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0

In general, I don't think so. Suppose, for example, that $A$ is a symmetric matrix with all diagonal entries equal to 0 (like the adjacency matrix of a simple, loopless graph), and let $X$ be the upper triangular matrix such that $A=X+X^T$. Then the only eigenvalue of $X$ will be 0, and likewise for $X^T$. The first standard basis vector is an ...


2

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


0

Well, I guess a simple solution would be as follows: Form the matrix D by merging A and B, $D_{n\times(m+p)}=[A|\sqrt{\lambda} B]$ then solve the PCA for D. $D\cong WK$ Then the solution to the original problem is: $W=W$, $[H|\sqrt{\lambda} G]=K$ In fact similar approach would be applicable to other matrix factorization methods such as NMF and ...


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) = ...


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 - ...


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 ...


0

If $H$ is Hermitian and negative definite, then all principal minors (the submatrices) are negative definite. Let $\tilde H$ be such a negative definite minor with size $m\times m$, then its determinant (as product of its $m$ negative eigenvalues) has sign $$ (-1)^m. $$


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

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

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 ...


0

suppose $\lambda$ is an eigenvalue of $T.$ we will first show that $\lambda$ must be real. $\lambda$ is an eigenvalue of $T$ means there is an $f \neq 0$ such that $\dfrac{f^\prime(x)}{x} = Tf(x) = \lambda f(x)$ since $f$ is not identically zero, there is an $x_0 \in (0,1)$ such that $f(x_0) \neq 0$ that is $\lambda =\dfrac{f^\prime(x_0)}{x_0f(x_0)}$ which ...


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 ...


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.


0

Hint: if $f$ is an eigenvector corresponding to $\lambda$, then $f$ satisfies $$ \frac{f'(x)}{x} = \lambda f(x) $$ For which $\lambda$ does this have a non-zero solution $f\in C^\infty(0,1)$?


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 $\|A\| = \|X\|^2$.


0

Yes, you must have a non-trivial solution for $\lambda=0$. For example, the SL problem $y^{\prime\prime}+\lambda y=0$, with boundary conditions $y^{\prime}(0)=0$, $y^{\prime}(1)=0$ has $\lambda=0$ for an eigenvalue with $y(x)=c$, where $c \neq 0$ is a constant for an eigenfunction.


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 $$ A-aI=\begin{bmatrix}0&-1\\0&0\end{bmatrix} $$ ...


0

Nope. This is a polynomial equation in $a$ of degree $2$, so that if there is only one $a\in\mathbb R$ with $a^2=0$ (and this is the case: any $a\neq 0$ will have $a^2>0$), then this $a$ will be a double root, that is, a root of multiplicity $2$. In the end, you can never have an eigenvalue of multiplicity greater than $n$ for an $n\times n$ matrix, ...


0

Well, $u$ would be a generalized eigenvector in your case. Then the solution of $\dot{x} = Ax$ would be $$ x(t) = \sum_{i=1}^{k} \sum_{j=1}^{m_i} c_{ij} v_{ij} \frac{t^{\ell_{ij}-1}}{(\ell_{ij}-1)!} e^{\lambda_i t}$$ where $k$ is the number of different eigenvalues of $A$, $m_i$ is the algebraic multiplicity of the eigenvalue $\lambda_i$, $v_{ij}$ is the ...


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

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 $\lambda \ge 0$.


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.


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 ...


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

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 ...


0

This won't quite work: e.g. you might have $A = I$, in which case your $T$ could be any invertible matrix, and there's no reason to think $T B T^{-1}$ will be in Jordan normal form. What you want to do is choose $T$ on each eigenspace of $A$ to put into Jordan form the restriction of $B$ to that eigenspace.


0

What did you try? You could use the well-known formula $$ \DeclareMathOperator{\Var}{Var} \Var(AX) = A \Var(X) A^T $$ You can find such formulas here: https://en.wikipedia.org/wiki/Multivariate_random_variable


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

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 $A$.


0

An $n\times n$ matrix with $n$ independent eigenvectors can be written as $A=P^{-1}DP$, where $D$ is the diagonal matrix $diag(\lambda_1\:\lambda_2\:...\lambda_n)$ and $P$ is the matrix $(\vec{v}_1\:|\:\vec{v}_2\:|...|\:\vec{v}_n)$ where $v_i$ is the corresponding eigenvector to $\lambda_i$. ...


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

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.


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

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

Hint: by Sylvester's crieterion, it is enough to check the principal minors. Note that $a - b^2 > 0$.


0

the characteristic polynomials of $AB$ and $BA$ are still $\lambda^{m-r}p(\lambda)$ and $\lambda^{n-r}p(\lambda), p(0) \neq 0$ the reason is $tr(AB)^k = tr(BA)^k$ for all $k$. showing the coefficients of the characteristic polynomials are the same.


