# Linear Operators: Identifying and Solving

• Axiom17
A(p ?)You have done it! You have shown that formally the definition of a linear operator is equivalent to the definition of the operator A that shifts the argument by 1. So your operator is linear. :!!:It is not that you should change the order of the terms or the letters. You should just use the same notation everywhere if you want to avoid confusion.So now you have learned to be careful with parenthesis. You need to use the distributive law (f+g)(x)=f(x)+g(x) carefully. You need to use the associativity of multiplication (cf)(x)=cf(x) carefully.You also have learned how to use definitions. You also have learned that linear operator
Axiom17

## Homework Statement

I have some operators, and need to figure out which ones are Linear (or not).

For example:

1. $$\hat{A} \psi(x) \equiv \psi(x+1)$$

## Homework Equations

I have defined the Linear Operator:

$$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$

Last edited:
Axiom17 said:

## Homework Statement

I have some operators, and need to figure out which ones are Linear (or not).

For example:

1. $$\hat{A} \psi(x) \equiv \psi(x+1)$$

## Homework Equations

I have defined the Linear Operator:

$$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$

## The Attempt at a Solution

$$(A\psi )(x)=\psi (x+1)$$

which you read: the effect of acting with A on $$psi$$ is a new function, denoted $$A\psi$$ whose value at every point x is equal to the value of the original function at the point x displaced by 1.

Now you look at your

$$\hat{A}[p\psi_{1}+q\psi_{2}]$$

Here you have A acting on the sum of functions. This sum of functions is a function. So from the definition of A:

$$(\hat{A}[p\psi_{1}+q\psi_{2}])(x)=(p\psi_{1}+q\psi_{2})(x+1)$$

Now, how are sums of functions defined? Isn't it true that they are defined like:

$$(f+g)(x)=f(x)+g(x)$$

$$(cf)(x)=cf(x)$$.

I understand a bit more but I still don't quite get it. This is what I have so far:

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x) = \left(AP\psi_{1}+AQ\psi_{2}\right)(x) = AP\psi_{1}(x)+AQ\psi_{2}(x)$$

.. but then I don't see how to relate that to $\psi(x+1)$

Axiom17 said:
I understand a bit more but I still don't quite get it. This is what I have so far:

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x) = \left(AP\psi_{1}+AQ\psi_{2}\right)(x) = AP\psi_{1}(x)+AQ\psi_{2}(x)$$

.. but then I don't see how to relate that to $\psi(x+1)$

You have missed an important step. You want to calculate

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x)$$

Here A acts on a certain function(which is a linear combination of two other functions) and you want to evaluate the result at x. For this you use the definition of A, this is definition that you wrote at the very beginning. Whatever is the function A acts on, it shifts its argument by 1. Here A acts on what is in the [] bracket. It will shifts its argument by 1. Thus:

$$\left(A\left[P\psi_{1}+Q\psi_{2}\right]\right)(x)=(P\psi_{1}+Q\psi_{2})(x+1)$$

You used just the definition of A, what A does to whatever it acts on. Now you continue using the definition of the sum of two functions and the product of a number and a function.

So, continue, and see where you will end. But first, let me know if you really-really understand what I wrote above.

I think I understand what you've wrote. I see that basically have:

$$(A\left[function])(x)=(function)(x+1)$$

which is in this case:

$$(A\left[P\psi_{1}+Q\psi_{2}\right])(x)=(P\psi_{1}+Q\psi_{2})(x+1)$$

Then using the relation $(f+g)(x)=f(x)+g(x)$ the RHS of the equation becomes:

$$(P\psi_{1})(x+1)+(Q\psi_{2})(x+1)$$

(which hopefully is correct).

Then using the other relation $(cf)(x)=cf(x)[/tex] this becomes: $$P\psi_{1}(x+1)+Q\psi_{2}(x+1)$$ .. but that's what I had 2 steps before! Now, look at your right hand side: $$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$ Which you want to calculate at x $$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)$$ First you notice that you have here a sum of two functions evaluated at x. Therefore the first step is to us general formula (f+g)(x)=f(x)+g(x). That is $$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)=(p\hat{A}\psi_{1})(x)+(q\hat{A}\psi_{2})(x).$$ Then you need to use (cf)(x)=cf(x) and at the end the definition of A (in two terms). See if you get the same as from the LHS (left hand side). All this exercise is nothing but careful application of definitions! It's pretty formal. Nothing really deep. But it teaches you to be careful with putting and moving parenthesis right. BTW: P and p, as well as Q and q should be the same. You better use p,q everywhere. I've got a bit confused now but perhaps I'm not too far away So this is the operator: $$\hat{A}\psi(x)=(A\psi)(x)=\psi(x+1)$$ This is the definition of a linear operator: $$\hat{A}[p\psi_{1}+q\psi_{2}]=p\hat{A}\psi_{1}+q\hat{A}\psi_{2}$$ Thus obtain this: $$\left(A\left[p\psi_{1}+q\psi_{2}\right]\right)(x) = \left(Ap\psi_{1}+Aq\psi_{2}\right)(x) = Ap\psi_{1}(x)+Aq\psi_{2}(x)$$ Then this step: $$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)=(p\hat{A}\psi_{1})(x)+(q\hat{A}\psi_{2})(x)$$ Now using [itex](cf)(x)=cf(x)$ can change the RHS of the above to:

