Linear Algebra: Linear Independence and writing Matrices as linear combinations

In summary: However, I think I may have accidentally reversed the signs of the a, b, c, and d coefficients when I reduced the 4x5 matrix.
  • #1
mattst88
29
0

Homework Statement



If linearly dependent, write one matrix as a linear combination of the rest.

[tex]\left[\begin{array}{cc} 1&1 \\ 2&1 \end{array}\right][/tex] [tex]\left[\begin{array}{cc} 1&0 \\ 0&2 \end{array}\right][/tex] [tex]\left[\begin{array}{cc} 0&3 \\ 2&1 \end{array}\right][/tex] [tex]\left[\begin{array}{cc} 4&6 \\ 8&6 \end{array}\right][/tex]

Homework Equations



The Attempt at a Solution



Choosing coefficients a, b, c, d for the matrices, respectively, I set up 4 equations and 4 unknowns:

row 1 column 1: [tex]a + b + 4d = 0[/tex]
row 1 column 2: [tex]a+ 3c + 6d = 0[/tex]
row 2 column 1: [tex]2a + 2c + 8d = 0[/tex]
row 2 column 2: [tex]a + 2b + c + 6d = 0[/tex]

From this, I create a 4x5 matrix and using Gauss-Jordan elimination arrive at:

Edit: this is wrong. See below

[tex]\left[\begin{array}{ccccc}
1&0&0&2&0 \\
0&1&0&2&0 \\
0&0&1&\frac{4}{3}&0 \\
0&0&0&0&0 \end{array}\right][/tex]

From this, it is clear that the 4 matrices are linearly dependent.

What I do not understand: how do I, given this last matrix, write one of the matrices as a linear combination of the others?

Thanks
 
Last edited:
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  • #2
Your last matrix means this:
a = -2d
b = -2d
c = -4/3 d

From this we can deduce that d is arbitrary, so let d = 1.
Then a = -2, b = -2, and c = -4/3.

Put these values into the equation you set up to determine linear dependence and you'll see that one of them is a particular linear combination of the other three matrices.

BTW, when you solved the 4 equations in 4 unknowns, you could have used a 4 x 4 matrix instead of a 4 x 5 augmented matrix. Your 5th column started as all zeroes and never changed.
 
  • #3
Mark44 said:
Your last matrix means this:
a = -2d
b = -2d
c = -4/3 d

From this we can deduce that d is arbitrary, so let d = 1.
Then a = -2, b = -2, and c = -4/3.

Put these values into the equation you set up to determine linear dependence and you'll see that one of them is a particular linear combination of the other three matrices.

I attempted to check this, but it appears to not work, as the equation generated is not true.

[tex]-2a - 2b + \frac{4}{3}c = d[/tex]

Have I reduced the 4x5 matrix incorrectly or is the problem more fundamental?
 
Last edited:
  • #4
edit: read the equations wrong
 
  • #5
It's possible you made a mistake in reducing your 4x5 matrix. If so, that will affect the values of a, b, c, and d.

Assuming for the moment that your work was correct, these values are solutions of the equation
a*M1 + b*M2 + c*M3 + d*M4 = 0, where M1 etc are the 2x2 matrices in your first post, and NOTof the equation -2a -2b + 4/3 c = d.

The a, b, c, and d values should tell you which matrix is a linear combination of the others.
 
  • #6
Going from the 4 equations/unknowns to the matrix I made a silly mistake causing the row reduced matrix to be incorrect.

The (I think) correct row reduced matrix is

[tex]
\left[\begin{array}{ccccc}
1&0&0&3 \\
0&1&0&1 \\
0&0&1&1 \\
0&0&0&0 \end{array}\right]
[/tex]

From here, I thought I would read out of the matrix the following equations:

[tex]a = -3d[/tex]
[tex]b = -d[/tex]
[tex]c = -d[/tex]
where d is any arbitrary constant.

Assigning d = 1, and checking to see if [tex]a * M1 + b * M2 + c * M3[/tex] was in fact [tex]d * M4[/tex], I end up with exactly the wrong signs:

[tex]-3\left[\begin{array}{cc}1&1\\2&1\end{array}\right] + -1\left[\begin{array}{cc}1&0\\0&2\end{array}\right] + -1\left[\begin{array}{cc}0&3\\2&1\end{array}\right] = \left[\begin{array}{cc}-4&-6\\-8&-6\end{array}\right][/tex]

This must mean that a should equal 3d, b should equal d, c should equal d, without the negatives, but I do not understand why.
 
  • #7
After thinking about it more, I guess that the a, b, c columns should be partitioned from the d column since I intend to write d * M4 in terms of the other matrices and their respective coefficients.

This would explain the sign error from before.

Can someone confirm?
 
  • #8
Yes, that's the problem. The a, b, c, and d are coefficients in this equation:
a*M1 + b*M2 + c*M3 + d*M4 = 0. I checked and they work out.
 

1. What is linear independence in linear algebra?

Linear independence is a concept in linear algebra that describes the relationship between vectors. It means that no vector in a set of vectors can be written as a linear combination of the other vectors in the set. In simpler terms, it means that the vectors in the set are not redundant and each one brings something unique to the table.

2. How do you determine if a set of vectors is linearly independent?

To determine if a set of vectors is linearly independent, you can use the technique of Gaussian elimination. This involves writing the vectors as columns in a matrix and performing row operations to reduce the matrix to row-echelon form. If the resulting matrix has a pivot position in every column, then the vectors are linearly independent. If there are any columns without a pivot, then the vectors are linearly dependent.

3. What is a linear combination of vectors?

A linear combination of vectors is the result of multiplying each vector by a scalar (a real number) and then adding them together. For example, if we have two vectors, v1 and v2, the linear combination of these vectors would be αv1 + βv2, where α and β are scalars.

4. Why is writing matrices as linear combinations useful in linear algebra?

Writing matrices as linear combinations is useful because it allows us to express a matrix as a combination of simpler matrices. This can help us solve systems of linear equations, find the inverse of a matrix, and perform other operations more easily. It also allows us to easily determine if a set of vectors is linearly independent, as mentioned in question 2.

5. Can any matrix be written as a linear combination of simpler matrices?

No, not every matrix can be written as a linear combination of simpler matrices. For example, a singular matrix (a matrix with no inverse) cannot be written as a linear combination of other matrices. Additionally, the number of simpler matrices needed to represent a matrix as a linear combination may vary depending on the matrix and the vectors used in the combination.

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