The properties of matrices

Associativity

Let be \(A \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})\), \(B \in \hspace{0.03em} \mathcal{M}_{p,q} (\mathbb{K})\) et \(C \in \hspace{0.03em} \mathcal{M}_{q,r} (\mathbb{K})\) three matrices.

1. Computation of \((A \times B) \times C\)

   By definition, we do have:

   $$ \forall (i, j) \in [\![1, n]\!] \times [\![1, r]\!],$$
   $$\Bigl( (A \times B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^q (ab)_{i,k} \times c_{k,j} $$

   But, the factor \((ab)_{i,k}\) is worth:

   $$ \forall (i, k) \in [\![1, n]\!] \times [\![1, q]\!],$$

   $$ (ab)_{i,k} = \sum_{l = 1}^p a_{i,l} \times b_{l,k} $$

   So, we replace it in the main expression and:

   $$\Bigl( (A \times B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^q \left[ \sum_{l = 1}^p a_{i,l} \times b_{l,k} \right] \times c_{k,j} $$

   Since the factor \(c_{k,j}\) is independent from \(l\), it can be considered as a constant, and integrated inside the inner sum.

   $$\Bigl( (A \times B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^q \sum_{l = 1}^p \Bigl[ a_{i,l} \times b_{l,k} \times c_{k,j} \Bigr] \qquad (1) $$

2. Computation of \( A \times (B \times C)\)

   Let us now calculate the product \(A \times (B \times C)\).

   $$ \forall (i, j) \in [\![1, n]\!] \times [\![1, r]\!],$$
   $$\Bigl( A \times (B \times C) \Bigr)_{i,j} = \ \sum_{k = 1}^p a_{i,k} \times (bc)_{k,j} $$

   In the same way, we replace \((bc)_{k,j}\) by its expression and:

   $$\Bigl( A \times (B \times C) \Bigr)_{i,j} = \ \sum_{k = 1}^p a_{i,k} \times \left[ \sum_{l = 1}^q b_{k,l} \times c_{l,j} \right] $$
   $$\Bigl( A \times (B \times C) \Bigr)_{i,j} = \ \sum_{k = 1}^p \sum_{l = 1}^q \Bigl[ a_{i,k} \times b_{k,l} \times c_{l,j} \Bigr] \qquad (2) $$

   In both expression \((1)\) and \((2)\), the variables \(k\) and \(l\) are free variables:

   $$\Bigl( (A \times B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^q \sum_{l = 1}^p \Bigl[ a_{i,l} \times b_{l,k} \times c_{k,j} \Bigr] \qquad (1) $$
   $$\Bigl( A \times (B \times C) \Bigr)_{i,j} = \ \sum_{k = 1}^p \sum_{l = 1}^q \Bigl[ a_{i,k} \times b_{k,l} \times c_{l,j} \Bigr] \qquad (2) $$

   Therefore, they can be interverterd, and now \((1)\) and \((2)\) are equal and:

   $$\Bigl( (A \times B) \times C \Bigr)_{i,j} = \Bigl( A \times (B \times C) \Bigr)_{i,j} = \ \sum_{k = 1}^p \sum_{l = 1}^q \Bigl[ a_{i,l} \times b_{l,k} \times c_{k,j} \Bigr] $$

And finally,

Distributivity

1. Left distributivity

   Let \(A \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})\) be a matrix and \((B, C) \in \hspace{0.03em} \mathcal{M}_{p,q} (\mathbb{K})^2\) two matrices.

   With the definition of the product matrix , we do have:

   $$ \forall (i, j) \in [\![1, n]\!] \times [\![1, q]\!],$$
   $$\Bigl( A \times (B + C) \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[ a_{i,k} \times (b + c)_{k,j}\Bigr] $$
   $$\Bigl( A \times (B + C) \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[ a_{i,k} \times (b_{k,j} + c_{k,j})\Bigr]$$
   $$\Bigl( A \times (B + C) \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[ a_{i,k} \times b_{k,j} + a_{i,k} \times c_{k,j} \Bigr]$$
   $$\Bigl( A \times (B + C) \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[ a_{i,k} \times b_{k,j}\Bigr] + \sum_{k = 1}^p \Bigl[a_{i,k} \times c_{k,j}\Bigr] $$

   And finally,

   $$ \forall A \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K}) , \ \forall (B, C) \in \hspace{0.03em} \mathcal{M}_{p,q} (\mathbb{K})^2, $$
   $$ A \times (B + C) = A \times B + A \times C $$

2. Right distributivity

   Let \((A, B) \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})^2 \) be two matrices and \( C \in \hspace{0.03em} \mathcal{M}_{p,q} (\mathbb{K})\) an other matrix.

