# Linear Algebra with Proof by Contradiction

• Devil Moo
In summary, the conversation discusses a linear algebra question about proving that if the union of two subspaces of a vector space is also a subspace, then one of the subspaces must be contained in the other. The conversation presents a proof by contradiction, showing that if both subspaces are not contained in each other, then their union cannot be a subspace. This leads to the conclusion that at least one of the original statements must be false.
Devil Moo
This is a linear algebra question which I am confused.

1. Homework Statement

Prove that "if the union of two subspaces of ##V## is a subspace of ##V##, then one of the subspaces is contained in the other".

## The Attempt at a Solution

Suppose ##U##, ##W## are subspaces of ##V##. ##U \cup W## is a subspace of ##V##. (statement A)

Suppose ##U## is not contained in ##W## and vice versa. (statement B)

Let ##u \in U, \not\in W## and ##w \in W, \not\in U##.

##u \in U \cup W, w \in U \cup W##
##u + w \in U \cup W##
##u + w \in U## or ##u + w \in W##

Suppose ##u + w \in U## (statement C)

##u + w + (-u) \in U##
##w \in U##

Which one does it conclude? (if A is true, then B is false) or (if A and B are true, then C is false)

It's the second one: (if A and B are true, then C is false)

Since you have proven C false, you can conclude that ##u+v\in W##. By similar steps to before you can also conclude that that is false.

You can then, with a few more steps, conclude that at least one of A or B must be false.

Arguments of this kind are much easier to understand and control if you grasp the notion of Conditional Proof or sub-proof. Every time you make a new assumption, you are opening a new sub-proof. When you reach a contradiction, you close that sub-proof, concluding the opposite of the assumption that you used to open it. It is not unusual to have several nested levels of proof. In your case there are three levels.

For the proof by contradiction about ##\sqrt 2## is a irrational number, we conclude that it is true once we find the contradiction.
In this case, why can't I conclude that ##u + v \in W## is true?

Is ##not (u + v \in U) = u + v \in W## wrong?

Last edited:
Devil Moo said:
In this case, why can't I conclude that ##u + v \in W## is true?
You can. Read the second line of my post again.

Then you go on to prove that it is also false, which is a second contradiction, which tells you that either A or B must be false.

## 1. What is linear algebra?

Linear algebra is a branch of mathematics that deals with the study of linear equations, vector spaces, and linear transformations. It involves the use of mathematical structures and techniques to solve problems related to systems of linear equations, geometric transformations, and data analysis.

## 2. What is a proof by contradiction?

A proof by contradiction is a method of mathematical proof used to show that a statement is true by assuming the opposite and then showing that it leads to a logical contradiction. This type of proof is based on the principle that if an assumption leads to a contradiction, then the assumption must be false and the original statement must be true.

## 3. How is proof by contradiction used in linear algebra?

In linear algebra, proof by contradiction is often used to prove the uniqueness of solutions to systems of linear equations. By assuming that there is another solution that contradicts the original solution, a proof by contradiction can show that the original solution is the only possible solution.

## 4. What are the benefits of using proof by contradiction in linear algebra?

Proof by contradiction is a powerful tool in linear algebra as it allows for the efficient and elegant proof of mathematical theorems. It also allows for the exploration of different solutions and helps to deepen understanding of the underlying concepts in linear algebra.

## 5. Are there any limitations to using proof by contradiction in linear algebra?

While proof by contradiction is a useful technique, it may not always be applicable in linear algebra. It relies on the ability to find a contradiction, which may not always be possible. In addition, it may not be the most efficient method for proving certain theorems, and other proof techniques may be more suitable.

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