New answers tagged

4

Suppose $A$ and $B$ are positive operators on a finite-dimensional inner product space $V$ and $\|A + B\| = \|A\| + \|B\|$. Because $A$ and $B$ are positive, $A+B$ is also a positive operator. Thus there exists $x \in V$ such that$$\|x\| = 1 \quad \text{and} \quad\|A+B\| = \langle (A+B)x, x \rangle. $$ Now \begin{align*} \|A+B\| &= \langle (A+B)x, x ...


0

Solve $\ker (A - \lambda I)$ for $\lambda$, which will be your eigenvalues. Then by defintion, the eigenvector for every eigenvalue solves $$ Av = \lambda v \iff (A - \lambda I)v = 0. $$ Solve this system for every $\lambda$ and you will obtain the columns for $P$. The inverse is calculated per usual.


1

Clearly (as the matrix is upper triangular) the characteristic polynomial is $\;(t-2)(t-3)^2\;$ , so the matrix is diagonizable iff $\;(t-2)(t-3)\;$ is its minimal polynomial, and indeed: ...


0

You almost solved your problem. The only thing that you forgot to do was a change a basis. Let $e_1,e_2,\ldots,e_{n+5}$ denote the canonical basis of ${\mathbb R}^{n+5}$. The identical columns in $C$ force the subspace $Z$ spanned by $e_{k}-e_{n+1} (n+2\leq k \leq n+5)$ to be included in the kernel of $C$. We construct an orthonormal basis in which $Z$ ...


1

Let $v$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, i.e. $Av=\lambda v$. Then $(\lambda I-B)^{-1}(A-B)v=(\lambda I-B)^{-1}(\lambda I -B)v=v $. Hence by definition of the operator norm, $\left\|(\lambda I-B)^{-1}(A-B)\right\|\geq 1$. Question: Where did we use that $\lambda$ is not an eigenvalue for $B$?


1

If $x$ is a unit eigenvector of $A$ associated with $\lambda$, we have $$ (\lambda I-B)^{-1}(A-B)x=(\lambda I-B)^{-1}(\lambda I-B)x=x $$ and hence $$ 1=\|x\|=\|(\lambda I-B)^{-1}(A-B)x\|\leq\|(\lambda I-B)^{-1}(A-B)\|. $$


0

The operator $L$ defined on a domain $\mathcal{D}(L)$ of twice continuously differentiable functions satisfying the two endpoint conditions is symmetric in the inner product $L^2[a,b]$ where $(h,k)=\int_{a}^{b}h(t)\overline{k(t)}dt$, which is to say that $$ (Lf,g) = (f,Lg),\;\;\; f,g\in\mathcal{D}(L). $$ Because of this, if $Lu = 0$ for ...


0

The characteristic polynomial is $ \text{det}(A−λI)=(2−λ)(3−λ)^2$ so the eigenvalues of your matrix are $2$ and $ 3$. Therefore $2$ is an eigenvalue with algebraic multiplicity $1,$ and $3$ is an eigenvalue with algebraic multiplicity $2$. Recall the geometric multiplicity can also be described as the dimension of the nullspace of $A-λI$. So for $λ = 2$ we ...


0

A matrix is diagonalizable if it is similar to a diagonal matrix. So in order to prove this matrix is diagonalizable, why don't we make it similar to a diagonal matrix. As it is upper triangular, we can just read the eigenvalues off of the diagonal (to convince yourself of this, try finding $det(A-\lambda I)$ for any diagonal A or remember that finding the ...


1

I'll just remind you of some standard facts that I hope will lead you in the right direction, but you should refresh these in the book/references. First the homogenous PDE, $ \partial_t u = \Delta u$. As mentioned, $u_{i,j}(r,t) := \psi_{i,j}(r) e^{\lambda_{i,j} t} $ solves the heat equation, with initial condition at time zero $ \psi_{i,j}(r) $. Since ...


0

To find the eigenvectors corresponding to an eigenvalue $\lambda$, compute the kernel of the matrix: $A - \lambda I$. Therefore to find the eigenvectors corresponding to $8$. Look at $$A - 8I = \begin{pmatrix} -2 & -2 & -1 \\ -2 & -2 & -1 \\ -1 & -1 & -3 \end{pmatrix}_.$$ Row reducing this matrix (this is not echelon form, but is ...


1

At first, to summarize this problem, let replace diagonal factor matrices as: $$ \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} \lambda-mk & 1 \\ \lambda-k & 1 \end{bmatrix} \begin{bmatrix} I \\ A \end{bmatrix} $$ where $\lambda$ is eigenvalue. Therefore, we can suppose the following determinant: $$ \begin{aligned} \det \begin{bmatrix} X ...