   As well as before:

   $$ \forall (i, j) \in [\![1, n]\!] \times [\![1, q]\!],$$
   $$\Bigl( (A + B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[(a + b)_{i,k} \times c_{k,j}\Bigr] $$
   $$\Bigl( (A + B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[(a_{i,k} + b_{i,k}) \times c_{k,j}\Bigr] $$
   $$\Bigl( (A + B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[a_{i,k} \times c_{k,j} + b_{i,k} \times c_{k,j}\Bigr] $$
   $$\Bigl( (A + B) \times C \Bigr)_{i,j} = \ \sum_{k = 1}^p \Bigl[a_{i,k} \times c_{k,j}\Bigr] + \sum_{k = 1}^p \Bigl[ b_{i,k} \times c_{k,j}\Bigr] $$

   And finally,

   $$ \forall (A,B) \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})^2 , \ \forall C \in \hspace{0.03em} \mathcal{M}_{p,q} (\mathbb{K}), $$
   $$ (A + B) \times C = A \times C + B \times C $$

Bilinearity

Let \(A \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})\) and \(B \in \hspace{0.03em} \mathcal{M}_{p,q} (\mathbb{K})\) be two matrices and \(\lambda \in \mathbb{R}\) a reel number.

With the definition of the product matrix , we do have:

In the same way:

And as a result,

Multiplication by the identity

Let \(A \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})\) be a matrix.

1. Computation of \(I_n \times A\)

   With the definition of the product matrix , we do have:

   $$ \forall (i, j) \in [\![1, n]\!] \times [\![1, p]\!],$$
   $$(I_n \times A)_{i,j} = \sum_{k = 1}^n (I_n)_{i,k} \times a_{k,j} $$

   But the factor \((I_n)_{i,k}\) is worth :

   $$ (I_n)_{i,k} = \Biggl \{ \begin{gather*} 1, \text{ if } (i = k) \\ 0 \ \text{otherwise} \end{gather*} $$

   So,

   $$(I_n \times A)_{i,j} = a_{i,j} = (A)_{i,j} $$

   It is the unchanged starting matrix.

2. Computation of \(A \times I_p\)

   Idem, on the other side:

   $$ \forall (i, j) \in [\![1, n]\!] \times [\![1, p]\!],$$
   $$(A \times I_p)_{i,j} = \sum_{k = 1}^p a_{i,k} \times (I_p)_{k,j} $$

   In the same way and for all \((i,j)\), in this sum of products, when \((k = j)\) we obtain \(a_{i,j}\) \since all other terms are worth \(0\) and therefore:

   $$(A \times I_p)_{i,j} = a_{i,j} = (A)_{i,j} $$

And as a result,

Diagonal product matrix

1. Product of two diagonal matrix

   Let \(\Bigl[ D_1 = diag(\lambda_1, \lambda_2, \ ..., \lambda_n), \ D_2 = diag(\mu_1, \mu_2, \ ..., \mu_n) \Bigr] \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})^2 \) be two diagonal matrix.

   With the definition of the product matrix , we do have:

   $$ \forall (i, j) \in [\![1, n]\!]^2,$$
   $$(D_1 \times D_2)_{i,j} = \sum_{k = 1}^n (d_1)_{i,k} \times (d_2)_{k,j} $$

   Retrieving the definition of a diagonal matrix , in each inner product of all these sums:

   $$ \Biggl \{ \begin{gather*} \forall (i, k) \in [\![1, n]\!]^2, \ (i \neq k) \Longrightarrow (d_1)_{i,k} = 0 \\ \forall (k, j) \in [\![1, n]\!]^2, \ (k \neq j) \Longrightarrow (d_2)_{k,j} = 0 \end{gather*} $$

