### Matrix calculus: Rank and inverse of a matrix

### Rank and column space of a matrix

We will study the subspaces spanned by the rows of a matrix and the columns of a matrix, respectively. Next we will take a look at the relation with systems of equations.

The column space was dealt with *before*, but will be introduced once more in conjunction with the row space:

Row and column spaceConsider the real #(m\times n)#-matrix #A#.

- Each row of #A# is of length #n#, so the rows belong to #\mathbb{R}^n#. The subspace of #\mathbb{R}^n# spanned by the rows is called the
**row space**of #A#. - Similarly, each column of #A# belongs to #\mathbb{R}^m#. The subspace of #\mathbb{R}^m# spanned by the columns is called the
**column space**of #A#.

In theory *Basis and echelon form* we have seen that the dimension of the row space is the same as the *rank* of the matrix. The notion of rank was introduced *before* as the number of rows distinct from the zero row in a row echelon form of the matrix. We will write #\text{rank}(A)# for the rank of #A#.

We will now show that the rank is equal to the dimension of the column space.

*Rank is dimension column space*For every matrix #A# the dimension of the row space is equal to the dimension of the column space. This number is equal to the *rank* of #A#.

*Previously* we concluded that the system of equations #A\vec{x}=\vec{b}# has a solution if and only if #\vec{b}# belongs to the column space of #A#, and that there is only one solution if the columns are independent. Thanks to the statement above we can conclude the following.

Unique solution and maximal rankLet #A# be an #(m\times n)#-matrix. The system of equations \[A\vec{x}=\vec{b}\] has no more than one solution for each vector #\vec{b}# in #\mathbb{R}^m# if and only if the rank of #A# is equal to #n#. In that case, we have #m\ge n#.

Determining the rank of a matrix is straightforward: we row reduce the matrix to an echelon form and count the number of rows distinct from the zero row. The statement above shows that we can also determine the rank by column reduction.

Determine the rank of the matrix \[A=\matrix{1 & 0 & 3 & -5 \\ 1 & 0 & 3 & -5 \\ 2 & 0 & 6 & -10 \\ }\]

#\text{rank}(A)=# #1#

With the aid of elementary row operations we reduce the matrix to the reduced echelon form: \[ \begin{array}{rcl}A = \matrix{1&0&3&-5\\1&0&3&-5\\2&0&6&-10\\}&\sim\matrix{1&0&3&-5\\0&0&0&0\\2&0&6&-10\\}&{\color{blue}{\begin{array}{c}\phantom{x} R_2-R_1\phantom{x}\end{array}}}\\&\sim\matrix{1&0&3&-5\\0&0&0&0\\0&0&0&0\\}&{\color{blue}{\begin{array}{c}\phantom{x} R_3-2R_1\end{array}}}\end{array}\] Because the *rank* is the number of non-null rows of this matrix, we conclude that the rank of the matrix #A# equals #1#.

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