Can a Finite Set of Vectors Form a Subspace?

[MienTommy]

Homework Statement

\begin{pmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix}

Is this set a subspace of ℝ³

Homework Equations

The set must be closed under addition.
The set must be closed under multiplication.
The set must contain the zero vector.

The Attempt at a Solution

1. It obviously contains the zero vector (column 3)

2. \begin{pmatrix}
2\\
0\\
0\\
\end{pmatrix}

is a multiple of column 1.
\begin{pmatrix}
1 & 1 & 0 & 0 & 2\\
0 & 1 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}

In order to check if this set contains the multiple of column 1, I set the 2 * column 1 equal to the matrix.

It row reduces to
\begin{pmatrix}
1 & 0 & 0 & 0 & 2\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}

Wouldn't this be closed under multiplication? Since this contains a solution?

3. Closed under addition
Column 2 + column 3 becomes
\begin{pmatrix}
1\\
2\\
1\\
\end{pmatrix}It row reduces to
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 1\\
0 & 0 & 1 & 0 & 1
\end{pmatrix}

And wouldn't this be closed under addition? Since this contains a solution?

Edit: I have attached the problem and solution to the thread. I'm not sure how my teacher came across this solution.

 

[Mark44]

Homework Statement

\begin{pmatrix}
1 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 1 & 0
\end{pmatrix}

Is this set a subspace of ℝ³

This is not a set -- it's a matrix.

What is the actual question?

Homework Equations

The set must be closed under addition.

In other words, if you add any two members of the set, do you get another vector that is also in the set?

The set must be closed under multiplication.

If you multiply any of the members of the set by a scalar, do you get another vector that is also a member of the set?

The set must contain the zero vector.

The Attempt at a Solution

1. It obviously contains the zero vector (column 3)

2. \begin{pmatrix}
2\\
0\\
0\\
\end{pmatrix}

What you have below is mostly gibberish. You should focus more on the meanings of the definitions, and less on working with matrices that don't mean anything.

is a multiple of column 1.
\begin{pmatrix}
1 & 1 & 0 & 0 & 2\\
0 & 1 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}

In order to check if this set contains the multiple of column 1, I set the 2 * column 1 equal to the matrix.

It row reduces to
\begin{pmatrix}
1 & 0 & 0 & 0 & 2\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0
\end{pmatrix}

Wouldn't this be closed under multiplication? Since this contains a solution?

3. Closed under addition
Column 2 + column 3 becomes
\begin{pmatrix}
1\\
2\\
1\\
\end{pmatrix}It row reduces to
\begin{pmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 1\\
0 & 0 & 1 & 0 & 1
\end{pmatrix}

And wouldn't this be closed under addition? Since this contains a solution?

 

[MienTommy]

This is not a set -- it's a matrix.

What is the actual question?

I've attached the problem as an image to the OP. I'm not sure how my teacher came across the solution for this problem. It doesn't make sense to me because the span of that set is a subspace correct?

 

[Mark44]

I've attached the problem as an image to the OP. I'm not sure how my teacher came across the solution for this problem. It doesn't make sense to me because the span of that set is a subspace correct?

I responded to what you posted before you edited your post.
The question you asked is very different from what is in the image.

In the image there are two parts. The first part asks whether the given three vectors form a basis for R³. The second part asks whether the given vectors are a subspace. In general, a finite set of vectors won't be a subspace of a vector space, because the axioms of closure under addition and closure under scalar multiplication won't be satisfied.

 

[HallsofIvy]

As Mark44 said, you ask if a set is a subspace but do not give a set of vectors, you give a matrix. Do you mean to treat the columns of the matrix as vectors? The first thing you should realize is that a subspace necessarily contains an infinite number of vectors and this matrix only has a finite number of vectors. Looking at your attachment, there is no matrix but a list of vectors so you simply wrote that incorrectly. But it still is only a finite set and so cannot be a subspace.