0

The eigenspace is the set $E=\{kv_3\mid k\in\Bbb R\}$ where the eigenvector: $$v_3=\left(\begin{array}{c}\frac{1}{4}\\\frac{1}{8}\\1\end{array}\right),$$ is performing as a base vector for $E$.


0

The vector you give is an eigenvector associated to the eigenvalue $\lambda = 3$. The eigenspace associated to the eigenvalue $\lambda = 3$ is the subvectorspace generated by this vector, so all scalar multiples of this vector. A basis of this eigenspace is for example this very vector (yet any other non-zero multiple of it would work too).


0

A polynomial $f$ has a multiple root iff $\gcd(f,f')\ne 1$.


0

With only a little extra work we can generalize this result to rectangluar matrices, as long as their sizes are such that cyclic permutation is well defined. Specifically, suppose $B$ is a wide $r$-by-$n$ matrix, and $B$ is a tall $n$-by-$r$ matrix. The nonzero eigenvalues of the $r$-by-$r$ matrix $AB$ are the same as the eigenvalues of a larger $n$-by-$n$ ...


2

The eigenvector you obtained and the one in the book differ only in sign, that is by multiplication by $-1$. Both vectors are eigenvectors for eigenvalue $0$ and equally correct answers. For the other eigenvalue, you just forgot to subtract in the last line. If you correct this you will get two free variables.


1

The eigenfunctions will not necessarily be orthogonal with respect to an unweighted space. For the sake of discussion, assume $P$, $q$ and $r$ are real; $P$ is continuously differentiable on $[0,l]$ and strictly positive; $r$ is continuous and strictly positive; Then this problem falls into a nice classical setting where you can multiply by a factor to ...


0

The bound also follows from an application of Weyl's inequalities (sometimes called the Courant-Weyl inequalities).


0

Let $A=\begin{bmatrix}2 & 1\\2 &3\end{bmatrix}$. The roots of the characteristic polynomial for the matrix coincide with its eigenvalues, so we first factor the polynomial: $$p_A(t)=det(tI-A)=det\begin{bmatrix}t-2 & -1\\-2 &t-3\end{bmatrix}=(t-2)(t-3)-(-1)(-2)=t^2-5t+6-2=t^2-5t+4=(t-4)(t-1)$$ Thus, the roots of $p_A(t)$ are $4$ and $1$, which ...


3

The answer to your question is no. That is, your friends are wrong. In general, applying row-reduction to your matrix will change its eigenvalues. We can see what goes wrong if we take this approach slightly differently. In particular: $$ \pmatrix{ 2 & 1 & 2 \\ 0 & 2 & -1 \\ 0 & 1 & 0 \\ } \to \pmatrix{ 2 & 1 & 2 \\ ...


1

Intro I'll present an incomplete interpretation framework. Let's being by considering two points in one dimension. We can easily generalize the results to more than one dimension. Our matrix is given by, $$M=\begin{bmatrix} 0 & |x_1-x_2| \\ |x_1-x_2| & 0 \end{bmatrix}$$ The eigenvalues are given by $$\lambda_1=-\lambda_2=|x_1-x_2|$$ The ...


0

I assume you have shown $R(h(T))=R(h(T)|_{K_{\lambda_1}})+\cdots+R(h(T)|_{K_{\lambda_n}})$, and attack the uniqueness part. Note that $K_i$ is invariant under $h(T)$ as $h(T)w_i=p(T-\lambda_i)w_i\in K_i, \forall w_i\in K_i$, where $p(x)$ is the polynomial defined by $p(x)=h(x+\lambda_i)$. Thus $R(h(T)|_{K_{_i}}\leq K_i$, and $V=\oplus K_i$ implies the ...


1

In char polynomial, coefficient of $\lambda^{n-i}$ is sum of all principal minors of order $i$. Since rank is one all minors of order $2$ or greater are zero. So only coefficient of $\lambda^{n-1}$ has possibility to be non zero. If it too is zero. all evalues will be zero, so rational. If it is non zero, again it has two be rational since char polynomial of ...


3

If it has rank $1$, then all but one eigenvalues are $0$, so the value of the nonzero eigenvalue equals the trace, which is rational.