   So, for any \(k\), the product \( \Bigl[ (d_1)_{i,k} \times (d_2)_{k,j} \Bigr] \neq 0 \) only if:

   $$ \Bigl[ (i = k) \land (k = j) \Bigr] \Longleftrightarrow (i = k = j) $$

   We then have:

   $$ \forall (i, j) \in [\![1, n]\!]^2, \ (D_1 \times D_2)_{i,j} = \Biggl \{ \begin{gather*} (d_1)_{i,j} \times (d_2)_{i,j}, \text{ si } (i = j) \\ 0 \ \text{otherwise} \end{gather*} $$

   So,

   $$ \forall (i, j) \in [\![1, n]\!]^2, \ (D_1 \times D_2)_{i,j} = \Biggl \{ \begin{gather*} (d_1)_{k,k} \times (d_2)_{k,k} = \lambda_k \ \mu_k, \text{ si } (i = j = k) \\ 0 \ \text{otherwise} \end{gather*} $$
   $$ (D_1 \times D_2) = \begin{pmatrix} \textcolor{rgb(118 139 240)}{\lambda_1 \ \mu_1} & 0 & 0 & \dots & 0 \\ 0 & \textcolor{rgb(118 139 240)}{\lambda_2 \ \mu_2} & 0 & \dots & 0 \\ 0 & 0 & \textcolor{rgb(118 139 240)}{\lambda_3 \ \mu_3} & \dots & 0 \\ \hspace{0.1em}\vdots & \hspace{0.1em} \vdots & \hspace{0.1em} \vdots & \textcolor{rgb(118 139 240)}{\ddots} & \hspace{0.1em} \vdots \\ 0 & 0 & 0 & \dots & \textcolor{rgb(118 139 240)}{\lambda_n \ \mu_n} \end{pmatrix} $$

   In the same way, if we perform the product on the other way round:

   $$ \forall (i, j) \in [\![1, n]\!]^2,$$
   $$(D_2 \times D_1)_{i,j} = \sum_{k = 1}^n (d_2)_{i,k} \times (d_1)_{k,j} $$

   This same reasoning leads us to the same result, that is to say that:

   $$ \forall (i, j) \in [\![1, n]\!]^2, \ (D_1 \times D_2)_{i,j} = \Biggl \{ \begin{gather*} (d_2)_{k,k} \times (d_1)_{k,k} = \mu_k \ \lambda_k, \text{ si } (i = j = k) \\ 0 \ \text{otherwise} \end{gather*} $$

   The product of numbers on the field \(\mathbb{K}\) having commutative law, both results of the products \((D_1 \times D_2)\) and \((D_2 \times D_1)\) are equal.

   And finally,

   $$ \forall \Bigl[ D_1 = diag(\lambda_1, \lambda_2, \ ..., \lambda_n), \ D_2 = diag(\mu_1, \mu_2, \ ..., \mu_n) \Bigr] \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})^2, $$
   $$ D_1 \times D_2 = D_2 \times D_1 = diag \left(\lambda_1 \mu_1, \lambda_2 \mu_2, \ ..., \lambda_n \mu_n \right) $$

2. A diagonal matrix raised to the power of \(n\)

   Let \(\Bigl[ D = diag(\lambda_1, \lambda_2, \ ..., \lambda_n) \Bigr] \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K}) \) be a diagonal matrix.

   Following the same reasoning as above, by a direct recurrence we do obtain as a result that:

   $$ \forall \Bigl[ D = diag(\lambda_1, \lambda_2, \ ..., \lambda_n) \Bigr] \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K}), $$
   $$ D^m = diag \left(\lambda_1^m, \lambda_2^m, \ ..., \lambda_n^m \right) $$

Link between matrix, comatrix, transposed and determinant

Let \(A \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})\) be a matrix.

