## Calculus: Matrix Calculus

In this section, we show some useful formulas that are used in the derivation of the forms of the stress matrices of hyperelastic materials.

### Statement 1:

Let be a deformation gradient satisfying . Let be the square of the right stretch tensor of the polar decomposition of the deformation gradient. Let and be the first and second invariants of respectively, which for simplicity, will be denoted by and . Then:

Note that the first two equalities hold for any matrix . The last formula is referred to as the Jacobi formula.

##### Proof:

For the first statement, we start by noting that the first invariant of can be written using the components of as follows:

By taking the derivative of with respect to the arbitrary component we get:

Therefore,

The second inequality can be obtained similarly, by writing in terms of the components of :

Therefore:

The second term can be evaluated using component form as follows:

i.e.,

Therefore, the derivative of the second invariant of has the following form:

The Jacobi formula can be shown by first writing the expression of the determinant of a matrix in component form as follows:

If is invertible, which is the case when represents a deformation gradient, then it is enough to show the following equivalent form of the Jacobi formula:

To prove the above, we start by the left hand side in component form:

Each term of the above equation represents the triple product of three row vectors in . Since the triple product of linearly dependent vectors is zero, the three terms can be rewritten as follows:

Substituting in the above equation leads to:

Therefore,

### Statement 2 (Component form):

Let be a deformation gradient satisfying . Let be the square of the right stretch tensor of the polar decomposition of the deformation gradient. Let be the singular value decomposition of the deformation gradient where are rotation matrices, the columns of are the eigenvectors of and is a diagonal matrix with entries:

Then:

##### Proof:

Rearranging the singular value decomposition equality:

Therefore, using component form we have:

By taking the derivatives with respect to the general component :

Similarly,

### Statement 2 (Tensor Form):

In the previous statement, the derivatives of the eigenvectors with respect to the components of were shown to be functions of the components of and . Since the columns of are the eigenvectors of then the derivatives are functions of the components of the eigenvectors. The complexity of the component form hides the simplicity of the actual relationship and so the tensor product form will be used to show how simple the relationship actually is. In addition, we are going to also show the simple relationship of the derivatives of the eigenvectors of with respect to the tensor as follows:

Let be the Right Cauchy-Green Deformation Tensor and be the Left Cauchy-Green Deformation Tensor. Let , and be the eigenvalues of and . Let , , and be the eigenvectors of and , , and be the eigenvectors of . Show that (no summation):

##### Proof:

First, assume that is a scalar valued function of . Therefore:

By taking the derivative of with respect to , we get the second order tensor:

Setting we have:

Since the eigenvalues of are independent of each other and of the eigenvectors, we have:

Therefore:

The same statement applies for and . Therefore :

The same proof can be used to show that :

#### Equivalence of the component and tensor forms of statement 2:

Note that since the columns of in the statement above are the eigenvectors of , . Also, we have . Also, . Therefore:

which is the component form obtained above.

### Statement 3:

Let be a deformation gradient satisfying . Let be the square of the right stretch tensor of the polar decomposition of the deformation gradient. Let . Let be the first and second invariants of respectively.

Then:

Where are the first and second invariants of respectively.

##### Proof:

From the properties of the first invariant of a matrix:

Therefore:

Similarly:

Therefore: