# Finding the basis of this subspace

• Dell
In summary, the basis for W={(x,y,z,t)|x+y=0, x+t=0} is (1,1,0,0) and (0,1,0,1), which are linearly independent and span W. The correct solution in the book is (1,1,0,0) and (0,-1,0,1). The reason for the discrepancy is that the matrices used in the book and the OP's method represent different equations.
Dell
in linear algebra, if i am told to find a basis for the following

W={(x,y,z,t)|x+y=0, x+t=0}

what i did was

1 1
1 0
0 0
0 1
after performing elementary actions on rows, i came to

1 0
0 1
0 0
0 0
from here i can see that they are linearly independant and they cleary span their own space so they are my basis.

basis= (1,1,0,0) (0,1,0,1)

the correct solution in my book was quite different, they put the vectors lying down like so

1 1 0 0
0 1 0 1

and eventually got to

1 0 0 -1
0 1 0 1

what is the reason that my way is wrong,

Last edited:
I think you might have started off with the wrong vectors. Based on your definition of W, I see that
x = -y
y = y
z = z
t = t

Since x = -y and x = -t, it's clear that y = t, so the four equations above simplify to:
x = -t
y = t
z = z
t = t
This set of equations can be written as
(x, y, z, t) = t(-1, 1, 0, 1) + z(0, 0, 1, 0)

So two vectors that span W are the two above, and they are clearly linearly independent, since no multiple of either could possibly produce the other. (It's very easy to check independence when you have only two vectors. With three or more you can't do it by inspection.)

Notice that these two vectors satisfy the equations that are part of your definition; your vectors don't.

Dell said:
in linear algebra, if i am told to find a basis for the following

W={(x,y,z,t)|x+y=0, x+t=0}

what i did was

1 1
1 0
0 0
0 1
after performing elementary actions on rows, i came to

1 0
0 1
0 0
0 0
from here i can see that they are linearly independant and they cleary span their own space so they are my basis.

basis= (1,1,0,0) (0,1,0,1)
Mark's work is absolutely correct. Let's just make sure you know what you're doing with those matrices.

The span of W is the null space of the matrix

$$A = \begin{pmatrix} 1 & 1 & 0 & 0\\ 1 & 0 & 0 & 1\end{pmatrix}$$

,i.e., is the solution set of Av=0 where v=(x,y,z,t).

By row reducing (just one operation) Av=0 we see that the null space of A is equal to the null space of

$$A' = \begin{pmatrix} 1&1&0&0\\0&-1&0&1\end{pmatrix}$$.

,i.e., the solution set of A'v = 0.

With v = (x,y,z,t), writing this out we have x+y=0 and -y+t=0. The solution set of this follows from Mark's work (thus illustrating that matrices in this particular example are quite unnecessary).

Dell said:
the correct solution in my book was quite different, they put the vectors lying down like so

1 1 0 0
0 1 0 1

and eventually got to

1 0 0 -1
0 1 0 1

what is the reason that my way is wrong,
Presuming these matrices are supposed to represent A (from our work above), check you're referring to the same question! Our set-up has x+t=0 whereas the row "0 1 0 1" suggests they have y+t=0 !

## 1. What is the basis of a subspace?

The basis of a subspace is a set of linearly independent vectors that span the entire subspace. This means that any vector within the subspace can be written as a linear combination of the basis vectors.

## 2. How do you find the basis of a subspace?

To find the basis of a subspace, you can use the following steps:
1. Determine the dimensionality of the subspace.
2. Choose a set of vectors that span the subspace.
3. Use Gaussian elimination or other methods to reduce the set of vectors to a set of linearly independent vectors.
4. The resulting set of linearly independent vectors is the basis of the subspace.

## 3. Can a subspace have more than one basis?

Yes, a subspace can have multiple bases. This is because there can be multiple sets of linearly independent vectors that span the same subspace. However, all of these bases will have the same number of vectors, known as the dimensionality of the subspace.

## 4. What is the significance of finding the basis of a subspace?

Finding the basis of a subspace is important because it allows us to understand the structure and properties of the subspace. It also helps us to easily represent vectors within the subspace and perform calculations on them.

## 5. Can the basis of a subspace change?

The basis of a subspace can change when the subspace itself changes. For example, if new vectors are added to the subspace, the basis may change. However, the dimensionality of the subspace will remain the same, and the new basis will still span the subspace.

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