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4

If $n=1$, there's nothing to prove, as all matrices are scalar multiples of the identity. If $n\ge 2$, take two linearly independent vectors $v_1,v_2$ with corresponding eigenvalues $\lambda_1$, $\lambda_2$. Since every vector is an eigenvector, so is $v_1-v_2$ with corresponding eigenvalue $\lambda_3$. So then $A(v_1-v_2) = \lambda_3 (v_1 -v_2) = ...


3

Let $\rho=\max_{\|x\|=1}\|Ax\|$ and $u=\arg\max_{\|x\|=1}\|Ax\|$. The case $\rho=0$ (i.e. $A=0$) is trivial. Suppose $\rho\ne0$. Then $Au=\rho v$ for some unit vector $v$ and $\|Av\|\le\rho$. Since $\langle u,Av\rangle=\rho$, we must have $Av=\rho u$. Therefore, $(-\rho,u)$ (when $v=-u$) or $(\rho,u+v)$ (when $u+v\ne0$) is an eigenpair of $A$. Normalize the ...


3

Note that $$ AW = \lambda BW \implies\\ (A - \lambda B)W = 0 $$ Thus, to find $W$, we should simply ensure that each column $x$ of $W$ is a solution to the homogeneous system of equations $$ (A - \lambda B)x = 0 $$ In Matlab, use null(A - lambda * B) to find a basis to this solutions space.


2

call the eigenvectors $u_1, u_2$ and $u_3$ the eigenvectors corresponding to the eigenvalues $1, -2, $ and $2.$ then $$A = 1\dfrac{u_1u_1^T}{u_1^Tu_1} - 2\dfrac{u_2u_2^T}{u_2^Tu_2} + 2\dfrac{u_3u_3^T}{u_3^Tu_3}$$ you can verify this by computing $Au_1, \cdots$. this expression for $A$ is called the spectral decomposition of a symmetric matrix.


2

Writing the matrix down in the basis defined by the eigenvectors is trivial. It's just $$ M=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 2 \end{array} \right). $$ Now, all we need is the change of basis matrix to change to the standard coordinate basis, namely: $$ S = \left( \begin{array}{ccc} 1 & 1 & ...


2

It seems to me that this is pretty much just bash-it-out linear algebra using definitions. I'll sketch the reasoning and let you fill in the deets: First, let $C=B^{-1}A$ for ease of notation. Now let's let $\mathbf v$ be an eigenvector of $C$ with eigenvalue $\eta$. First, $(I+k\theta C)\mathbf v = (1+k\theta\eta)\mathbf v$; multiply by the inverse of the ...


2

Hint: let $\mathbf {S := (B)^{-1}A}$, then the eigenvalues of $\mathbf{I}+k\theta \mathbf S$ are the roots of the following polynomial: \begin{align} \lambda(\mathbf I + k\theta \mathbf S) &= \mathrm{det} (\lambda \mathbf I - (\mathbf I + k\theta \mathbf S)),\\ &= \mathrm{det} ([\lambda-1] \mathbf I - k\theta \mathbf S). \end{align} Let $\eta := ...


2

They are always non-negative. Suppose $\lambda$ is an eigenvalue of $B^TB$ corresponding to an unit eigenvector $v$, then $\langle v, B^T B v \rangle = \lambda = \langle Bv, B v \rangle = \|Bv\|^2$. Hence $\lambda \ge 0$.


2

Let $x$ an eigenvector of $B^TB$ associated to the eigenvalue $\lambda$, which is real since $B^TB$ is symmetric, then $$\lambda ||x||^2=\langle B^TB x,x\rangle=\langle Bx,Bx\rangle=||Bx||^2\implies \lambda\ge0$$


2

You're computing the determinant of $A$, which is not the characteristic polynomial. The characteristic polynomial is rather $$ \det(A-XI)=\det\begin{bmatrix}a-X&-1\\0&a-X\end{bmatrix}=(a-X)^2 $$ which has one root (equal to $a$) with multiplicity $2$. The matrix is not diagonalizable, because $$ A-aI=\begin{bmatrix}0&-1\\0&0\end{bmatrix} $$ ...


