Chapter 3.5, Problem 1PS

To determine

(a)

The dimensions of the four subspaces if a 7 by 9 matrix has rank 5 and what is the sum of all four dimensions.

The four subspaces are:

   $\begin{array}{l}\text{dim}\left(\overrightarrow{C}\left(A\right)\right)=5\\ \text{dim}\left(\overrightarrow{C}\left(A^{\text{T}}\right)\right)=5\\ \text{dim}\left(\overrightarrow{N}\left(A\right)\right)=4\\ \text{dim}\left(\overrightarrow{N}\left(A^{\text{T}}\right)\right)=2\end{array}$

The sum of all four subspaces is $16$.

Given:

A 7 by 9 matrix has rank 5.

Calculation:

A 7 by 9 matrix has rank 5, row and column space dimension is 5 and left nullspace is 4.

Thus, the $\text{dim}\left(\overrightarrow{C}\left(A\right)\right)$is 5 and $\text{dim}\left(\overrightarrow{C}\left(A^{\text{T}}\right)\right)$is also 5

Calculate the value of $\text{dim}\left(\overrightarrow{N}\left(A\right)\right)$by subtracting 5 from 9

   $\begin{array}{l}\text{dim}\left(\overrightarrow{N}\left(A\right)\right)=9-5\\ \text{dim}\left(\overrightarrow{N}\left(A\right)\right)=4\end{array}$

Calculate the value of $\text{dim}\left(\overrightarrow{N}\left(A^{\text{T}}\right)\right)$ by subtracting 7 from 9

   $\begin{array}{l}\text{dim}\left(\overrightarrow{N}\left(A\right)\right)=9-7\\ \text{dim}\left(\overrightarrow{N}\left(A\right)\right)=2\end{array}$

Therefore, the four subspaces are

   $\text{dim}\left(\overrightarrow{C}\left(A\right)\right)=5,\text{dim}\left(\overrightarrow{C}\left(A^{\text{T}}\right)\right)=5,\text{dim}\left(\overrightarrow{N}\left(A\right)\right)=4,\text{dim}\left(\overrightarrow{N}\left(A^{\text{T}}\right)\right)=2$

Sum of all the subspaces is:

   $\begin{array}{c}5+5+4+2=16\\ \text{Also,}9+7=16\end{array}$

Therefore, the sum of all spaces is $16$.