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

[karush]

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.

 

[jvicens]

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.