2

You can find a computation method for the eigenvalues and eigenvectors in the following recent document: "Eigendecomposition of Block Tridiagonal Matrices" by A. Sandryhaila and J.M.F. Moura (arxiv.org/pdf/1306.0217)


1

You have $$A^{20}x=\left(-\frac{1}{3}\right)^{20}V_1-\left(\frac{1}{3}\right)^{20}V_2+2V_3,$$ and this is $$\left(-\frac{1}{3}\right)^{20}\left(\begin{array}{c}1\\0\\0\end{array}\right)-\left(\frac{1}{3}\right)^{20}\left(\begin{array}{c}1\\1\\0\end{array}\right)+2\left(\begin{array}{c}1\\1\\1\end{array}\right)= ...


1

You have a mistake in your calculation of the eigenvalue, but I still think the problem is not right. The eigenvalues of $M$ are $\frac a2\pm |b|$. One then finds easily that a corresponding pair of eigenvectors is given by $$ (e^{-i\phi/2},-e^{i\phi/2})^T,\ \ (e^{-i\phi/2},e^{i\phi/2})^T. $$ The eigenspaces are spanned by each of these two vectors, so ...


2

Here's a counterexample. Let $V$ be the space of functions $\mathbb{Z}\to K$ which have finite support. Define $T:V\to V$ by $(T\varphi)(n)=\varphi(n+1)$ (i.e., thinking of an element of $V$ as a "bi-infinite sequence", shift the terms in the sequence by $1$). Then $T$ is an isomorphism, but if $T\varphi=\lambda\varphi$, then ...


3

The answer is no. Consider the space of infinite real sequences and let $T(x)=(0,x_1,x_2,\dots)$. Suppose $T(x)=\lambda x$. If $\lambda=0$, then $x$ has to be zero, so $0$ is not an eigenvalue. If $\lambda \neq 0$, then $0=\lambda x_1$, so $x_1=0$, but then $x_1=0=\lambda x_2$. Continuing by induction we again conclude that $x=0$. So $\lambda$ is not an ...


2

No, the shift homomorphism defined on $(e_1,...,e_n,...)$ by $f(e_i)=e_{i+1}$ does not have an eigenvalue.


4

If $A$ is a a multiple of $I_n$, then we may take $B=A$. So suppose that $A$ is not a multiple of $I_n$. Clearly the problem is invariant by conjugation (i.e. for any invertible matrix $P$, $(A,B)$ satisfies the conditions iff $(P^{-1}AP,P^{-1}BP)$ does). Lemma. If $A$ is not a multiple of $I_n$, then there is a conjugate of $A$ whose lower leftmost ...


1

Call the matrix $A$. Over $\mathbb{R}$, the matrix $A$ is diagonalizable and so is $A^2$, whose eigenvalues are $\lambda_i^2$. We have $$ \operatorname{tr}(A^2) = \lambda_1^2 + \lambda_2^2 + \lambda_3^2 + \lambda_4^2 $$ but $\operatorname{tr}(A^2) = \sum_{i=1}^4 (A^2)_{ii} = \sum_{i=1}^4 \left( \sum_{j = 1}^4 a_{ij} a_{ji} \right)$ is a sum of product of ...


1

First remark, your matrix is not symmetric unless $A$ is so. For simplicity I will assume at least that $A$ is diagonalisable. Let $B$ be the matrix obtained for $k=0$; it is the Kronecker product of an $m'\times m'$ matrix$~P$ with all its entries equal to$~1$, and $-A$, for $m'=m+1$ (what was the point of the shift in $m$ anyway?). $P$ has rank$~1$ and ...


1

I'm not claiming this is the easiest proof, but it is what came to mind. Write $e_1,\ldots,e_n$ for the canonical basis. Since $A$ is selfadjoint, we can write $$ A=\sum_{j=1}^n\lambda_j P_j,$$ where $P_1,\ldots,P_n$ are (pairwise orthogonal) rank-one projections that add to the identity. Now, if $A_{ii}=\lambda_1$, $$ \lambda_1=A_{ii}=\langle ...


4

If $x$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$, then your computation of $$(A + I)x = (\lambda + 1) x$$ is exactly the definition that states $x$ is an eigenvector of $A + I$ corresponding to eigenvalue $\lambda + 1$.


3

Hint: If $A^2 = A$, then $A$ is diagonalizable with eigenvalues $0$ and $1$. However, by the Perron theorem, the multiplicity of the eigenvalue $1$ is just $1$. Similarly, if $A^k = A$, note By the Perron theorem that $A$ has the eigenvalue $1$ with multiplicity $1$ and no other eigenvalues of magnitude $1$.


1

@joriki's proof considered the case $\lambda\ne 0$. Now, for the case that $\lambda=0$, we have \begin{align} \left\vert \begin{matrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{matrix} \right\vert=0\quad\mbox{and}\quad \left\vert \begin{matrix} 0 & 0 & 1 &0\\ 0 & 0 & 0 &1\\ 1 & 0 & 0 &1\\ 0 ...


