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0 \end{pmatrix} $$ $$ ...


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

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

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 ...


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 $BA(Bv) = B(AB v) = \lambda B v$.


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 $(AA')^{n+1}=A(A'A)^nA'$


0

Suppose $A$ is an $n\times n$ symmetric matrix and $x\in{\mathbb R}^n$. Let $f(x)=x^TAx.$ When you use Lagrange multipliers to find the minimum and maximum of $f(x)$ subject to $\|x\|^2=1$, the eigenvalues show up as feasible Lagrange multipliers, and eigenvectors show up as vectors corresponding to the feasible Lagrange multipliers (in fact, you get the ...


0

Here is a start -- perhaps you can fill in the details: We have $w = (w_1,\ldots,w_n)$ with every entry positive. Let $v = (\frac{1}{w_1},\ldots,\frac{1}{w_n})$, inverting each entry. Observe that, as a linear transformation, $A$ is given by $A(x) = (x \cdot v) w$, where "$\cdot$" is the Euclidean dot product. Find the kernel of $A$ and its dimension. ...


0

Here's the $2 \times 2$ case you can generalize: Suppose $M$ is an arbitrary $2 \times 2$ matrix with two eigenvalues, $\lambda_1, \lambda_2$, and hence two corresponding eigenvectors, $v_1$ and $v_2$. Define a matrix $V$ whose columns are those two eigenvectors, \begin{equation} V =\left( \begin{matrix} v_1 & v_2 \\ | & | \end{matrix} \right) ...


0

If you are taking the logarithm of the elements, then you get a skew-symmetric matrix, since $$\ln 1 = 0$$ and $$\ln \left(1/a_{i,j}\right) = - \ln a_{i,j}.$$ By the way it is a known trick in some AHP-related method.


0

Since $M$ is full rank it admits a base of auto-vectors $v_1,..,v_n$ for its auto-values $\lambda_1,..,\lambda_n$. Let $a_1,..,a_n \in \mathbb{R}$ be the coordinates such that: $\vec{1} = a_1v_1 + .. + a_nv_n$ and $b_1,..,b_n \in \mathbb{R}$ such that: $\theta = b_1v_1 + .. + b_nv_n$ $M\theta = \vec{1} = b_1Mv_1 + .. + b_nMv_n = \lambda_1b_1v_1 + .. + ...


0

Building on another answer, it's not just the eigenvalues that change when you change one entry -- you also change the eigenvectors. And the way they change when you change a single entry will depend on the eigenvalues and the eigenvectors. You could start with the old eigenvalue and eigenvector pair and optimize to find the modified eigenvector and ...


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 $\lambda^2 + 2\lambda + 3.$


0

The problem of finding eigenvalues of a matrix (especially for large non-sparse matrices) is highly unstable; for symmetric matrices the problem is a bit better, but even there the situation is far from ideal. You have continuous dependence of eigenvalues on the coefficients of the matrix. Unfortunately, in general case the estimations on this dependence ...


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 ...


1

Substituting $C$ with $D$ you get $ D S x = \lambda S x$, so $S x$ is an eigenvector of $D$.


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!


1

Suppose $v$ is an eigenvector of $A$ with eigenvalue $\lambda$. What is $A^2 v$?


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

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 ...


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

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 $f_A$. ...


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 ...


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.


0

Though Jibounet's answer covers what is needed, let me add something from a general group-theoretic viewpoint: In any group $G$ if $x,y\in G$ are two elements then the elements $xy$ and $yx$ are conjugate (could be the same), because $ y(xy)y^{-1}= yx$. Now apply this to group of non-singular matrices, then $ST$ and $TS$ are conjugate: and in matrices ...


0

Considering nonzero eigenvalues first, suppose $ST$ has a nonzero eigenvalue $\lambda$, so that there's some nonzero $v$ such that $$STv = \lambda v.$$ What can we say about $TS$? Since $S(Tv) = \lambda v$, let's see what happens when we apply $TS$ to $Tv$: \begin{align*}TS(Tv) &= T(\lambda v)\\ &= \lambda Tv \end{align*} where the latter holds ...


0

You did half the work (and you're right) : if $T$ or $S$ in an element of $\mathrm{GL}_{n}(\mathbb{C})$ (i.e. an invertible matrix), the following equality holds : $\mathrm{P}_{TS}=\mathrm{P}_{ST}$. Here, $\mathrm{P}_{TS}$ (resp. $\mathrm{P}_{ST}$) denotes the characteristic polynomial of $TS$ (resp. $ST$). If $T$ (or $S$) is no invertible, you can use the ...


0

\begin{align} M &= D-ADA \\ &= D^{1/2}(I-D^{-1/2}ADAD^{-1/2})D^{1/2} \end{align} $M$ is positive definite only when eigenvalues of $D^{-1/2}ADAD^{-1/2}$ are less than one. Let $x$ be the non-zero eigenvector of $A$, so that $A=xx^\top$. Because the non-zero eigenvalue of $D^{-1/2}ADAD^{-1/2} = x^\top D^{-1}xx^\top Dx$, you want to prove that $$ ...


0

We know that min. eigen value is : min ${x^T Mx}$; x is of unit norm. Now $x^TMx = x^TDx-x^TA^TDAx= x^TDx-(Ax)^TD(Ax)$ Take x to be unit eigen vector of A corresponding to e.v. $1$ we have $Ax=x$, So $x^TMx=0 $ implies min. eigen value is non positive. NOTE: It might be zero even when $D$ is not equal to $kI$ HINT: Take $A=Diag(1,0,0)$ and $D= ...


0

We want to show that $T$ is some multiple of the identity, i.e. its matrix in some (and hence every) basis is of the form. $$[T]B=\begin{pmatrix} a&0&\dots&0\\ 0&a&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&a \end{pmatrix}.$$ So assume for contradiction that two of the entries on its diagonal are ...


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 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 ...


0

In data analysis, the eigenvectors of a covariance (or correlation matrix) are usually calculated. Eigenvectors are the set of basis functions that are the most efficient set to describe data variability. They are also the coordinate system that the covariance matrix becomes diagonal allowing the new variables referenced to this coordinate system to be ...


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

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 $


0

Hint: The discriminant has to be non-negative to have real roots of a quadratic.



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