Octave for Chapter 3
Examples in this section provide sample Octave code for answering questions about linear combinations, span, and linear independence. You can access our code through the link at the bottom of each template. Feel free to modify the code and experiment to learn more!
You can write your own code using Octave software or online Octave cells. To access Octave cells online, go to the Sage Math Cell Webpage, select OCTAVE as the language, enter your code, and press EVALUATE.
To ”save” or share your online code, click on the Share button, select Permalink, then copy the address directly from the browser window. You can store this link to access your work later or share this link with others. You will need to get a new Permalink every time you modify the code.
Octave Tutorial
If possible, express each of and as a linear combination of vectors in .
For each of and we will attempt to solve the system
where , , are vectors of .
We will use the following code.
% Let A_b1 and A_b2 be augmented matrices
% whose columns are the vectors of W
% together with vectors b1 and b2, respectively.
w1=[3; 6;-1; 0; 4];
w2=[1; -2; 3; 5; 1];
w3=[-1; 0; 4; -2; 3];
b1=[0; 4; 1; -1; 7];
b2=[-4; 26; -11; -41; 10];
A_b1=[w1 w2 w3 b1];
A_b2=[w1 w2 w3 b2];
% Reduced row-echelon form
rref_A_b1=rref(A_b1)
rref_A_b2=rref(A_b2)
The reduced row-echelon form of looks like this:
rref_A_b1 =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0
The row indicates that the system is infeasible, so cannot be written as a linear combination of vectors in .
The reduced row-echelon form of looks like this:
rref_A_b2 =This shows that the system of equations has a solution: , , . We write,
1 0 0 2
0 1 0 -7
0 0 1 3
0 0 0 0
0 0 0 0
Determine whether the given vectors are linearly independent.
We will have to determine whether the equation \begin{equation} \label{eq:oct_lin_comb} a\vec{w}_1+b\vec{w}_2+c\vec{w}_3=\vec{0} \end{equation}
has a non-trivial solution. To do this we will consider an augmented matrix . In the
following code, we will skip the zero vector.
% Define the vectors individually
w1=[38; 15; -11];
w2=[64; 82; 14];
w3=[-3; 13; 9];
A=[w1 w2 w3];
% Reduced row-echelon form
rref_A=rref(A)
We get the following output
rref_A =We can see that there are infinitely many solutions , , . Letting , we write, Since (eq:oct_lin_comb) has a non-trivial solution, we conclude that the vectors are linearly dependent.
1 0 -0.5
0 1 0.25
0 0 0
From the set of vectors below, construct a linearly independent set that has the
same span as .
We will refer to vectors of as , in order of appearance. We will build a linearly
independent set from these vectors by adding one vector to our set at a time, and
checking for linear independence every time a vector is added. If, after adding one
vector, the resulting collection is linearly independent, we will keep the newly
added vector, if the collection is not linearly independent we will discard
the new vector and move on to the next. The following code accomplishes
this.
% Define the vectors individually
w1=[12; 10; -20; 5];
w2=[3; -14; 21; 11];
w3=[-6; -38; 62; 17];
w4=[4; -4; 23; 19];
w5=[-6; 6; 47; 30];
w6=[2; 11; 17; 8];
% Clearly {w1} is linearly independent
% Let’s add w2 to the set and check.
rref([w1 w2]) % Looks like w1 and w2 are linearly independent
rref([w1 w2 w3]) % w3 is redundant - this set is linearly dependent. We won’t add w3 to the set
rref([w1 w2 w4]) % these are linearly independent
rref([w1 w2 w4 w5]) % w5 is redundant; don’t add
rref([w1 w2 w4 w6])
Explain why
See discussion below Definition def:redundant for guidance.
By starting with , adding vectors to our set one at a time, and checking for linear independence, we determined that and are redundant and can be removed from the set without changing the span.
Octave Exercises
If possible, express vector as a linear combination of the vectors in . If you find a
linear combination, verify its correctness by hand.
- (a)
-
is in . is NOT in .
- (b)
-
is in . is NOT in .
In Example ex_oct_redundant, we constructed a linearly independent set from vectors of by starting
with and adding one vector at a time to our set moving from left to right and
checking for linear independence at every step. Our resulting set has the same span
as the original set . What if we start this process with and move from right to left to
identify and remove redundant vectors? Would we end up with the same final set?
Modify the code of Example ex_oct_redundant to find out. Does our final set have the same span as ?
Prove your claim.
2024-09-28 14:38:31