# Nonlinear finite elements/Matrices

Much of finite elements revolves around forming matrices and solving systems of linear equations using matrices. This learning resource gives you a brief review of matrices.

## Matrices

Suppose that you have a linear system of equations
    ${\begin{aligned}a_{11}x_{1}+a_{12}x_{2}+a_{13}x_{3}+a_{14}x_{4}&=b_{1}\\a_{21}x_{1}+a_{22}x_{2}+a_{23}x_{3}+a_{24}x_{4}&=b_{2}\\a_{31}x_{1}+a_{32}x_{2}+a_{33}x_{3}+a_{34}x_{4}&=b_{3}\\a_{41}x_{1}+a_{42}x_{2}+a_{43}x_{3}+a_{44}x_{4}&=b_{4}\end{aligned}}~.$
Matrices provide a simple way of expressing these equations. Thus, we can instead write
    ${\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\a_{41}&a_{42}&a_{43}&a_{44}\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}={\begin{bmatrix}b_{1}\\b_{2}\\b_{3}\\b_{4}\end{bmatrix}}~.$
An even more compact notation is
    $\left[{\mathsf {A}}\right]\left[{\mathsf {x}}\right]=\left[{\mathsf {b}}\right]~~~~{\text{or}}~~~~\mathbf {A} \mathbf {x} =\mathbf {b} ~.$
Here $\mathbf {A}$ is a $4\times 4$ matrix while $\mathbf {x}$ and $\mathbf {b}$ are $4\times 1$ matrices. In general, an $m\times n$ matrix $\mathbf {A}$ is a set of numbers arranged in $m$ rows and $n$ columns.
    $\mathbf {A} ={\begin{bmatrix}a_{11}&a_{12}&a_{13}&\dots &a_{1n}\\a_{21}&a_{22}&a_{23}&\dots &a_{2n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&a_{m3}&\dots &a_{mn}\end{bmatrix}}~.$

## Practice Exercises

Practice: Expressing Linear Equations As Matrices

## Types of Matrices

Common types of matrices that we encounter in finite elements are:
* a row vector that has one row and $n$ columns.
    $\mathbf {v} ={\begin{bmatrix}v_{1}&v_{2}&v_{3}&\dots &v_{n}\end{bmatrix}}$
* a column vector that has $n$ rows and one column.
    $\mathbf {v} ={\begin{bmatrix}v_{1}\\v_{2}\\v_{3}\\\vdots \\v_{n}\end{bmatrix}}$
* a square matrix that has an equal number of rows and columns.
* a diagonal matrix which is a square matrix with only the
diagonal elements ($a_{ii}$) nonzero.
    $\mathbf {A} ={\begin{bmatrix}a_{11}&0&0&\dots &0\\0&a_{22}&0&\dots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\dots &a_{nn}\end{bmatrix}}~.$
* the identity matrix ($\mathbf {I}$) which is a diagonal matrix and
with each of its nonzero elements ($a_{ii}$) equal to 1.
    $\mathbf {A} ={\begin{bmatrix}1&0&0&\dots &0\\0&1&0&\dots &0\\\vdots &\vdots &\vdots &\ddots &\vdots \\0&0&0&\dots &1\end{bmatrix}}~.$
* a symmetric matrix which is a square matrix with elements
such that $a_{ij}=a_{ji}$.
    $\mathbf {A} ={\begin{bmatrix}a_{11}&a_{12}&a_{13}&\dots &a_{1n}\\a_{12}&a_{22}&a_{23}&\dots &a_{2n}\\a_{13}&a_{23}&a_{33}&\dots &a_{3n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{1n}&a_{2n}&a_{3n}&\dots &a_{nn}\end{bmatrix}}~.$
* a skew-symmetric matrix which is a square matrix with elements
such that $a_{ij}=-a_{ji}$.
    $\mathbf {A} ={\begin{bmatrix}a_{11}&a_{12}&a_{13}&\dots &a_{1n}\\-a_{12}&a_{22}&a_{23}&\dots &a_{2n}\\-a_{13}&-a_{23}&a_{33}&\dots &a_{3n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\-a_{1n}&-a_{2n}&-a_{3n}&\dots &a_{nn}\end{bmatrix}}~.$
Note that the diagonal elements of a skew-symmetric matrix have to be zero: $a_{ii}=-a_{ii}\Rightarrow a_{ii}=0$.