2

Hint: look at the vector $$ \begin{bmatrix} v\\ cv \end{bmatrix} $$ where $v$ is an eigenvector of $B$ with eigenvalue $\lambda$, and $c$ is some constant. Under what condition on $c$ is this an eigenvector for your matrix?


1

Add $\frac1\lambda$ of all except the first three rows to the second and third rows to get rid of the columns of $1$s. That leaves $(-\lambda)^n$ from the diagonal times $$ \begin{vmatrix} -\lambda&0&1\\ 0&-\lambda+\frac n\lambda&\frac n\lambda\\ 1&\frac n\lambda&-\lambda+\frac n\lambda \end{vmatrix} =-\lambda\left(-\lambda+\frac ...


1

First we diagonalize $A = SDS^{-1}$ with $S^TS=I$ and $D$ diagonal. Then $$ \pmatrix{S^{-1}&0\\0&S^{-1}} M \pmatrix{S&0\\0&S} =\pmatrix{D & D+I \\D+I & 0}. $$ Applying a suitable permutation $P$, we obtain the block diagonal structure $$ P\pmatrix{D & D+I \\D+I & 0}P^{-1} = \pmatrix{ \lambda_1 & \lambda_1+1& \\ ...


0

To find the eigenvalues you can use the characteristic polynom : $$det \left( \begin{bmatrix} 1-X & 1 & 0 \\ 1 & 0-X & -1\\ 0 & -1 & 1-X \end{bmatrix}\right)=det \left( \begin{bmatrix} 1-X & 0 & 1-X \\ 1 & -X & -1\\ 0 & -1 & 1-X \end{bmatrix}\right)=(1-X)det \left( \begin{bmatrix} ...


0

Here's an example. Consider the matrix $$A = \begin{bmatrix} 3&0&-1\\ 1&2&-1\\ 2&0&0\\ \end{bmatrix}$$ which has eigenvalues $\lambda_1=1$ and $\lambda_2=2$. Let's find the eigenspace for $v_2$; to do so, we solve the system $(A-\lambda_2I)\mathbf v = \mathbf0$. In other words, we want the null space of $(A-\lambda_2I)$. $$A - 2I = ...


0

You just solve the linear system of equations $(A - \lambda_1 I)\mathbf{v} = \mathbf{0}$ (using Gaussian elimination for example) and find a basis of linearly independent solutions which will be eigenvectors associated to the eigenvalue $\lambda_1$.


1

Your first equation is not correct. If you plug $\theta = 0$, the matrix becomes $\operatorname{diag}(-3,-1,1,3)$ and so the characteristic equation should be $(\lambda - 3)(\lambda + 3)(\lambda - 1)(\lambda + 1) = 0$. This is indeed consistent with the second equation but from the first equation you get $$ -(1 + \lambda)(1 - \lambda)(3 - \lambda) = 0 $$ ...


0

@copper.hat 's comment was the correct answer -- since det$(A-I\lambda)$ and det$(D-I\lambda)$ must be invertible, they must have a non-zero determinant. Therefore, they cannot possibly be the solution to the eigenmode equation.


2

I think this is a very good example where the slightly more abstract proof using linear maps (which you use in any case) instead of matrices simplifies the argument and shows more clearly what is going on. Let $(W, \left< \cdot, \cdot \right>)$ be a finite dimensional real inner product space and let $T \colon W \rightarrow W$ be a symmetric operator ...


1

$V$ needs to be an invariant subspace under $A$ so that $T$ goes from $V$ to $V$. Otherwise, if $V$ was not invariant, $T$ would go from $V$ to $\Bbb{R}^n$ and then it would not have a matrix representation in $\Bbb{R}^{(n-1)x(n-1)}$. Here's a proof that $w_j$ is an eigenvector of $A$: $$Bu_j=\lambda_ju_j$$ This is because $u_j$ is an eigenvector of ...


1

The problem is that the matrix $\left( \begin{matrix} \lambda & 0 \\ 0 & 0 \end{matrix}\right)$ has two eigenvalues: $\lambda$ and $0$.


1

If $0$ is an eigenvalue for the linear transformation $T:V\rightarrow V$, then by the definitions of eigenspace and kernel you have $$V_0=\{v\in V| T(v)=0v=0\}=\text{ker}T.$$ If you have only one eigenvalue, which is $0$ the dimension of $\text{ker}T$ is equal to the dimension of $V_0$. For instance: consider the endomorphism, whose associated matrix ...



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