1. Computation of \( \bigl[A \times com(A)^T \bigr]\)

   If we do perform the product \(A \times com(A)^T\), we obtain that:

   $$ \forall (i, j) \in [\![1, n]\!]^2,$$
   $$ \left(A \times com(A)^T \right)_{i,j} = \sum_{k = 1}^n a_{i,k} \times C_{j,k} $$
   • \( \alpha) \) if \( (i = j) \)

     In the case of \( (i = j) \), alors :

     $$ (i = j) \Longrightarrow \left[ \left(A \times com(A)^T \right)_{i,j} = \sum_{k = 1}^n a_{i,k} \times C_{i,k} \right] $$

     It is the definition of the determinant of \(A_{i,j}\) according to the line \(i\). thus:

     $$ \forall (i, j) \in [\![1, n]\!]^2, \ (i = j), $$
     $$ \left(A \times com(A)^T \right)_{i,j} = det(A) $$

     The product matrix will be then worth \(det(A)\) all along the diagonal.

   • \( \beta) \) if \( (i \neq j) \)

     In the case of \( (i \neq j) \), then by default:

     $$ (i \neq j) \Longrightarrow \left[ \left(A \times com(A)^T \right)_{i,j} = \sum_{k = 1}^n a_{i,k} \times C_{j,k} \right] \qquad (3) $$

     Let us imagine a new matrix \(B\), which is a copy of \(A\), and such as the \(j\) line be replaced by the \(i\) line.

     $$ \forall (i, j) \in [\![1, n]\!]^2,$$
     $$ A = \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} & \dots & a_{1, n} \\ a_{2,1} & a_{2,2} & a_{2,3} & \dots & a_{2,n} \\ \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \vdots & \hspace{0.5em} \vdots \\ \textcolor{rgb(93 183 129)}{a_{i,1}} & \textcolor{rgb(93 183 129)}{a_{i,2}} & \textcolor{rgb(93 183 129)}{a_{i,3}} & \textcolor{rgb(93 183 129)}{\dots} & \textcolor{rgb(93 183 129)}{a_{i, n}} \\ \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \vdots & \vdots \\ \textcolor{rgb(232 124 124)}{a_{j,1}} & \textcolor{rgb(232 124 124)}{a_{j,2}} & \textcolor{rgb(232 124 124)}{a_{j,3}} & \textcolor{rgb(232 124 124)}{\vdots} & \textcolor{rgb(232 124 124)}{a_{j, n}} \\ \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & a_{n,3} & \dots & a_{n, n} \end{pmatrix} $$

     $$ B = \begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} & \dots & a_{1, n} \\ a_{2,1} & a_{2,2} & a_{2,3} & \dots & a_{2,n} \\ \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \vdots & \hspace{0.5em} \vdots \\ \textcolor{rgb(93 183 129)}{a_{i,1}} & \textcolor{rgb(93 183 129)}{a_{i,2}} & \textcolor{rgb(93 183 129)}{a_{i,3}} & \textcolor{rgb(93 183 129)}{\dots} & \textcolor{rgb(93 183 129)}{a_{i, n}} \\ \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \vdots & \vdots \\ \textcolor{rgb(93 183 129)}{a_{i,1}} & \textcolor{rgb(93 183 129)}{a_{i,2}} & \textcolor{rgb(93 183 129)}{a_{i,3}} & \textcolor{rgb(93 183 129)}{\dots} & \textcolor{rgb(93 183 129)}{a_{i, n}} \\ \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \hspace{0.5em} \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & a_{n,3} & \dots & a_{n, n} \end{pmatrix} $$

     The aim is to compute the determinant of \(B\) on the line \(i\), where for all \(k\), all elements \(\textcolor{rgb(93 183 129)}{a_{i,k}}\) will replace the \(\textcolor{rgb(232 124 124)}{a_{j,k}}\) ones.

     In this way, if we do compute \(det(B)\) on the line\(i\), we obtain that:

     $$ det(B) = \sum_{k = 1}^n a_{i,k} \times C_{j,k} $$

     But, when we calculate any determinant of a matrix having two equal lines, this will worth zero.