## Matrix addition

Let $\mathbf {A}$ and $\mathbf {B}$ be two $m\times n$ matrices with components $a_{ij}$ and $b_{ij}$, respectively. Then
    $\mathbf {C} =\mathbf {A} +\mathbf {B} \implies c_{ij}=a_{ij}+b_{ij}$

## Multiplication by a scalar

Let $\mathbf {A}$ be a $m\times n$ matrix with components $a_{ij}$ and let $\lambda$ be a scalar quantity. Then,
    $\mathbf {C} =\lambda \mathbf {A} \implies c_{ij}=\lambda a_{ij}$

## Multiplication of matrices

Let $\mathbf {A}$ be a $m\times n$ matrix with components $a_{ij}$. Let $\mathbf {B}$ be a $p\times q$ matrix with components $b_{ij}$.
The product $\mathbf {C} =\mathbf {A} \mathbf {B}$ is defined only if $n=p$. The matrix $\mathbf {C}$ is a $m\times q$ matrix with components $c_{ij}$. Thus,
    $\mathbf {C} =\mathbf {A} \mathbf {B} \implies c_{ij}=\sum _{k=1}^{n}a_{ik}b_{kj}$
Similarly, the product $\mathbf {D} =\mathbf {B} \mathbf {A}$ is defined only if $q=m$. The matrix $\mathbf {D}$ is a $p\times n$ matrix with components $d_{ij}$. We have
    $\mathbf {D} =\mathbf {B} \mathbf {A} \implies d_{ij}=\sum _{k=1}^{m}b_{ik}a_{kj}$
Clearly, $\mathbf {C} \neq \mathbf {D}$ in general, i.e., the matrix product is not commutative.
However, matrix multiplication is distributive. That means
    $\mathbf {A} (\mathbf {B} +\mathbf {C} )=\mathbf {A} \mathbf {B} +\mathbf {A} \mathbf {C} ~.$
The product is also associative. That means
    $\mathbf {A} (\mathbf {B} \mathbf {C} )=(\mathbf {A} \mathbf {B} )\mathbf {C} ~.$

## Transpose of a matrix

Let $\mathbf {A}$ be a $m\times n$ matrix with components $a_{ij}$. Then the transpose of the matrix is defined as the $n\times m$ matrix $\mathbf {B} =\mathbf {A} ^{T}$ with components $b_{ij}=a_{ji}$. That is,
    $\mathbf {B} =\mathbf {A} ^{T}={\begin{bmatrix}a_{11}&a_{12}&a_{13}&\dots &a_{1n}\\a_{21}&a_{22}&a_{23}&\dots &a_{2n}\\a_{31}&a_{32}&a_{33}&\dots &a_{3n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&a_{m3}&\dots &a_{mn}\end{bmatrix}}^{T}={\begin{bmatrix}a_{11}&a_{21}&a_{31}&\dots &a_{m1}\\a_{12}&a_{22}&a_{32}&\dots &a_{m2}\\a_{13}&a_{23}&a_{33}&\dots &a_{m3}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{1n}&a_{2n}&a_{3n}&\dots &a_{mn}\end{bmatrix}}$
An important identity involving the transpose of matrices is
    ${(\mathbf {A} \mathbf {B} )^{T}=\mathbf {B} ^{T}\mathbf {A} ^{T}}~.$

## Determinant of a matrix

The determinant of a matrix is defined only for square matrices.
For a $2\times 2$ matrix $\mathbf {A}$, we have
    $\mathbf {A} ={\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}}\implies \det(\mathbf {A} )={\begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}}=a_{11}a_{22}-a_{12}a_{21}~.$
For a $n\times n$ matrix, the determinant is calculated by expanding into minors as
    ${\begin{aligned}&\det(\mathbf {A} )={\begin{vmatrix}a_{11}&a_{12}&a_{13}&\dots &a_{1n}\\a_{21}&a_{22}&a_{23}&\dots &a_{2n}\\a_{31}&a_{32}&a_{33}&\dots &a_{3n}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&a_{n3}&\dots &a_{nn}\end{vmatrix}}\\&=a_{11}{\begin{vmatrix}a_{22}&a_{23}&\dots &a_{2n}\\a_{32}&a_{33}&\dots &a_{3n}\\\vdots &\vdots &\ddots &\vdots \\a_{n2}&a_{n3}&\dots &a_{nn}\end{vmatrix}}-a_{12}{\begin{vmatrix}a_{21}&a_{23}&\dots &a_{2n}\\a_{31}&a_{33}&\dots &a_{3n}\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n3}&\dots &a_{nn}\end{vmatrix}}+\dots \pm a_{1n}{\begin{vmatrix}a_{21}&a_{22}&\dots &a_{2(n-1)}\\a_{31}&a_{32}&\dots &a_{3(n-1)}\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&\dots &a_{n(n-1)}\end{vmatrix}}\end{aligned}}$
In short, the determinant of a matrix $\mathbf {A}$ has the value
    ${\det(\mathbf {A} )=\sum _{j=1}^{n}(-1)^{1+j}a_{1j}M_{1j}}$
where $M_{ij}$ is the determinant of the submatrix of $\mathbf {A}$ formed by eliminating row $i$ and column $j$ from $\mathbf {A}$.
Some useful identities involving the determinant are given below.
* If $\mathbf {A}$ is a $n\times n$ matrix, then
    $\det(\mathbf {A} )=\det(\mathbf {A} ^{T})~.$
* If $\lambda$ is a constant and $\mathbf {A}$ is a $n\times n$ matrix, then
    $\det(\lambda \mathbf {A} )=\lambda ^{n}\det(\mathbf {A} )\implies \det(-\mathbf {A} )=(-1)^{n}\det(\mathbf {A} )~.$
* If $\mathbf {A}$ and $\mathbf {B}$ are two $n\times n$ matrices, then
    $\det(\mathbf {A} \mathbf {B} )=\det(\mathbf {A} )\det(\mathbf {B} )~.$
If you think you understand determinants, take the quiz.

## Inverse of a matrix

Let $\mathbf {A}$ be a $n\times n$ matrix. The inverse of $\mathbf {A}$ is denoted by $\mathbf {A} ^{-1}$ and is defined such that
    ${\mathbf {A} \mathbf {A} ^{-1}=\mathbf {I} }$
where $\mathbf {I}$ is the $n\times n$ identity matrix.
The inverse exists only if $\det(\mathbf {A} )\neq 0$. A singular matrix does not have an inverse.
An important identity involving the inverse is
    ${(\mathbf {A} \mathbf {B} )^{-1}=\mathbf {B} ^{-1}\mathbf {A} ^{-1},}$
since this leads to: ${(\mathbf {A} \mathbf {B} )^{-1}(\mathbf {A} \mathbf {B} )=(\mathbf {B} ^{-1}\mathbf {A} ^{-1})(\mathbf {A} \mathbf {B} )=\mathbf {B} ^{-1}\mathbf {A} ^{-1}\mathbf {A} \mathbf {B} =\mathbf {B} ^{-1}(\mathbf {A} ^{-1}\mathbf {A} )\mathbf {B} =\mathbf {B} ^{-1}\mathbf {I} \mathbf {B} =\mathbf {B} ^{-1}\mathbf {B} =\mathbf {I} .}$
Some other identities involving the inverse of a matrix are given below.
* The determinant of a matrix is equal to the multiplicative inverse of the
determinant of its inverse.
    $\det(\mathbf {A} )={\cfrac {1}{\det(\mathbf {A} ^{-1})}}~.$
* The determinant of a similarity transformation of a matrix
is equal to the original matrix.
    $\det(\mathbf {B} \mathbf {A} \mathbf {B} ^{-1})=\det(\mathbf {A} )~.$
We usually use numerical methods such as Gaussian elimination to compute the inverse of a matrix.

## Eigenvalues and eigenvectors

A thorough explanation of this material can be found at Eigenvalue, eigenvector and eigenspace. However, for further study, let us consider the following examples:
* Let :$\mathbf {A} ={\begin{bmatrix}1&6\\5&2\end{bmatrix}},\mathbf {v} ={\begin{bmatrix}6\\-5\end{bmatrix}},\mathbf {t} ={\begin{bmatrix}7\\4\end{bmatrix}}~.$
Which vector is an eigenvector for $\mathbf {A}$ ?
We have $\mathbf {A} \mathbf {v} ={\begin{bmatrix}1&6\\5&2\end{bmatrix}}{\begin{bmatrix}6\\-5\end{bmatrix}}={\begin{bmatrix}-24\\20\end{bmatrix}}=-4{\begin{bmatrix}6\\-5\end{bmatrix}}$ , and $\mathbf {A} \mathbf {t} ={\begin{bmatrix}1&6\\5&2\end{bmatrix}}{\begin{bmatrix}7\\4\end{bmatrix}}={\begin{bmatrix}31\\43\end{bmatrix}}~.$
Thus, $\mathbf {v}$ is an eigenvector.
* Is $\mathbf {u} ={\begin{bmatrix}1\\4\end{bmatrix}}$ an eigenvector for $\mathbf {A} ={\begin{bmatrix}-3&-3\\1&8\end{bmatrix}}$ ?
We have that since $\mathbf {A} \mathbf {u} ={\begin{bmatrix}-3&-3\\1&8\end{bmatrix}}{\begin{bmatrix}1\\4\end{bmatrix}}={\begin{bmatrix}-15\\33\end{bmatrix}}$ , $\mathbf {u} ={\begin{bmatrix}1\\4\end{bmatrix}}$ is not an eigenvector for $\mathbf {A} ={\begin{bmatrix}-3&-3\\1&8\end{bmatrix}}~.$