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0

Let $\mathrm A \in [0,1)^{n \times n}$ be a symmetric, positive definite, nonnegative matrix whose diagonal entries are equal to $1$. Using the Gershgorin circle theorem, the minimum eigenvalue of $\mathrm A$ is bounded by $$\lambda_{\min} (\mathrm A) \geq 1 - \max_{1 \leq i \leq n} \, \sum_{j \neq i} a_{ij} = 1 - \| \mathrm A - \mathrm I_n\|_{\infty}$$ ...


1

If it's an $n \times n$ matrix with all diagonal entries $1$, the trace is $n$, so that won't help. The Frobenius norm is bounded by $n$. Since the smallest eigenvalue can be arbitrarily close to $0$, I don't see how you could possibly get a nonzero lower bound in terms of the Frobenius norm.


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Counter-example: take $M_1 = I$, the identity matrix, and $M_2$ as any matrix of the same dimension (not a scalar multiple of the identity). You can see the above statement fails to hold.


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By $λ^2 + 4λ - 12 = 0$, $λ = 2 $ or $λ = -6 $ Through diagonalization, A is similar to D= \begin{bmatrix} 2 & 0\\ 0 & -6 \end{bmatrix} $A^3$ is similar to $D^3$ = \begin{bmatrix} 8 & 0\\ 0 & -216 \end{bmatrix} So eigenvalue for $A^3$ is $8 $ and $-216$ .


3

I presume the statement is that $A$ is a 2 by 2 matrix with $p(\lambda)=\det(\lambda I_2 -A) = \lambda^2+4 \lambda -12$? The roots ($-6$ and $2$) are the eigenvalues. Eigenvalues behave nice when taking powers so if $\lambda$ is an eigenvalue of $A$ then $\lambda^3$ is an eigenvalue of $A^3$, etc... Do you need a proof? The last part is less trivial, I ...


3

Suppose that $v$ is an eigenvector of $A$ with eigenvalue $\lambda$. If you apply the given equation to the vector $v$, you find $$ \begin{align} 0 &= (A^2 - 3A + 2I)v \\ &= A^2v - 3Av + 2v \\ &= A(\lambda v) - 3\lambda v + 2v \\ &= \lambda^2v - 3\lambda v + 2v \\ &= (\lambda^2 - 3\lambda + 2)v \end{align} $$ So what does this tell you ...


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You can take any eigenvector for $v_2$. To find the columns of $U$ corresponding to $0$, you proceed as follows: In this case, $AA^T=\begin{bmatrix} 10&-20&-20\\ -20&40&40\\ -20&40&40\\ \end{bmatrix}$. Row reducing, we get $\begin{bmatrix} 1&-2&-2\\ 0&0&0\\ 0&0&0\\ \end{bmatrix}$. So, the eigenvectors ...


1

It's because $-A^2 = A^T A$ has only real nonnegative eigenvalues: if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$ then $$-\lambda^2 \|v\|^2 = -v^T A^2 v = v^T A^T A v = \|A v \|^2.$$ Skew-symmetric matrices do not have to have nonzero determinant, the zero matrix is a counterexample.


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For $12$, $u_1=\begin{bmatrix} 1/\sqrt{2}\\ 1/\sqrt{2}\end{bmatrix}$ is a normalised eigenvector. Now, compute the first column of $V$, $$v_1=\frac{A^T}{\sqrt{12}}u_1= \begin{bmatrix} 2/\sqrt{24}\\ 4/\sqrt{24}\\ 2/\sqrt{24}\end{bmatrix}.$$Corresponding to $10$, a nomalised eigenvector is $u_2=\begin{bmatrix}1/\sqrt{2}\\ -1/\sqrt{2}\end{bmatrix}$. ...


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Exactly like in the link you did. The upper left hand 2 by 2 block will be the one for the first pair, the lower right hand 2 by two block will be the one for the second pair, and the upper right and lower left will be all 0s.


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Let $\lambda \in \mathbb C$ be an eigenvalue with eigenvector $0 \neq x \in \mathbb C^n$. Then $$(x,x) = (A^tAx,x) = (Ax,Ax) = \lvert\lambda \rvert^2 (x,x).$$ Thus $\lvert \lambda \rvert = 1$ for all eigenvalues. Since $A$ is real, eigenvalues come in conjugate pairs, so either all three are real, or two are complex and one is real, but the two which are ...


2

Assuming $A$ is a real matrix, using singular value decomposition we can write $$ A = U S V^T$$ where $S$ is a real valued diagonal matrix (i.e., $S=S^T$); $U$ is the left Eigenvector and $V$ the right Eigenvector. Then, you can write $$A^TA = V S^T U^T U S V^T = VS^2V^T$$. However, if $A$ is positive symmetric, then $U=V$ and you can use eigenvalue ...


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If we compute the characteristic equation we find that the eigenvalues are the solutions to $$\begin{align} -\lambda(\pi^2t^2-\lambda)+36^2 &= \lambda^2-\lambda \pi^2t^2 +36^2\\ &= 0 \end{align}$$ Using the quadratic formula we find that $$\lambda=\frac{\pi^2t^2\pm\sqrt{\pi^4t^4-4\cdot 36^2}}{2}$$ The eigenvalues are real if and only if $$\pi^4t^4-4\...


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Hint: calculate the determinant of \begin{bmatrix} π^2t^2 - \lambda & 36\\ -36 & -\lambda \\ \end{bmatrix}, the result will be a second order polynomial in $\lambda$, set it to zero, and try ABC formula (specificly, check when the determinant > 0).


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There is no way to "see" the eigenvectors directly, no more than it is easy to predict the solution of any given linear system of equations. Because this is exactly what the problem is; solving $$(A - \lambda I)x = 0.$$ Of course, in some cases, it can be done, e.g. if $A - \lambda I$ turns out to be a matrix with some special structure. But to a general ...


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Suppose $\lambda$ is an eigenvalue of $A$ with associated eigenvector $x\neq0$. Then we have $$|\lambda|\cdot|x|=|\lambda x|=|Ax|>|x|, $$ and thus $$|\lambda|>1.$$ Since the eigenvalue $\lambda$ was arbitrary, we can conclude that $|\lambda|>1$ for all eigenvalues of $A$


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What about $\textbf{A}:=\begin{bmatrix}8&12\\-3&-4\end{bmatrix}$? This matrix has only one eigenvalue $2$, which is greater than $1$. However, under the standard norm on $\mathbb{R}^2$, the size of $\textbf{A}\,\textbf{u}$ is less than the size of $\textbf{u}$, if $\textbf{u}:=\begin{bmatrix}3\\-2\end{bmatrix}$. You can even require that the ...


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The above approaches are correct, but just to make the link with your solution, here is another way of writing things: $$ \begin{cases} \nabla_x \mathcal{L} = 0\\ \nabla_{\lambda} \mathcal{L} = 0\\ \end{cases} \quad \Rightarrow\quad \begin{cases} Ax = \lambda x\\ x^{T}x=1\\ \end{cases} $$ In other words, unitary eigenvectors of $A$ satisfy the Lagrangian ...


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Hint: Let $\lambda$ be an eigenvalue of $A$ such that $|\lambda| = \max_i |\lambda_i| = \rho(A)$, and $v$ an associated eigenvector. One has $$Av = \lambda v \Longrightarrow \rho(A) ||v|| = ||Av|| \leqslant |||A||| \centerdot ||v||$$ Since $||v|| \neq 0$, this simplifies to $\boxed{\rho(A) \leqslant |||A|||}.$ All you have to do is prove $|||A||| \...


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First, show that for any $x$ with $\|x\| \leq 1$, we have $\|xx^T\| \leq 1$. Then, if $A_i = x_ix_i^T$, it suffices to note by the triangle inequality and homogeneity of a norm that $$ \left\| \frac 1n \sum_{i=1}^n A_i \right\| \leq \frac 1n \sum_{i=1}^n \|A_i\| $$


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$$\mathrm x^T \mathrm A \mathrm x \geq \lambda_{\min} (\mathrm A) \|\mathrm x\|_2^2 = \lambda_{\min} (\mathrm A) > 0$$ because $\|\mathrm x\|_2 = 1$ and $\mathrm A \succ \mathrm O$. The minimum is attained at the intersection of the eigenspace of $\lambda_{\min} (\mathrm A)$ with the unit Euclidean sphere.


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Sketch: Since $A$ is positive-definite, it is symmetric and hence has an orthonormal eigenbasis. Let $v_1,\cdots,v_n$ be the eigenvectors with corresponding eigenvalues $\lambda_1,\cdots,\lambda_n$. Therefore we can write $x=\sum a_iv_i$ where $\sum a_i^2=1$. The sum condition comes from the fact that $\langle\sum a_iv_i,\sum a_iv_i\rangle=\sum a_ia_j\...


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Given any real analytic function in a neighborhood of $0,$ we get $f(A)$ defined as long as some induced norm for $A$ is smaller than the radius of convergence for the Taylor series of $f.$ Or, of course, if the radius is infinite, as in $e^x.$ However, the Cayley Hamilton Theorem says that $f(A)$ can be rewritten as a polynomial in $A,$ of degree no ...


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As far as I understood your question, you have found out the eigen-values of your matrix $$A=\begin{bmatrix}1&2 \\ 4&3\end{bmatrix}.$$ Now to find out the eigen-vectors, solve $Ax=\lambda x\implies (A-\lambda I)x=0 .$ Solve for $x$, that'll give you the eigen-vector corresponding to the eigenvalue $\lambda$. So the characteristic equation of the ...


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The characteristic polynomial of $A=\left[\begin{array}{rr} 18 & 12 \\ -40 & -26 \end{array}\right]$ is $$ \chi_A(t)=\det(tI-A)=\det\left[\begin{array}{rr} t - 18 & -12 \\ 40 & t + 26 \end{array}\right]=t^{2} + 8t + 12=(t + 2) \cdot (t + 6) $$ The eigenvalues of $A$ are thus $\lambda=-2$ and $\mu=-6$. To compute a basis of $E_\lambda$, note ...


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Hints For 1: Consider $A^4(v) = A(A(A(A(v))))$. What do you know about $A(v)$ and the properties of linear transformations that allow you to compute $A^4(v)$?. For 2: $A(v) = 4v$. If we assume $A^{-1}$ exists, we can rewrite this as: $v = A^{-1}(4v) = 4A^{-1}(v)$. What can we conclude from this? For 3: If $A(v) = 4v$, then what is $(A+4I)(v) = A(v) + ...


2

Assuming $av_1+bv_2+cv_3=v$, we have $$\begin{cases} b+2c=0\\ -2a+2b=0\\ 2a-c=5 \end{cases}$$ and by solving the sysytem, you have got $a=2$, $b=2$, and $c=-1$. To find $Av$, as you mentioned, we have $$Av=A(2v_1+2v_2-v_3)=-2v_1+2v_2-4v_3= \begin{bmatrix}-6\\8\\0\end{bmatrix}$$


1

Hint: Try this. Assume arbitrary scalars $\alpha,\beta,\gamma$ and take $$v=\alpha v_1+\beta v_2+\gamma v_3.$$ Now solve for $\alpha,\beta,\gamma$.


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$A$ is only diagonal with respect to a basis of eigenvectors, namely $v_1$ and $v_2$, not the standard basis, which is why your computation fails. Try using linearity and the definition of eigenvectors instead: $A(v_1+v_2)=Av_1+Av_2=\lambda_1 v_1+\lambda_2 v_2$ Similarly for $A(3v_1)$.


0

Well basically the matrix $$ \begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} $$ works. The eigenvalue of this matrix is $\lambda^2 + 1$ and the eigenvalue of this matrix $A^2$ is $(\lambda +1)(\lambda + 1)$ where both algebraic and geometric multiplicity 2


1

Going off your formula for the eigenvalues: If $b^2 - 4k \leq 0$, then the real part of the eigenvalues will be $-\frac{b}{2}$. So in this case, the eigenvalues will have real part less than $0$ if $-\frac{b}{2} \leq 0$, or, equivalently, if $b \geq 0$ If $b^2 - 4k \geq 0$, then the eigenvalues are all real numbers of the form: $$\frac{-b \pm \sqrt{b^2 -...


0

Not exactly. By row reductions, the correct row echelon form should be $$ \left[\begin{array}{ccc|c} a_{11}-1 & a_{12} & a_{13} & 0\\ 0&0&0&0 \\ 0&0&0&0 \end{array}\right] $$ instead. Since you don't know the exact values of the $a_{ij}$s, you cannot use as denominators in divisions. All you can say is that the eigenspace ...


2

With inner product do you mean $\langle x, y \rangle = x^T y $ ? Because in that case $ \langle Ax, y \rangle = (Ax)^T y = x^T A^T y = x^T (A^* y) = \langle x, A^* y \rangle$


1

$$\begin{array}{ll} \text{maximize} & x^T A x\\ \text{subject to} & x^T x = 1\end{array}$$ where $A \in \mathbb{R}^{n \times n}$ is symmetric and, thus, has real eigenvalues. We define the Lagrangian $$\mathcal{L} (x,\lambda) := x^T A x - \lambda (x^T x - 1)$$ Taking the partial derivatives of $\mathcal{L}$ and finding where they vanish, $$(A - \...


1

To find the eigenvalues of $A$, we want to find $\lambda$ such that $A - \lambda I$ has a nontrivial null space (equivalent to the determinant being 0). $$A-\lambda I = \begin{bmatrix} -\lambda & -4 & -6 \\ -1 & -\lambda & -3 \\ 1 & 2 & 5-\lambda \end{bmatrix}$$. By either row reduction or taking the determinant, we can compute ...


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reference on the subject Isospectral vs Isometry of the problem of Carolyn Gordon, David L. Webb and Scott Wolpert. The problem is from an article by Marc Kac and is known as "can you hear the shape of a drum". The negative solution in high dimension was found by Milnor before Kac's paper, but getting examples for Kac's problem about planar domains was ...


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You can reformulate the problem by considering a reduced phase space. In your example the full phase space is $M= \{(X,Y,Z) \in \mathbb{R}^3 \}$. Introduce the reduced phase space $M_{Z=c}= \{(X,Y,Z) \in \mathbb{R}^3 : Z=c \} $. Observe there is no more motion tangential to your eigenvector $(0,0,1)$ so you do not have to deal with the associated center ...


2

Note that your matrix is the block diagonal matrix $$ M = \left[\begin{array}{rr} A & 0\\ 0 & B \end{array}\right] $$ where \begin{align*} A &= \left[\begin{array}{rr} -12 & -6 \\ 8 & 2 \end{array}\right] & B &= \left[\begin{array}{rr} -14 & -9 \\ 42 & 25 \end{array}\right] \end{align*} Next, note that the eigenvalues and ...


2

Expanding on stewbasic's comment: Your matrix can be decomposed as $2I + 2X$, where $X$ is simply filled with $1$s. We are looking for scalars $\lambda$ and vectors $v$ satisfying $$\lambda v = (2I+2X)v = 2v + 2Xv \iff (\lambda/2 - 1)v = Xv.$$ The eigenvalues of $X$ are easy to compute: the eigenvalues of any $n \times n$ matrix of rank $1$ are given by its ...


1

Try $$ \pmatrix{1/2 & 1/3 & 1/6\cr 0 & 1/3 & 2/3\cr 1/2 & 1/3 & 1/6\cr}$$ Or, if you want the entries strictly positive, $$ \pmatrix{s & 1/3 & 2/3 - s\cr 1-2s & 1/3 & 2s - 1/3\cr s & 1/3 & 2/3-s\cr}\ \text{for} \ \frac{1}{6} < s < \frac{1}{2},\ s \ne \frac{1}{3}$$


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By definition, an eigenvector $v$ with eigenvalue $\lambda$ satisfies $Av = \lambda v$, so we have $Av-\lambda v =Av - \lambda I v = 0$, where $I$ is the identity matrix. Thus, $$(A-\lambda I)v = 0,$$ and $v$ is in the nullspace of $A-\lambda I$. Since the eigenvalue in your example is $\lambda = 8$, to find the eigenspace related to this eigenvalue we ...


2

The dimension of the eigenspace is given by the dimension of the nullspace of $A - 8I = \left(\begin{matrix} 1 & -1 \\ 1 & -1 \end{matrix} \right)$, which one can row reduce to $\left(\begin{matrix} 1 & -1 \\ 0 & 0 \end{matrix} \right)$, so the dimension is $1$. Note that the number of pivots in this matrix counts the rank of $A-8I$. ...


4

Note that the characteristic polynomial of $ A= \left[\begin{array}{rr} -6 & k \\ -1 & -2 \end{array}\right] $ is $$ \chi_A(t) =\det(tI-A) =\det \left[\begin{array}{rr} t + 6 & -k \\ 1 & t + 2 \end{array}\right] =t^{2} + 8 \,t + k + 12 $$ This polynomial has a real repeated root if and only if $$ 8^2-4(k+12)=0 $$ Hence $A$ has one real ...


3

$\lambda^2 - 11\lambda + 24 - 8k = 0\\ \lambda = \dfrac {11 \pm \sqrt{25 +32k}}{2} $ $A$ has 2 distinct real egeinvalues if $25 +32k>0$


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The determinant is $$(3 - \lambda)(8 - \lambda) - 8k = \lambda^2 - 11\lambda + 24 - 8k.$$ In order to find the eigenvalues, you have to solve the equation $$\lambda^2 - 11\lambda + 24 - 8k = 0.$$ The solutions are: $$\lambda = \frac{11 \pm \sqrt{11^2 - 4(24-8k)}}{2} = \frac{11 \pm \sqrt{25+32k}}{2}$$ which of course depend on $k$. The two solutions are ...


3

Note that the characteristic equation of this matrix is given by $$ \lambda^2 - 11\lambda + (24 - 8k) = 0 $$ The question is to find which values of $k$ produce two distinct real roots $\lambda$ to this equation. It is helpful to consider the discriminant of this quadratic equation, namely $$ b^2 - 4ac = 121 - 4(24 - 8k) $$ recall that a quadratic equation ...


2

You need to solve the equation with $\lambda$ the variable and $k$ a parameter. The solutions are $$ -\frac{11}{2} \pm\sqrt{\frac{11^2}{4} - (24 - 8k) } $$ These are the eigenvalues of the matrix in dependence of $k$. These two are distinct and real if and only if $$\frac{11^2}{4} - (24 - 8k)> 0.$$ Solve this to get your $k$.


0

According to MathSciNet those three people have two co-authored papers: Isospectral plane domains and surfaces via Riemannian orbifolds. Invent. Math. 110 (1992), no. 1, 1–22 One cannot hear the shape of a drum. Bull. Amer. Math. Soc. (N.S.) 27 (1992), no. 1, 134–138. A link to the latter paper can be found on this page.


4

You have: $$Au_0=c_1\lambda_1x_1+....c_n\lambda_nx_n,$$ then $$A^2u_0=A(Au_0)=c_1\lambda_1A(x_1)+....c_n\lambda_nA(x_n)=c_1\lambda_1^2x_1+....c_n\lambda_n^2x_n.$$ Now repeat the same argument other $98$ times.


3

The way you're working the easiest way to go is by induction: $$A^2u_0=A(c_1\lambda_1x_1+...+c_n\lambda_nx_n)=\sum_{k=1}^nc_k\lambda_k Ax_k=\sum_{k=1}^nc_k\lambda_k^2x_k$$ and etc. Using similarity we could show the above with diagonal matrices and stuff.



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