2

I'm not sure this helps. Note that $A$ is symmetric positive semidefinite. The biggest eigenvalue will be $\max_{\|v\|=1} \langle v , Av\rangle $. Since $\langle v , Av\rangle = \sum_k (x_k^T v)^2$, we can let $X=\begin{bmatrix} x_1^T \\ \vdots \\ x_n^T \end{bmatrix}$ and then we have $\langle v , Av\rangle = \| X v \|^2$ and so $\|A\| = \|X\|^2$.


2

AFTERTHOUGHT: it occurs to me that there is something that does not need determinant, although the concept is implicit. As you can easily confirm, we have a matrix $A$ such that $\color{red}{A^2 = I}.$ Now, if you are willing to accept the proposition that every square matrix has an eigenvalue (possibly complex) then we can write $$ Av = \lambda v $$ for ...


2

The problem is that you forgot to normalize your vector. The answer should indeed be $$ \pmatrix{1/3\\1/3\\1/3} $$


1

Just to clarify for those who may see this question in the future, the actual matrix was $$A = \begin{pmatrix} -a & -b & -b\\ c & -d & 0\\ 0 & d & 0 \end{pmatrix}$$ so that the polynomial to solve was $$\lambda^3 + (a+d)\lambda^2 + (ad + bc)\lambda + bcd,$$ which has an obvious root $\lambda = -d$, and hence factors out.


1

For $n=4$, the eigenvalues are $4, -4, 0, 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 2, 2, -2, -2, -2, -2, -2, -2, -2, -2$. For $n=5$, the eigenvalues are the roots of $(x - 5) \cdot (x + 5) \cdot (x - 1)^5 \cdot (x + 1)^5 \cdot x^{24} \cdot (x^2 - 5)^6 \cdot (x^2 - 5\cdot x + 5)^8 \cdot (x^2 + 5\cdot x + 5)^8 \cdot (x^2 - 2\cdot x - 4)^{10} \cdot (x^2 + 2\cdot x - ...


1

Note that $A^*A$ is necessarily positive semidefinite. To show this, note that for any $x \in \Bbb C^n$, $$ x^*(A^*A)x = (x^*A^*)Ax = \|Ax\|^2 \geq 0 $$ The result you showed regarding $(x,Mx)$ is called Rayleigh's theorem. See also the min-max theorem (AKA the Courant-Fischer principle).


1

If the distinct eigenvalues of the adjacency matrix of a $k$-regular graph are $k=\theta_1\ge\cdots\ge\theta_m$, the eigenvalues of its Laplacian in non-decreasing order are \[ 0, k-\theta_2,\ldots,k-\theta_m. \] So in this case the algebraic connectivity and the spectral gap coincide. If the graph is not regular then, in general, there is no simple or ...


1

Use Jordan canonical form. For a $k \times k$ Jordan block $B = \lambda I + N$, where $N^k = 0$, $\exp(tB) = \exp(t\lambda) \exp(tN)$ where $\exp(tN) = I + tN + \ldots + t^{k-1} N^{k-1}/(k-1)!$ is polynomial in $t$. Thus if $\text{Re}\; \lambda < 0$, $\exp(tB) \to 0$ as $t \to \infty$.


1

You know that the matrix of the endomorphism is $$ A=\begin{bmatrix} 1 & 2 & a \\ 0 & 3 & b \\ -1 & 1 & c \end{bmatrix} $$ for some $a,b,c$. Elimination gives $$ \begin{bmatrix} 1 & 2 & a \\ 0 & 3 & b \\ 0 & 3 & c+a \end{bmatrix} $$ and, since we know that the rank must be $2$, we can conclude that $b=c+a$. We ...