1

$$\begin{align} \bigg(\frac{d^{4}}{dt^{4}} - \lambda \bigg) y &= \bigg( \frac{d^{2}}{dt^{2}} + \sqrt{\lambda} \bigg) \bigg(\frac{d^{2}}{dt^{2}} - \sqrt{\lambda} \bigg) y\\ &= \bigg(\frac{d}{dt} + i \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - i\lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} + \lambda^{\frac{1}{4}} \bigg) \bigg(\frac{d}{dt} - ...


2

(Too long for a comment.) I don't know why we "should" apply the Courant-Fischer minimax principle. The statement can be proved pretty easily by Sylvester's law of inertia (which is a much weaker statement than the Courant-Fischer minimax principle). Since $B$ has full rank, by matrix congruence (consider $\pmatrix{I&-\frac12A(B^T)^{-1}\\ 0&I}$), ...


1

Since $P^T=P^{-1}$, then $P$ is an orthogonal matrix, which indicates that If $$E=[\epsilon_1,\epsilon_2,\cdots\epsilon_n]$$ is an orthonormal basis in $\mathbb R^n$. Then $$T(E)=[T(\epsilon_1),T(\epsilon_2),\cdots T(\epsilon_n)]=[\epsilon_1,\epsilon_2,\cdots\epsilon_n]P=[\epsilon'_1,\epsilon'_2,\cdots\epsilon'_n]=E'$$ is also an orthonormal ...


1

Hint: The standard dot product on $\mathbb{R}^n$ can be written as $$x \cdot y = y^T x.$$


1

Here is an inductive proof: Let ${\cal C}$ be the commuting family of operators. Let me call an operator $A$ on a subspace $S$ a multiplier operator on $S$ iff there exists some $\lambda$ such that $Ax=\lambda x$ for all $x \in S$. Let me call a subspace $S$ invariant iff $AS \subset S$ for all $A \in {\cal C}$. Note that if a one dimensional space $S$ ...


1

Let's say your vector space is $\mathbb{C^n}$. Then $M_n(\mathbb{C})$, i.e. all $n$ by $n$ matrices are the bounded operators on $\mathbb{C^n}$. Now, $M_n(\mathbb{C})$ is finite dimensional. So, even if you have infinitely many operators, say $A_1,A_2,\cdots$, there will exist $i_1,i_2,\cdots,i_k$ such that $A_{i_1},\cdots,A_{i_k}$ will span the rest of the ...


1

One answer is that finding the eigenvalues will work for classifying critical points of functions $\mathbb{R}^n \to \mathbb{R}$, while the discriminant trick only works when $n=2$. The reason is that the discriminant is the determinant of the Hessian, and the determinant is the product of the eigenvalues. In dimension 2, then, you can tell whether the ...


2

Suppose that $\lambda$ is a eigenvalue of $A$ with $n$ linearly independent eigenvectors associated to it. This means that $A$ is diagonalizable. Hence there exists an invertible matrix $B$ such that $$A=B^{-1}DB$$ where $D$ is the diagonal matrix with $\lambda$ as each diagonal element. Then we can write $$A=B^{-1}DB=B^{-1}(\lambda I)B=\lambda ...


1

$A=\sum_{i=1}^n \lambda_i v_iv_i^\top$ where $(\lambda_i,v_i)$ are the eigenvalue/vector pairs of $A$. So, if you know that $|\lambda_1| \ge |\lambda_2| \ge \dots \ge |\lambda_n|$, then you can obtain $\lambda_2$ by computing the largest eigenvalue/vector pair in absolute value of $$B=A-\lambda_1 v_1v_1^\top$$


1

Here is a theorem that you can use: where $$\vec{x}=\frac{1}{\lambda_1 v_{1,k}}\begin{pmatrix} a_{k1}\\ a_{k2}\\ ...\\ a_{kn} \end{pmatrix}$$ $v_{1,k}$ is the $k$th component of $\vec{v}_1$, $a_{ki}$ is the $ki$th element of $A$. The row $k$ is smallest index such that $v_{1,k}$ is the infinity norm of $\vec{v}$, i.e., the largest component. Then you ...


4

The answer is no. Take the identity matrix and the matrix $$a_{11}=a_{12}=a_{22}=1, a_{21}=0.$$


0

This is a quick exercise in matrix algebra: $$ D^{-1/2} (D- A) D^{-1/2} = I - D^{-1/2} A D^{-1/2} $$ Then notice $\lambda$ is an eigenvalue of $M$ if and only if $1-\lambda$ is an eigenvalue of $I-M$. In fact the eigenvectors corresponding to $\lambda$ for $M$ are the same as the eigenvectors corresponding to $1-\lambda$ for $I-M$, and vice versa.


2

First, there is a useful relation between eigenvalues, singular values, and the Frobenius norm$\color{red}{^*}$: $$ \sum_i|\lambda_i|^2\leq\sum_i\sigma_i^2=\|A\|_F^2, $$ where $\lambda_i$ and $\sigma_i$ ($i=1,\ldots,n$) are, respectively, the eigenvalues and singular values of $A$. Since for any nonsingular $X$, $X^{-1}AX$ has the same eigenvalues as $A$, we ...



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