### 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

Yes

We will approach this just like inverting a matrix: we augment the matrix with an identity matrix and apply

\[\begin{aligned}\left(\begin{array}{ccc|ccc} 1&1&5&1 & 0 & 0 \\ -1&15&-13&0 &1 &0\\ 1&5&-13&0 &0 &1\\ \end{array} \right)&\sim \left( \begin{array}{ccc|ccc} 1 &1&5&1 & 0 & 0\\ 0 &16&-8&1& 1 & 0 \\ 0 &4&-18&-1& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to R_2 +R_1\\ R_3 \to R_3 - R_1 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &1&5 &1 & 0 & 0 \\ 0 &1 &-\frac{1}{2}&\frac{1}{16}&\frac{1}{16} &0 \\ 0 &4&-18&-1& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to \frac{1}{16}R_2\\ \mbox{} \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{11}{2}&\frac{15}{16}&-\frac{1}{16}&0 \\ 0 &1 &-\frac{1}{2}&\frac{1}{16}&\frac{1}{16}&0 \\ 0 &0 &-16&-\frac{5}{4}&-\frac{1}{4}&1\\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1\to R_1-R_2\\ \mbox{}\\ R_3\to R_3 -4R_2 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{11}{2}&\frac{15}{16}&-\frac{1}{16}&0\\ 0 &1 &-\frac{1}{2}&\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &\frac{5}{64}&\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}\\ {}\\ R_3\to -\frac{1}{16}R_3 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &0 &\frac{65}{128}&-\frac{19}{128}&\frac{11}{32}\\ 0 &1 &0 &\frac{13}{128}&\frac{9}{128}&-\frac{1}{32}\\ 0 &0 &1 &\frac{5}{64}&\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1 \to R_1 - \frac{11}{2}R_3\\ R_2 \to R_2 +\frac{1}{2}R_3\\ {} \end{array}}} \end{aligned} \] The left-hand matrix of the result has rank 3. Hence, the answer is: Yes.

We will approach this just like inverting a matrix: we augment the matrix with an identity matrix and apply

*Gaussian elimination*:\[\begin{aligned}\left(\begin{array}{ccc|ccc} 1&1&5&1 & 0 & 0 \\ -1&15&-13&0 &1 &0\\ 1&5&-13&0 &0 &1\\ \end{array} \right)&\sim \left( \begin{array}{ccc|ccc} 1 &1&5&1 & 0 & 0\\ 0 &16&-8&1& 1 & 0 \\ 0 &4&-18&-1& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to R_2 +R_1\\ R_3 \to R_3 - R_1 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &1&5 &1 & 0 & 0 \\ 0 &1 &-\frac{1}{2}&\frac{1}{16}&\frac{1}{16} &0 \\ 0 &4&-18&-1& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to \frac{1}{16}R_2\\ \mbox{} \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{11}{2}&\frac{15}{16}&-\frac{1}{16}&0 \\ 0 &1 &-\frac{1}{2}&\frac{1}{16}&\frac{1}{16}&0 \\ 0 &0 &-16&-\frac{5}{4}&-\frac{1}{4}&1\\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1\to R_1-R_2\\ \mbox{}\\ R_3\to R_3 -4R_2 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &\frac{11}{2}&\frac{15}{16}&-\frac{1}{16}&0\\ 0 &1 &-\frac{1}{2}&\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &\frac{5}{64}&\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}\\ {}\\ R_3\to -\frac{1}{16}R_3 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &0 &\frac{65}{128}&-\frac{19}{128}&\frac{11}{32}\\ 0 &1 &0 &\frac{13}{128}&\frac{9}{128}&-\frac{1}{32}\\ 0 &0 &1 &\frac{5}{64}&\frac{1}{64}&-\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1 \to R_1 - \frac{11}{2}R_3\\ R_2 \to R_2 +\frac{1}{2}R_3\\ {} \end{array}}} \end{aligned} \] 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.

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