Problem 40 - Differential Equations and Linear Algebra

Question

Let \(A=\left[\begin{array}{rrrr}1 & 1 & -1 & 1 \\ 2 & -3 & 5 & -6 \\ 5 & 0 & 2 & -3\end{array}\right],\) and let \(\mathbf{v}_{1}=\) (-2,7,5,0) and \(\mathbf{v}_{2}=(3,-8,0,5)\). (a) Show that \(\left\\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\\}\) is a basis for the null space of \(A\). (b) Using the basis in part (a), write an expression for an arbitrary vector \((x, y, z, w)\) in the null space of \(A\).

Step-by-Step Solution

Answer

In this exercise, we showed that the vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ form a basis for the null space of the given matrix $A$. We verified that both A * v1 = 0 and A * v2 = 0, and checked that v1 and v2 are linearly independent. Then, we found an expression for an arbitrary vector (x, y, z, w) in the null space of A as (x, y, z, w) = \( \begin{bmatrix} -2c1 + 3c2 \\ 7c1 -8c2 \\ 5c1 \\ 5c2 \end{bmatrix} \), where c1 and c2 are scalars.

Step 1: Show that v1 and v2 belong to the null space of A

To show that both vectors v1 and v2 belong to the null space of A, we have to verify that A * v1 = 0 and A * v2 = 0: A * v1 = \[\begin{bmatrix} 1 & 1 & -1 & 1 \\ 2 & -3 & 5 & -6 \\ 5 & 0 & 2 & -3 \end{bmatrix}\] * \[\begin{bmatrix} -2 \\ 7 \\ 5 \\ 0 \end{bmatrix}\] = \[\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\] A * v2 = \[\begin{bmatrix} 1 & 1 & -1 & 1 \\ 2 & -3 & 5 & -6 \\ 5 & 0 & 2 & -3 \end{bmatrix}\] * \[\begin{bmatrix} 3 \\ -8 \\ 0 \\ 5 \end{bmatrix}\] = \[\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\] We can see that both A * v1 and A * v2 give the zero vector, thus both v1 and v2 belong to the null space of A.

Step 2: Check if v1 and v2 are linearly independent

In order to check if v1 and v2 are linearly independent, we have to make sure that there is no scalar c so that v1 = c * v2. Since both v1 and v2 are non-zero vectors, we can directly compare their components, and we can see that the ratio between each corresponding component is not constant, therefore, v1 and v2 are linearly independent. Now, we have shown that v1 and v2 both belong to the null space of A and are linearly independent, so they form a basis for the null space of A.

Step 3: Write an expression for an arbitrary vector in the null space of A

Let (x, y, z, w) be a vector in the null space of A, then it can be written as a linear combination of v1 and v2: (x, y, z, w) = c1 * v1 + c2 * v2 = (c1 * \[\begin{bmatrix} -2 \\ 7 \\ 5 \\ 0 \end{bmatrix}\] + c2 * \[\begin{bmatrix} 3 \\ -8 \\ 0 \\ 5 \end{bmatrix}\]) = \( \begin{bmatrix} -2c1 + 3c2 \\ 7c1 -8c2 \\ 5c1 \\ 5c2 \end{bmatrix} \) Where c1 and c2 are scalars, and (x, y, z, w) can take any value provided that it is in the null space of A.

Key Concepts

Linear Independence

Linear independence is an essential concept in linear algebra that helps determine the relationships among vectors. When two or more vectors are linearly independent, this means that no vector in the set can be expressed as a linear combination of the others. In simpler terms, each vector in the set adds a new direction to the vector space without being redundant or overlapping with other vectors.
In the context of the exercise, vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) were checked to ensure they are linearly independent. This was done by confirming that there are no scalars \(c_1\) and \(c_2\), not both zero, such that:

• \(\mathbf{v}_1 = c_1 \mathbf{v}_2\)
• or equivalently, \(c_1\mathbf{v}_1 + c_2 \mathbf{v}_2 = \mathbf{0}\)

When we compared the components of these vectors, they didn't form a constant ratio, signifying independence. If they had been linearly dependent, a constant ratio among their corresponding components would exist, or one vector would be a scalar multiple of the other.

Basis for the Null Space

The null space, also known as the kernel, of a matrix \(A\) consists of all the vectors that, when multiplied by \(A\), result in the zero vector. A basis for the null space is a set of linearly independent vectors that spans the null space.
In the exercise, the vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) were shown to belong to the null space of \(A\) because:

• \(A\mathbf{v}_1 = \mathbf{0}\)
• \(A\mathbf{v}_2 = \mathbf{0}\)

To confirm that these vectors form a basis, they must not only be members of the null space but also linearly independent, as demonstrated in the previous section. With these criteria satisfied, \(\{\mathbf{v}_1, \mathbf{v}_2\}\) serves as a basis for the null space of \(A\), providing the smallest set of vectors needed to describe all vectors in the null space.

Linear Combination

Understanding a linear combination of vectors is fundamental in expressing vectors within a vector space. A linear combination involves taking scalar multiples of vectors and adding them together. This concept helps describe how any vector in a space can be created from others in a basis.
Given vectors \(\mathbf{v}_1\) and \(\mathbf{v}_2\) from the exercise as the basis of the null space of \(A\), any vector \((x, y, z, w)\) in this space can be represented as:

• \((x, y, z, w) = c_1\mathbf{v}_1 + c_2\mathbf{v}_2\)
• Where \(c_1\) and \(c_2\) are scalars.

This representation is important because it shows that every vector in the null space is a combination of just the two basis vectors. For example, if we calculate:

• \[-2c_1 + 3c_2\]
• \[7c_1 - 8c_2\]
• \[5c_1\]
• \[5c_2\]

These components constitute the vector \((x, y, z, w)\) based on \(c_1\) and \(c_2\). Each specific choice of scalars will give a unique vector in the null space, highlighting the richness of solutions possible from just two basis vectors.