1

Let $\Lambda_A$ and $\Lambda_B$ be the spectra of $A$ and $B$, respectively. If $\mathrm{tr}(AB)$ was dependent only on the spectra of $A$ and $B$, that is, $$ \mathrm{tr}(AB)=f(\Lambda_A,\Lambda_B), $$ then for any orthogonal matrices $U$ and $V$, $\tilde{A}:=U^TAU$ and $\tilde{B}:=V^TBV$ would be still SPD with the spectra $\Lambda_A$ and $\Lambda_B$ and ...


1

Suppose $Av=\lambda v$. We can scale $v$ to have unit norm, so that $v=(\cos\alpha,\sin\alpha)$. Writing out $Av=\lambda v$ gives $$ (\cos\theta\cos\alpha+\sin\theta\sin\alpha,\sin\theta\cos\alpha-\cos\theta\sin\alpha) = \lambda(\cos\alpha,\sin\alpha). $$ We can simplify the LHS: $$ (\cos(\theta-\alpha),\sin(\theta-\alpha)) = \lambda(\cos\alpha,\sin\alpha). ...


1

That transformation $A$ is a reflection $P_x$ along the $x$-axis followed by a rotation $R_\theta$ by an angle $\theta$ : $$ A = \left( \begin{matrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{matrix} \right) = \left( \begin{matrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{matrix} \right) \left( \begin{matrix} 1 ...


1

the matrix $\pmatrix{\cos \theta & \sin \theta\\ \sin \theta & -\cos \theta}$ represents the reflection on a mirror along the line $y = \tan (\frac{\theta}{2})\ x.$ therefore $\pmatrix{\cos (\theta/2) \\\sin(\theta/2)}$ is an eigenvector corresponding to the eigenvalue $1$ and $\pmatrix{\sin (\theta/2) \\-\cos(\theta/2)}$ is an eigenvector ...


1

For any $n \times n$ matrix $A$, the characteristic polynomial is $$c_A(\lambda) := \det(\lambda I - A) = \lambda^n + \text{(lower order terms in $\lambda$)} ,$$ which in particular has degree $n$. In general, the degree of the minimal polynomial $c_A$ has degree $\leq n$, and is at least the number of distinct eigenvalues of $A$.


1

A minimal polynomial of a matrix and its characteristic polynomial have the same irreducible factors So the minimal polynomial of the matrix will be $(x-a)^k(x-b)^l$ where $1\leq k\leq p;1\leq l\leq q$ also note that for two distinct eigen values their eigen vectors are linearly independent So the dimension will be $k+l$ where $k,l$ are in the given ...


1

Try this : Define $T:\mathbb R^3\rightarrow \mathbb R^3$ by $T(1,0,0)=0;T(0,1,0)=(0,1,0);T(0,0,1)=(0,0,-1)$ NOTE:So the transformation becomes $T(c_1,c_2,c_3)=(0,c_2,-c_3)$


1

Suppose that $B^{-1}Av = \eta v$. Then, first \begin{align} (I-k(1-\theta)B^{-1}A)v & = v - k(1-\theta)B^{-1}Av\\ & = v-k(1-\theta)\eta v \\ & = (1-k(1-\theta))\eta v. \end{align} On the other hand, similarly, you can prove that $$(I+k \theta B^{-1}A) v = (1+k\theta\eta)v.$$ This, in turn, implies $$(1+k\theta\eta)^{-1}v = (I+k \theta ...


1

Let $f(x)=\langle Ax,x\rangle$. The gradient is better computed as a directional derivative: $$ \langle\nabla f(x),u\rangle=\lim_{t\to 0}\frac{f(x+tu)-f(x)}{t}=\cdots=\langle Ax,u\rangle+\langle Au,x\rangle=\langle 2Ax,u\rangle, $$ where the dots are straightforward and the last equality follows because $A$ is symmetric. Thus, $\nabla f(x)=2Ax$. Now the ...



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