     Which is obviously our case, hence:

     $$ det(B) = 0 \Longrightarrow \sum_{k = 1}^n a_{i,k} \times C_{j,k} = 0 \qquad (4) $$

     Then, giving an overview of both expressions \((3)\) and \((4)\):

     $$ \forall (i, j) \in [\![1, n]\!]^2, \ (i \neq j), $$
     $$ \left(A \times com(A)^T \right)_{i,j} = \sum_{k = 1}^n a_{i,k} \times C_{j,k} \qquad (3) $$

     $$ \sum_{k = 1}^n a_{i,k} \times C_{j,k} = 0 \qquad (4) $$

     This means that:

     $$ \forall (i, j) \in [\![1, n]\!]^2, \ (i \neq j), $$
     $$ \left(A \times com(A)^T \right)_{i,j} = 0 $$

     Our product matrix will be worth zero everywhere else.

     $$ A \times com(A)^T = \begin{pmatrix} \textcolor{rgb(118 139 240)}{det(A)} & 0 & 0 & \dots & 0 \\ 0 & \textcolor{rgb(118 139 240)}{det(A)} & 0 & \dots & 0 \\ 0 & 0 & \textcolor{rgb(118 139 240)}{det(A)} & \dots & 0 \\ \hspace{0.1em}\vdots & \hspace{0.1em} \vdots & \hspace{0.1em} \vdots & \textcolor{rgb(118 139 240)}{\ddots} & \hspace{0.1em} \vdots \\ 0 & 0 & 0 & \dots & \textcolor{rgb(118 139 240)}{det(A)} \end{pmatrix} $$

   And as a result:

   $$ A \times com(A)^T = det(A) \times I_n $$

2. Computation of \( \bigl[com(A)^T \times A\bigr]\)

   If we now commute both factors in the previous product such as:

   $$ \left(com(A)^T \times A \right)_{i,j} = \sum_{k = 1}^n C_{k,i} \times a_{k,j} $$

   Then, in the same way as before:

   • \( \alpha) \) if \( (i = j) \)

     In the case of \( (i = j) \), alors :

     $$ (i = j) \Longrightarrow \left[ \left(com(A)^T \times A \right)_{i,j} = \sum_{k = 1}^n C_{k,i} \times a_{k,i} \right] $$

     This is now the definition of the determinant of \(A_{i,j}\), but according to the column \(i\). Thus:

     $$ \forall (i, j) \in [\![1, n]\!]^2, \ (i = j), $$
     $$ \left(com(A)^T \times A \right)_{i,j} = det(A) $$

     As well as before, the product matrix will be worth \(det(A)\) all along the diagonal.

   • \( \beta) \) if \( (i \neq j) \)

     Again, in the cas of \( (i \neq j) \), then by default:

     $$ (i \neq j) \Longrightarrow \left[ \left(com(A)^T \times A \right)_{i,j} = \sum_{k = 1}^n C_{k,i} \times a_{k,j} \right] \qquad (3^*) $$

     And in the same way as before, we will now replace the \(j\) column by the \(i\) column in our new matrix \(B'\), and ths compute \(det(B')\) on the \(k\) line:

     $$ det(B') = 0 \Longrightarrow \sum_{k = 1}^n C_{k,i} \times a_{k,j} = 0 \qquad (4^*) $$

     But,

     $$ \forall (i, j) \in [\![1, n]\!]^2, \ (i \neq j), $$
     $$ \left(com(A)^T \times A \right)_{i,j} = \sum_{k = 1}^n C_{k,i} \times a_{k,j} \qquad (3^*) $$

     $$ \sum_{k = 1}^n C_{k,i} \times a_{k,j} = 0 \qquad (4^*) $$

     This means that:

     $$ \forall (i, j) \in [\![1, n]\!]^2, \ (i \neq j), $$
     $$ \left(com(A)^T \times A \right)_{i,j} = 0 $$

   And as well as above, the product matrix will be worth zero everywhere else.

   $$ com(A)^T \times A = \begin{pmatrix} \textcolor{rgb(118 139 240)}{det(A)} & 0 & 0 & \dots & 0 \\ 0 & \textcolor{rgb(118 139 240)}{det(A)} & 0 & \dots & 0 \\ 0 & 0 & \textcolor{rgb(118 139 240)}{det(A)} & \dots & 0 \\ \hspace{0.1em}\vdots & \hspace{0.1em} \vdots & \hspace{0.1em} \vdots & \textcolor{rgb(118 139 240)}{\ddots} & \hspace{0.1em} \vdots \\ 0 & 0 & 0 & \dots & \textcolor{rgb(118 139 240)}{det(A)} \end{pmatrix} $$

   And as a result:

   $$ com(A)^T \times A = det(A) \times I_n $$

We finally proved that :

Linearity of transposition

Let \((A,B) \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})^2\) be two matrices of the same size and \((\lambda, \mu) \in \hspace{0.04em} \mathbb{R}^2\) two real numbers.