$$p\hat{A}\psi_{1}(x)+q\hat{A}\psi_{2}(x)$$

Now using $(f+g)(x)=f(x)+g(x)$ can change the above (the RHS) to:

$$(p\hat{A}\psi_{1}+q\hat{A}\psi_{2})(x)$$

.. which gives the LHS!

SO that means that the operator $\hat{A} \psi(x) \equiv \psi(x+1)$ is a linear operator.. right?

*note1: not sure what should be the correct notation as far as to use $\hat{A}$ or A throughout.*

*note2: I see I've changed around the order of terms, does this matter? I.e. is Ap the same as pA?*

All right. See how it should look like:

Definition of A: for any $$\psi$$ and any x:
$$(A\psi)(x)=\psi(x+1)$$

Theorem: A is linear

Proof: What it means A is linear? It means I need to prove that for any two functions $$\psi_1,\psi_2$$ and any numbers p,q we have
$$A(p\psi_1+q\psi_2)=pA\psi_1+qA\psi_2.$$
The above equations is eqaulity of two functions. To check that two functions are equal I need to check that there values at every point are equal. So I calculate the LHS:

$$(A(p\psi_1+q\psi_2))(x)=(p\psi_1+q\psi_2)(x+1)$$

(from the definition of A).

$$(p\psi_1+q\psi_2)(x+1)=p\psi(x+1)+q\psi_2(x+1)$$

(from the definition of sum and products of functions.

Now I calculate the RHS at x:

$$(pA\psi_1+qA\psi_2)(x)$$

First I use the defintion of sum and products
$$(pA\psi_1+qA\psi_2)(x)=p(A\psi_1)(x)+q(A\psi_2)(x)$$
And now I use definition of A in each of these terms:
$$p(A\psi_1)(x)+q(A\psi_2)(x)=p\psi_1(x+1)+q\psi_2(x+1)$$

I see that LHS=RHS.

Calculating LHS I first used definition of A (once) and then sum and products of functions.
Calculating RHS I first used sum and product definitions, and the definition of A (twice).

Think of it. Contemplate until you get it. You will use similar reasonings many times in the future.

Last edited:

## 1. What is a linear operator?

A linear operator is a mathematical function that maps one vector space to another while preserving the addition and scalar multiplication operations. In other words, the output of a linear operator is always a linear combination of its input.

## 2. How do you identify a linear operator?

To identify a linear operator, you must first check if it follows the properties of linearity. These properties include homogeneity, additivity, and preservation of multiplication. Additionally, a linear operator must also be well-defined and preserve the zero vector.

## 3. What is the difference between a linear operator and a matrix?

A linear operator is a function that operates on vector spaces, while a matrix is a rectangular array of numbers. A linear operator can be represented by a matrix, but not all matrices are linear operators. Additionally, linear operators can operate on infinite-dimensional vector spaces, while matrices are limited to finite dimensions.

## 4. How do you solve a system of equations using linear operators?

To solve a system of equations using linear operators, you can represent the equations as a matrix equation and then use methods such as Gaussian elimination or matrix inversion to obtain the solution. Another approach is to use eigenvalues and eigenvectors of the linear operator to solve the system.

## 5. What are some real-life applications of linear operators?

Linear operators have numerous applications in physics, engineering, and computer science. They are used in quantum mechanics to describe physical systems, in signal processing to analyze and manipulate signals, and in machine learning to model and solve problems. They are also essential in the study of differential equations and in the analysis of linear systems in control theory.

Replies
24
Views
1K
Replies
1
Views
444
Replies
10
Views
1K
Replies
1
Views
1K
Replies
1
Views
771
Replies
1
Views
1K
Replies
2
Views
1K
Replies
5
Views
2K