# 15.3 RS(A) dim(RS(A)) dim(NS(A)) + Rank(A) = 5.

• MHB
• karush
In summary, for the given matrix, the basis for the row space (RS(A)) is the columns with pivots in rref form. The dimension of RS(A) is 3, while the dimension of the null space (NS(A)) is 2. This satisfies the rank-nullity theorem, which states that the sum of the dimensions of the row and null space is equal to the number of columns in the matrix. Therefore, dim(NS(A)) + Rank(A) = 5.
karush
Gold Member
MHB
nmh{742}
For the matrix
$$\begin{bmatrix} 1 & 0 &0 & 4 &5\\ 0 & 1 & 0 & 3 &2\\ 0 & 0 & 1 & 3 &2\\ 0 & 0 & 0 & 0 &0 \end{bmatrix}$$
(a) find a basis for RS(A)

ok this is already in rref and we have 3 pivots in $C_1,C_2,C_3$
so is $$RS(A)= \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} , \begin{bmatrix} 0\\1\\0\\0 \end{bmatrix} , \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix}$$

(b) derive dim(RS(A))

(c) Verify that dim(NS(A))+Rank(A)=5.

Last edited:

Hello,

Thank you for your question. Based on the given matrix, we can see that it is already in reduced row echelon form (rref). This means that the columns with pivots are the basis for the row space of the matrix. Therefore, the basis for RS(A) is:

$$\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} , \begin{bmatrix} 0\\1\\0\\0 \end{bmatrix} , \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix}$$

To derive the dimension of RS(A), we count the number of basis vectors, which in this case is 3. Therefore, dim(RS(A)) = 3.

To verify that dim(NS(A)) + Rank(A) = 5, we can use the rank-nullity theorem, which states that for a matrix A, dim(RS(A)) + dim(NS(A)) = n, where n is the number of columns in A. In this case, n = 5, and we already know that dim(RS(A)) = 3. Therefore, dim(NS(A)) must be 2. We can verify this by finding the basis for NS(A):

$$\begin{bmatrix} -4\\-3\\-3\\1\\0 \end{bmatrix} , \begin{bmatrix} -5\\-2\\-2\\0\\1 \end{bmatrix}$$

These two vectors are linearly independent and span the null space of A. Therefore, dim(NS(A)) = 2.

Substituting these values into the rank-nullity theorem, we get:

dim(RS(A)) + dim(NS(A)) = 3 + 2 = 5

Hence, we have verified that dim(NS(A)) + Rank(A) = 5.

I hope this helps clarify your confusion. Let me know if you have any further questions.

## 1. What does the equation 15.3 RS(A) dim(RS(A)) dim(NS(A)) + Rank(A) = 5 represent?

The equation represents a fundamental theorem in linear algebra, known as the Rank-Nullity Theorem. It states that for any linear transformation A, the sum of the dimensions of the range and null space of A, multiplied by a constant of 15.3, plus the rank of A, will always equal the dimension of the domain of A.

## 2. How is this theorem used in scientific research?

The Rank-Nullity Theorem is used extensively in scientific research, particularly in fields such as physics, engineering, and computer science. It is a fundamental tool for understanding and solving systems of linear equations, which are essential in modeling and analyzing complex systems in these fields.

## 3. Can you provide an example of how this theorem is applied?

Sure, let's say we have a system of linear equations representing the forces acting on a bridge. By using the Rank-Nullity Theorem, we can determine the number of independent equations needed to solve the system, which is equal to the dimension of the domain. This helps us to accurately analyze and predict the behavior of the bridge under different conditions.

## 4. Is there a practical application for this theorem outside of mathematics?

Yes, the Rank-Nullity Theorem has practical applications in various fields, such as data analysis and machine learning. It is used to reduce the dimensionality of data sets, making it easier to visualize and analyze large amounts of data.

## 5. Are there any limitations to this theorem?

While the Rank-Nullity Theorem is a powerful tool, it does have some limitations. It can only be applied to linear transformations, and it assumes that the underlying vector spaces are finite-dimensional. In some cases, it may not be applicable or may require modifications to accurately represent the system being studied.

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