We saw with the multiplication of a matrix by a scalar all its elements were affected.

Moreover, the matrix sum (of the same size) is the addition of elements of both matrix having index \((i,j)\) together.

Now, with these two properties, we can build a linear combination such as:

Now taking its transposed matrix , in reverses all \(i\) and \(j\) index:

And as a result,

Transposed of a product

Let \(A \in \hspace{0.03em} \mathcal{M}_{n,p} (\mathbb{K})\) and \(B \in \hspace{0.03em} \mathcal{M}_{p,q} (\mathbb{K})\) be two matrices.

With the definition of the product matrix , we do have:

Now taking its transposed matrix , in reverses all \(i\) and \(j\) index:

But, the product of transposed is worth the same value:

So,

Computation of the inverse

Starting from the following expression \((5)\) :

By taking only the right part of the expression, we do have this:

Then, by multiplying each side by \(A^{-1}\):

Finally, by dividing each memmber by \(det(A)\), we do obtain tahat:

And finally,

Inverse of the inverse

Let \(A \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})\) a squared matrix of size \(n\).

Si, if \(A\) is inversible, then \(A^{-1}\) also is. We now have the following relation:

But also :

By multiplying each member of this expression by \(A\) from the left, we do obtain:

And finally,

Inverse of a transposed matrix

Let \(A \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})\) a squared matrix of size \(n\).

Matrix\(A\) is inversible if and only if \(det(A) \neq 0\). But both matrix \(A\) and \(A^T\) have the same determinant:

Then, if \(A\) is inversible, then \(A^T\) also is.

So, applied to our case:

Now, the transposed matrix of the matrix identity is invariant. Therefore:

By multiplying each member of this expression by \(\left(A^T \right)^{-1}\) from the right, we do obtain:

And as a result,

Inverse of a product

Let \((A,B) \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})\) be two squared matrices of the same size \(n\).

If both matrix \(A\) and \(B\) are inversible, then:

Combining both expressions \((7)\) and \((8)\), we now have:

Therefore, the product \((A \times B)\) is also inversible and:

By multiplying each member of this expression by \((B^{-1} A^{-1})\) from the left, we do obtain:

So finally,

Both expressions \((9)\) and \((10)\) have the same behaviour:

So, the order of transposition or inversion have no importance.

We deduce of it that:

Linearity of the trace

Let \((A,B) \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})^2\) be two squared matrices of the same size \(n\) and \((\lambda, \mu) \in \hspace{0.04em} \mathbb{R}^2\) two real numbers.

By the definition of the trace and by linear combination , we do have this:

So, we directly obtain two sums:

And finally,

Trace of a product

Let \((A,B) \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})^2\) be two squared matrices of the same size \(n\).

By the definition of the trace , we do have:

So, replacing it by its value in the previous expression:

Variables \(k\) and \(l\) are free variables, therefore they can be switched:

The product of scalar on the filed \(\mathbb{K}\) being commutative, we can transform it into:

But, this is the same thing as \(Tr(B \times A)\), is we look at the expression \((10)\) and switching \(A\) and \(B\) positions.

As a result we do obtain,

Powers of the matrix of ones

Let \(J_n\) bet the matrix of ones of size \(n\).

Starting from the operation \((J_n)^2\):

Therefore, the resulting matrix of the same size is \(n\) everywhere:

Likewise, if we now take the cube of \(J_n\), we do have now:

Thus,

Therefore, it can easily be shown by a direct recurrence that:

Newton's binomial and Geometrical identity

Let \((A,B) \in \hspace{0.03em} \mathcal{M}_{n}(\mathbb{K})^2\) be two matrices of the same size \(n\) that commutes .

We know that distributivity does apply on matrices. Thus:

Moreover, in our case of matrices \(A\) and \(B\) are commuting, so: \(\bigl(AB = BA \bigr)\).

Therefore, we do have:

So, we can prove by a recurrence that the Newton's binomial and the Geometrical identity do also apply in this specific case.