In mathematics, a linear map (also called a linear mapping, linear transformation, vector space homomorphism, or in some contexts linear function) is a mapping
V
→
W
{\displaystyle V\to W}
between two vector spaces that preserves the operations of vector addition and scalar multiplication. The same names and the same definition are also used for the more general case of modules over a ring; see Module homomorphism.
If a linear map is a bijection then it is called a linear isomorphism. In the case where
V
=
W
{\displaystyle V=W}
, a linear map is called a (linear) endomorphism. Sometimes the term linear operator refers to this case, but the term "linear operator" can have different meanings for different conventions: for example, it can be used to emphasize that
V
{\displaystyle V}
and
W
{\displaystyle W}
are real vector spaces (not necessarily with
V
=
W
{\displaystyle V=W}
), or it can be used to emphasize that
V
{\displaystyle V}
is a function space, which is a common convention in functional analysis. Sometimes the term linear function has the same meaning as linear map, while in analysis it does not.
A linear map from V to W always maps the origin of V to the origin of W. Moreover, it maps linear subspaces in V onto linear subspaces in W (possibly of a lower dimension); for example, it maps a plane through the origin in V to either a plane through the origin in W, a line through the origin in W, or just the origin in W. Linear maps can often be represented as matrices, and simple examples include rotation and reflection linear transformations.
In the language of category theory, linear maps are the morphisms of vector spaces.
In finite dimensions, a matrix can be decomposed into the sum of rank-1 matrices. This got me thinking - in what situations can a bounded linear operator mapping between infinite dimensional spaces be written as an (infinite) sum of rank-1 operators?
eg, let A be a bounded linear operator...
Homework Statement
1. Let T be a linear operator on an inner product space V. Let U = TT*. Prove that U = TU*.
2. For a linear operator T on an inner product space V, prove that T*T = T0implies T = T0.
Homework Equations
The Attempt at a Solution
1. This appeared at the...
Homework Statement
This is from a linear algebra textbook I'm reading. I don't know whether there is universal agreement as to how notation should read for this, so this thread may well be meaningless. But I'll just post it to see if anyone can decipher what the question means. I'm asked to...
Homework Statement
Show that the range of the linear operator defined by the equations is not all of R3, and find a vector that is not in the range
Homework Equations
w1 = x - 2y + z
w2 = 5x - y + 3z
w3 = 4x + y + 2z
The Attempt at a Solution
can I just show that it does not have...
Hi,
I am looking for eigen functions of the linear operator L defined by
L=(-2i(\nablaf).\nabla -i\nabla^2f +(\nablaf)^2)
and here f is an abitary function of x,y,z
[SOLVED] linear algebra determinant of linear operator
Homework Statement
Let T be a linear operator on a finite-dimensional vector space V.
Define the determinant of T as: det(T)=det([T]β) where β is any ordered basis for V.
Prove that for any scalar λ and any ordered basis β for V...
Let T:V -> V be a linear operator on a finite-dimensional inner product space V.
Prove that rank(T) = rank(T*).
So far I've proven that rank (T*T) = rank(T) by showing that ker(T*T) = ker(T). But I can't think of how to go from there.
Homework Statement
If I e.g. want to find the kernel and range of the linear opertor on P_3:
L(p(x)) = x*p'(x),
then we can write this as L(p'(x)) = x*(2ax+b). What, and why, is the kernel and range of this operator?
The Attempt at a Solution
The kernel must be the x's where L(p'(x))...
Q: Suppose V is a finite dimensional inner product space and T:V->V a linear operator.
a) Prove im(T*)=(ker T)^(|)
b) Prove rank(T)=rank(T*)
Note: ^(|) is orthogonal complement
For this question, I don't even know how to start, so it would be nice if someone can give me some hints. Thank...
Q) Let V be an inner product space and T:V->V a linear operator. Prove that if T is normal, then T and T* have the same image. (i.e. imT=imT*)
My Attempt:
<T(v),T(v)>
=<T*T(v),v>
=<TT*(v),v>
=<T*(v),T*(v)>
=>||T(v)|| = || T*(v)||
But this doesn't seem to help...
Thanks!
Definition from my textbook: For each linear operator T on a inner product space V, the adjoint of T is the mapping T* of V into V that is defined by the equation <T*(v),w> = <v,T(w)> for all v, w E V.
My instructor defined it by <T(v),w> = <v,T*(w)> and he said that these 2 definitions are...
I understand the definition of trace and linear operator individually but I don't seem to understand as to what does it mean by trace of a linear operator on a finite dimensional linear space.
What I have found out is that trace of a linear operator on a finite dimensional linear space is the...
Consider the linear operator T on \mathcal{C}^2 with the matrix
\bmatrix 2 && -3\\3 && 2 \endbmatrix
in the standard basis. With the basis vectors
\frac{1}{\sqrt{2}} \bmatrix i \\ 1 \endbmatrix, \quad \frac{1}{\sqrt{2}} \bmatrix -i \\ 1 \endbmatrix
this operator can be written
\bmatrix 2+3i...
Let be L and G 2 linear operators so they have the same set of Eigenvalues, then:
L[y]=-\lambda _{n} y and G[y]=-\lambda _{n} y
then i believe that either L=G or L and G are related by some linear transform or whatever, in the same case it happens with Matrices having the same...
let be the linear operator: (Hermitian ??)
L = -i(x\frac{d}{dx}+1/2)
then the "eigenfunctions" are y_{n} (x)=Ax^{i\lambda _{n} -1/2
then my question is how would we get the energies imposing boundary conditions? (for example y(0)=Y(L)=0 wher L is a positive integer )...:smile: :smile:
Let T: R^3 -> R^3 be a linear operator and
T(x;y;z)= (x -y + 3z;2x-y+8z;3x -5y+5z).
Find the rule, for the linear operator S: R^3 -> R^3 such that ker S= I am T and I am S=Ker T.
I'm not really sure how I should start this problem. Also I would like to know if I can assume S is the...