### 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

\[\begin{aligned}\left(\begin{array}{ccc|ccc} 1&-4&-3&1 & 0 & 0 \\ -1&20&3&0 &1 &0\\ 0&4&16&0 &0 &1\\ \end{array} \right)&\sim \left( \begin{array}{ccc|ccc} 1 &-4&-3&1 & 0 & 0\\ 0 &16&0&1& 1 & 0 \\ 0 &4&16&0& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to R_2 +R_1\\ \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &-4&-3 &1 & 0 & 0 \\ 0 &1 &0&\frac{1}{16}&\frac{1}{16} &0 \\ 0 &4&16&0& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to \frac{1}{16}R_2\\ \mbox{} \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &-3&\frac{5}{4}&\frac{1}{4}&0 \\ 0 &1 &0&\frac{1}{16}&\frac{1}{16}&0 \\ 0 &0 &16&-\frac{1}{4}&-\frac{1}{4}&1\\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1\to R_1+4R_2\\ \mbox{}\\ R_3\to R_3 -4R_2 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &-3&\frac{5}{4}&\frac{1}{4}&0\\ 0 &1 &0&\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &-\frac{1}{64}&-\frac{1}{64}&\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}\\ {}\\ R_3\to \frac{1}{16}R_3 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &0 &\frac{77}{64}&\frac{13}{64}&\frac{3}{16}\\ 0 &1 &0 &\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &-\frac{1}{64}&-\frac{1}{64}&\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1 \to R_1 +3R_3\\ \phantom{X}\\ {} \end{array}}} \end{aligned} \] 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*:\[\begin{aligned}\left(\begin{array}{ccc|ccc} 1&-4&-3&1 & 0 & 0 \\ -1&20&3&0 &1 &0\\ 0&4&16&0 &0 &1\\ \end{array} \right)&\sim \left( \begin{array}{ccc|ccc} 1 &-4&-3&1 & 0 & 0\\ 0 &16&0&1& 1 & 0 \\ 0 &4&16&0& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to R_2 +R_1\\ \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &-4&-3 &1 & 0 & 0 \\ 0 &1 &0&\frac{1}{16}&\frac{1}{16} &0 \\ 0 &4&16&0& 0 & 1 \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} \mbox{}\\ R_2 \to \frac{1}{16}R_2\\ \mbox{} \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &-3&\frac{5}{4}&\frac{1}{4}&0 \\ 0 &1 &0&\frac{1}{16}&\frac{1}{16}&0 \\ 0 &0 &16&-\frac{1}{4}&-\frac{1}{4}&1\\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1\to R_1+4R_2\\ \mbox{}\\ R_3\to R_3 -4R_2 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &-3&\frac{5}{4}&\frac{1}{4}&0\\ 0 &1 &0&\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &-\frac{1}{64}&-\frac{1}{64}&\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}\\ {}\\ R_3\to \frac{1}{16}R_3 \end{array}}}\\ &\sim \left( \begin{array}{ccc|ccc} 1 &0 &0 &\frac{77}{64}&\frac{13}{64}&\frac{3}{16}\\ 0 &1 &0 &\frac{1}{16}&\frac{1}{16}&0\\ 0 &0 &1 &-\frac{1}{64}&-\frac{1}{64}&\frac{1}{16} \\ \end{array} \right) &{\color{blue}{\begin{array}{ccc} {}R_1 \to R_1 +3R_3\\ \phantom{X}\\ {} \end{array}}} \end{aligned} \] 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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