Invertibility and rank

Matrix calculus: Rank and inverse of a matrix

Invertibility and rank

Previously we have seen some invertibility criteria for linear maps. Thanks to the theorem Linear map determined by the image of a basis this also provides invertibility criteria for matrices. We will add another criterium, in terms of the rank.

Invertiblity and rank

Let #n# be a natural number. For each #(n\times n)#-matrix #A# the following statements are equivalent:

The rank of #A# is #n#
The rows of #A# are independent
The columns of #A# are independent
The reduced echelon form of #A# is the identity matrix
The matrix #A# is invertible

#1\Rightarrow 2# and #1\Rightarrow 3#: If the rank of #A# is equal to #n#, then both the #n# rows as well as the #n# columns span an #n#-dimensional space, and this means they are independent.

#2\Rightarrow 1# and #3\Rightarrow 1#: If the #n# rows (or columns) are independent, then the rows (or columns) span an #n#-dimensional space, and the rank of the matrix is equal to #n#. Moreover, the column space (or row space) is then also #n#-dimensional and hence the columns (or rows) are independent.

Hence, statements 1, 2 and 3 are equivalent.

#1\Leftrightarrow 4#: The matrix #A# and the reduced echelon form of #A# have the same rank. From the structure of the reduced echelon form, we can see right away that the matrix has rank #n# if and only if it is an identity matrix.

#1\Leftrightarrow 5#: By the theorem Linear map determined by the image of a basis, #L_A# is invertible if and only if #\im{L_A} =\mathbb{R}^n#, meaning, if and only if the rank of #A# is equal to #n#.

Supplement

The given list can be supplemented as follows:

#A# is injective
#A# is surjective
#\ker{A}=\{\vec{0}\}#
#\im{A}=\mathbb{R}^n#
#\det(A)\neq0#

According to theorem Invertibility with the same dimensions for domain and codomain statements 6 and 7 are equivalent to statement 5.

According to the Criteria for injectivity en surjectivity statements 6 and 7 are equivalent to statements 8 respectively 9. Besides, statements 8 and 9 are equivalent to \[\dim{\ker{A}}=0\qquad\text{respectively}\qquad\dim{\im{A}}=n\] in accordance with the rank-nullity theorem. According to Rank is dimension column space, the last equation is equivalent to #\text{rank}(A)=n#, which confirms the equivalence of statements 1 and 9 directly.

The equivalence of statements 10 and 5 will be proven later in Invertibility in terms of determinant. In that statement we will also prove \[\det\left(A^{-1}\right)=\frac{1}{\det(A)}\] on the condition that #A^{-1}# exists. Together with statement 10 it then follows that every statement in the list for #A# is equivalent to the corresponding statement for #A^{-1}#, if the inverse exists.

Later in Determinant of transpose and product we prove #\det(A)=\det(A^\top)#. Together with statement 10 it then follows that every statement in the list for #A# is equivalent to the corresponding statement for #A^\top#.

According to the dependence criterion statements 2 and 3 are equivalent to the statements that there does not exists a non-trivial relation between the rows respectively columns of #A#.

Dimension 1 A #(1\times1)#-matrix #A=\matrix{a}# satisfies the conditions if and only if #a\ne0#.

Dimension 2

Let #A=\matrix{a&b\\ c & d}#. The columns of #A# are linearly independent if and only if there exists a non trivial linear combination resulting in the zero vector, hence if there are scalars #\lambda# and #\mu#, not both equal to #0#, such that

\[\lambda\rv{a,c}+\mu\rv{b,d} = \rv{0,0}\]

#\lambda \cdot a=- \mu\cdot b# and #\lambda \cdot c=- \mu\cdot d#

#\lambda \cdot a \cdot d = -\mu\cdot b \cdot d = \lambda\cdot c\cdot b#

so #\lambda=0# or #a\cdot d-b\cdot c = 0#. Just as: #\mu=0# or #a\cdot d-b\cdot c = 0#.

Conclusion: The columns of #A# are linearly independent if and only if #a\cdot d-b\cdot c \ne 0#. This condition is the same for #A^\top# as for #A#. This explains that the rank of #A# is equal to #2# if and only if the rank of #A^\top# would be equal to #2#. This shows once again that both statement 2 and 3 are equivalent with statement 1.

Statement 5 can even be illustrated by means of the following, easy to calculate, formula

\[\text{For }B = \matrix{d&-b\\ -c&a}\text{ we have } A\,B = (a\cdot d -b\cdot c)\cdot I_2\]

If #a\cdot d -b\cdot c\ne0#, then #A^{-1} =\frac{1}{a\cdot d -b\cdot c}\cdot B# is the inverse of #A#. Now assume that #a\cdot d -b\cdot c = 0#. If #A# is the zero matrix, then #A# is not invertible. If #A# is unequal to the zero matrix, then some column of #B# is unequal to #\vec{0}# and belongs to the kernel of #A#, so #A# is not invertible. We conclude that #A# is invertible if and only if #a\cdot d -b\cdot c\ne0#.

The expression #a\cdot d -b\cdot c# is known as the determinant of #A#, which we will see later.

Is the following matrix invertible?
\[
\matrix{
1 &2 & 0 \\
-1 &14 &-4 \\
-1 &2 &-17}
\]

Yes

We will approach this just like inverting a matrix: we augment the matrix with an identity matrix and apply Gaussian elimination:
\[\begin{aligned}\left(\begin{array}{ccc|ccc} 1&2&0&1 & 0 & 0 \\ -1&14&-4&0 &1 &0\\ -1&2&-17&0 &0 &1\\ \end{array} \right)&\sim \left( \begin{array}{ccc|ccc} 1 &2&0&1 & 0 & 0\\ 0 &16&-4&1& 1 & 0 \\ 0 &4&-17&1& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to R_2 +R_1\\ R_3 \to R_3 +R_1 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &2&0 &1 & 0 & 0 \\ 0 &1 &-\frac{1}{4}&\frac{1}{16}&\frac{1}{16} &0 \\ 0 &4&-17&1& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to \frac{1}{16}R_2\\ \mbox{} \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{1}{2}&\frac{7}{8}&-\frac{1}{8}&0 \\ 0 &1 &-\frac{1}{4}&\frac{1}{16}&\frac{1}{16}&0 \\ 0 &0 &-16&\frac{3}{4}&-\frac{1}{4}&1\\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1\to R_1-2R_2\\ \mbox{}\\ R_3\to R_3 -4R_2 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{1}{2}&\frac{7}{8}&-\frac{1}{8}&0\\ 0 &1 &-\frac{1}{4}&\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &-\frac{3}{64}&\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}\\ {}\\ R_3\to -\frac{1}{16}R_3 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &0 &\frac{115}{128}&-\frac{17}{128}&\frac{1}{32}\\ 0 &1 &0 &\frac{13}{256}&\frac{17}{256}&-\frac{1}{64}\\ 0 &0 &1 &-\frac{3}{64}&\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1 \to R_1 - \frac{1}{2}R_3\\ R_2 \to R_2 +\frac{1}{4}R_3\\ {} \end{array}}} \end{aligned} \] The left-hand matrix of the result has rank 3. Hence, the answer is: Yes.

Row reduction of #A# augmented with the #(3\times3)#-identity matrix not only shows that #A# is invertible, but also that the inverse is equal to the right-hand #(3\times3)#-matrix of the result.

New example