Find a basis of a linear system

[Kavorka]

Find a basis for the solution space of the linear system

x₁-x₂-2x₃+x₄ = 0
-3x₁+3x₂+x₃-x₄ = 0
2x₁-2x₂+x₃ = 0

I created a matrix (not augmented, will be 0 on right side no matter what row operations) and brought it to reduced echelon form. x₂ and x₄ were free variables and I set them to the variables t and s respectively. I found that:

x₁ = t - (1/5)s
x₃ = (-2/5)s

All of this I'm fairly certain I did right, I just wanted to make sure I am stating the basis correctly because this is the first problem I've done where I've found a basis for a linear system. Stating all the x's in terms of s and t I get (forgive weird formatting not sure how to do matrices on here):

[x₁]\\\\\\[1]\\\\\\\[-1/5]
[x₂]\\=\t\[1]\+\s\[0]
[x₃]\\\\\\[0]\\\\\\\[-2/5]
[x₄]\\\\\\[0]\\\\\\\[1]

Forming the 2 dimensional basis:
[1] [-1]
[1] [0]
[0] [-2]
[0], [5]

 

[Mark44]

Find a basis for the solution space of the linear system

x₁-x₂-2x₃+x₄ = 0
-3x₁+3x₂+x₃-x₄ = 0
2x₁-2x₂+x₃ = 0

I created a matrix (not augmented, will be 0 on right side no matter what row operations) and brought it to reduced echelon form. x₂ and x₄ were free variables and I set them to the variables t and s respectively. I found that:

x₁ = t - (1/5)s
x₃ = (-2/5)s

All of this I'm fairly certain I did right, I just wanted to make sure I am stating the basis correctly because this is the first problem I've done where I've found a basis for a linear system. Stating all the x's in terms of s and t I get (forgive weird formatting not sure how to do matrices on here):

[x₁]\\\\\\[1]\\\\\\\[-1/5]
[x₂]\\=\t\[1]\+\s\[0]
[x₃]\\\\\\[0]\\\\\\\[-2/5]
[x₄]\\\\\\[0]\\\\\\\[1]

Forming the 2 dimensional basis:
[1] [-1]
[1] [0]
[0] [-2]
[0], [5]

You can check this for yourself. If A is the matrix of coefficients of your system above, and x₁ and x₂ are your two basis vectors, verify that Ax₁ = 0 and that Ax₂ = 0. If either of these doesn't come out to 0, then definitely you have made a mistake. If they both come out to zero, then A times any linear combination of the two vectors will also be zero.

Edit: you have a sign error in your second basis vector.

 

[HallsofIvy]

You have four equation in three unknowns so one would initially expect the solution space to be 4- 3= 1 dimensional. Of course, if the equations are not independent that may not be true. However, I believe these are independent so your answer cannot be correct.

 

[Mark44]

However, I believe these are independent so your answer cannot be correct.

The three equations are dependent. The OP has a sign error in one of his basis vectors.

 

[HallsofIvy]

Oops. Thanks.

 

[Mark44]

Oops. Thanks.

Happens to us all...

 

[BvU]

From latexhelp:

Matrices: A matrix can easily be typed using the matrix, pmatrix, bmatrix or vmatrix environments. For example, the pmatrix environment starts with \bеgin{pmatrix} and ends with \еnd{pmatrix}. Columns are separated by & and rows are separated by \\ . Here is an example:

https://www.physicsforums.com/file://\\begin{pmatrix }
1 & 2 & 3 & 4\\
a & b & c & d\\
x & y & z & w
\end{pmatrix}

This yields
$$
\begin{pmatrix}
1 & 2 & 3 & 4\\
a & b & c & d\\
x & y & z & w
\end{pmatrix}
$$
The matrix, bmatrix and vmatrix environments produce similar results. Only the delimiters are different. See this web page for more examples.

 

[Mark44]

I like the bmatrix environment, which uses brackets around the matrix rather than big parentheses.

https://www.physicsforums.com/file:////begin%7Bpmatrix }
1 & 2 & 3 & 4\\
a & b & c & d\\
x & y & z & w
\end{bmatrix}

This yields
$$
\begin{bmatrix}
1 & 2 & 3 & 4\\
a & b & c & d\\
x & y & z & w
\end{bmatrix}
$$

The vmatrix environment is useful for rendering determinants.
\begin{vmatrix]
1 & 2 & 3\\
a & b & c\\
x & y & z
\end{vmatrix}
which renders as
$$
\begin{vmatrix}1 & 2 & 3\\
a & b & c\\
x & y & z
\end{vmatrix}$$

 

[Kavorka]

Thanks everyone! I noticed the sign error, and when multiplying my basis vectors by the original matrix I get a null matrix in both cases. I am just wondering, in back-checking your answer to say positively that your basis is correct, is this the only step you need to perform?

 

[BvU]

I seem to remember you also need to show the basis vectors are independent, which in this case is a piece of cake.