For questions involving matrices of infinite size, often identified with bounded linear operators on infinite-dimensional separable Hilbert spaces.
Questions tagged [infinite-matrices]
114 questions
27
votes
2 answers
Can linear maps between infinite-dimensional spaces be represented as matrices?
Any linear map between two finite-dimensional vector spaces can be represented as a matrix under the bases of the two spaces.
But if one or all of the vector spaces is infinite dimensional, is the linear map still represented as a matrix under…
Tim
- 49,162
16
votes
0 answers
What properties do the rings of infinite, upper-triangular matrices have?
I'm very familiar with the ring of $n\times n$ matrices over a field, the ring of $n\times n$ upper triangular matrices over a field, and the ring of infinite column-finite matrices over a field.
But until now, I haven't asked myself about infinite…
rschwieb
- 160,592
13
votes
1 answer
Do eigenvalues of a linear transformation over an infinite dimensional vector space appear in conjugate pairs?
While attempting to answer a question here (namely, the finite dimensional case of the title question: Prove that if $\lambda$ is an eigenvalue of $T$, a linear transformation whose matrix representation has all real entries, then…
Christian
- 2,529
12
votes
1 answer
Non-existence of an infinite-dimensional determinant on $\text{Aut}(V)$
In a recent thread I show that if $V$ is any infinite-dimensional vector space, then (assuming the axiom of choice) the monoid $\text{End}(V)$ has trivial abelianization. It follows in particular that any multiplicative function
$$\det :…
Qiaochu Yuan
- 468,795
12
votes
3 answers
Derivative as a matrix: $\mathbf{D}=\dfrac{\mathrm{d}}{\mathrm{d}x}$
I have a strange question, it is possible to consider the derivative as a matrix? (Both are linear transformation technically).
I thought about this example, since $1, x,x^2,...,x^n$ can be thought of as the basis of a vector space, I can consider a…
Math Attack
- 5,343
11
votes
2 answers
How can I get eigenvalues of infinite dimensional linear operator?
What I want to prove is that for infinite dimensional vector space, $0$ is the only eigenvalue doesn't imply $T$ is nilpotent.
But I am not sure how to find eigenvalues of infinite dimensional linear operator $T$. Since we normally find eigenvalues…
Esther Jacob
- 509
10
votes
0 answers
Matrix exponential, containing a thermal state
Define an infinite matrix $$ M =
\begin{bmatrix}
0 & -1 & 0 & 0 & \cdots \\
1 & 0 & -2 & 0 & \cdots \\
0 & 2 & 0 & -3 & \cdots \\
0 & 0 & 3 & 0 & \cdots \\
\vdots & \vdots & \vdots & \vdots & \ddots \\
\end{bmatrix}$$
Numerically, I've found that…
Yly
- 15,791
8
votes
2 answers
Infinite matrix which cannot be represented by bounded linear operator
Let ${\cal V}$ be n Hilbert space over $\mathbb{R}$ or $\mathbb{C}$, with an orthonormal basis $(e_n)_{n=1}^{\infty}$. For every bounded linear operator $A: {\cal V} \to {\cal V}$, we can associate $A$ with an matrix $(a_{ij})_{i,j=1}^{\infty}$…
user370220
7
votes
1 answer
Representing linear operators on infinite sequences as infinite matrices
So this question arose while I was working on a homework assignment to find the adjoint operator of a continuous operator on $l^2$. Anyway, it seemed like I could find it if I thought about the respective operators as infinite matrices. However,…
Fractal20
- 1,569
7
votes
2 answers
Invertibility of infinite-dimensional matrix
I have a matrix $M \in \mathbb{R}^{n \times n}$ whose columns are linearly independent. Hence, $M$ is invertible.
How to extend this conclusion to the case where $n$ is infinite?
Specifically, given that $n\in\mathbb{N}$, let $X$ and $Y$ be Banach…
lulu
- 111
6
votes
1 answer
Taking the $n$th power of an infinite dimensional matrix
I have a certain matrix $$M:=\begin{bmatrix}
0&-2&0&0&0\\
1&0&-4&0&0\\
0&1&0&-6&0\\
0&0&1&0&-8\\
0&0&0&1&0\\
&&&&&\ddots\end{bmatrix}$$
where I would like to find a closed form for $M^n$. (The reason is a bit convoluted; but I'm mostly concerned…
Habeeb M
- 337
- 1
- 10
6
votes
1 answer
Does Schmidt decomposition exist on infinite dimensional Hilbert spaces?
A pure quantum state in a bipartite system, which is an operator $$\rho = \langle\psi \,,\, \cdot \,\rangle \, \psi \in \mathcal{L}(H_1 \otimes H_2)$$ for some $\psi \in H_1 \otimes H_2$, is factorizable (i.e. not entangled) iff the reduced density…
Sahdo
- 95
6
votes
1 answer
How to invert $\infty \times \infty$ matrix?
I have an equation $AX=B$, where $A$ is $\infty \times \infty$ matrix, $X$ is $\infty \times 1$ vector and $B$ is $\infty \times 1$ vector.
$A$ and $B$ are known and I need to determine $X$.
For this, I think that I should calculate inverse of $A$…
Grešnik
- 1,812
6
votes
3 answers
Can we approximate any eigenvalue of an infinite matrix via eigenvalues of some sequence of submatrices which approximates the matrix?
Let $T:\ell^2\to\ell^2$ be a compact linear operator. Let $[T]=(a_{i,j})_{i,j=1}^{\infty}$ be the representing infinite matrix of $T$ with respect to the canonical base. Let $T_n$ be the finite rank operator defined by the matrix…
Chilote
- 4,367
- 1
- 26
- 51
6
votes
1 answer
If $\sum\limits_i\sum\limits_j\alpha_{ij}x_iy_j$ converges for every square integrable $(x_n)$ and $(y_n)$, then the order of the sums commutes
Let $(\alpha_{ij})$ be a square infinite matrix such that for all $x=(\xi_{n}),y=(\eta_{n}) \in \ell ^{2}$ we have that ${\displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\alpha_{ij}\xi_{i}\eta_{j}}$ converges.
The question: Is it correct to say…
Diego Fonseca
- 3,126