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

$\begin{array}{cc}{a}_{11}{x}_1+a_{12}{x}_2+a_{13}{x}_3+a_{14}{x}_4& =b_1\\ a_{21}{x}_1+a_{22}{x}_2+a_{23}{x}_3+a_{24}{x}_4& =b_2\\ a_{31}{x}_1+a_{32}{x}_2+a_{33}{x}_3+a_{34}{x}_4& =b_3\\ a_{41}{x}_1+a_{42}{x}_2+a_{43}{x}_3+a_{44}{x}_4& =b_4\end{array}.$

Matrices provide a simple way of expressing these equations. Thus, we can instead write

$\left[\begin{array}{cccc}{a}_{11}& a_{12}& a_{13}& a_{14}\\ a_{21}& a_{22}& a_{23}& a_{24}\\ a_{31}& a_{32}& a_{33}& a_{34}\\ a_{41}& a_{42}& a_{43}& a_{44}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}{b}_1\\ b_2\\ b_3\\ b_4\end{array}\right].$

An even more compact notation is

$\left[𝖠\right]\left[𝗑\right]=\left[𝖻\right]\text{or}𝐀𝐱=𝐛.$

Here $𝐀$ is a $4\times 4$ matrix while $𝐱$ and $𝐛$ are $4\times 1$ matrices. In general, an $m\times n$ matrix $𝐀$ is a set of numbers arranged in $m$ rows and $n$ columns.

$𝐀=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a_{m1}& a_{m2}& a_{m3}& \dots & a_{mn}\end{array}\right].$

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.

$𝐯=\left[\begin{array}{ccccc}{v}_1& v_2& v_3& \dots & v_n\end{array}\right]$

• a column vector that has $n$ rows and one column.

$𝐯=\left[\begin{array}{c}{v}_1\\ v_2\\ v_3\\ ⋮\\ v_n\end{array}\right]$

• 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.

$𝐀=\left[\begin{array}{ccccc}{a}_{11}& 0& 0& \dots & 0\\ 0& a_{22}& 0& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & a_{nn}\end{array}\right].$

• the identity matrix ($𝐈$) which is a diagonal matrix and

with each of its nonzero elements ($a_{ii}$) equal to 1.

$𝐀=\left[\begin{array}{ccccc}1& 0& 0& \dots & 0\\ 0& 1& 0& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& \dots & 1\end{array}\right].$

• a symmetric matrix which is a square matrix with elements

such that $a_{ij}=a_{ji}$.

$𝐀=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{12}& a_{22}& a_{23}& \dots & a_{2n}\\ a_{13}& a_{23}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a_{1n}& a_{2n}& a_{3n}& \dots & a_{nn}\end{array}\right].$

• a skew-symmetric matrix which is a square matrix with elements

such that $a_{ij}=-a_{ji}$.

$𝐀=\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ -a_{12}& a_{22}& a_{23}& \dots & a_{2n}\\ -a_{13}& -a_{23}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& -a_{2n}& -a_{3n}& \dots & a_{nn}\end{array}\right].$

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 $𝐀$ and $𝐁$ be two $m\times n$ matrices with components $a_{ij}$ and $b_{ij}$, respectively. Then

$𝐂=𝐀+𝐁⟹c_{ij}=a_{ij}+b_{ij}$

Multiplication by a scalar

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

$𝐂=\lambda 𝐀⟹c_{ij}=\lambda a_{ij}$

Multiplication of matrices

Let $𝐀$ be a $m\times n$ matrix with components $a_{ij}$. Let $𝐁$ be a $p\times q$ matrix with components $b_{ij}$.

The product $𝐂=𝐀𝐁$ is defined only if $n=p$. The matrix $𝐂$ is a $m\times q$ matrix with components $c_{ij}$. Thus,

$𝐂=𝐀𝐁⟹c_{ij}=\sum _{k=1}^n{a}_{ik}{b}_{kj}$

Similarly, the product $𝐃=𝐁𝐀$ is defined only if $q=m$. The matrix $𝐃$ is a $p\times n$ matrix with components $d_{ij}$. We have

$𝐃=𝐁𝐀⟹d_{ij}=\sum _{k=1}^m{b}_{ik}{a}_{kj}$

Clearly, $𝐂\ne 𝐃$ in general, i.e., the matrix product is not commutative.

However, matrix multiplication is distributive. That means

$𝐀(𝐁+𝐂)=𝐀𝐁+𝐀𝐂.$

The product is also associative. That means

$𝐀(𝐁𝐂)=(𝐀𝐁)𝐂.$

Transpose of a matrix

Let $𝐀$ be a $m\times n$ matrix with components $a_{ij}$. Then the transpose of the matrix is defined as the $n\times m$ matrix $𝐁={𝐀}^T$ with components $b_{ij}=a_{ji}$. That is,

$𝐁={𝐀}^T={\left[\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ a_{31}& a_{32}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a_{m1}& a_{m2}& a_{m3}& \dots & a_{mn}\end{array}\right]}^T=\left[\begin{array}{ccccc}{a}_{11}& a_{21}& a_{31}& \dots & a_{m1}\\ a_{12}& a_{22}& a_{32}& \dots & a_{m2}\\ a_{13}& a_{23}& a_{33}& \dots & a_{m3}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a_{1n}& a_{2n}& a_{3n}& \dots & a_{mn}\end{array}\right]$

An important identity involving the transpose of matrices is

$(𝐀𝐁{)}^T={𝐁}^T{𝐀}^T.$

Determinant of a matrix

The determinant of a matrix is defined only for square matrices.

For a $2\times 2$ matrix $𝐀$, we have

$𝐀=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]⟹\det (𝐀)=\left|\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right|=a_{11}{a}_{22}-a_{12}{a}_{21}.$

For a $n\times n$ matrix, the determinant is calculated by expanding into minors as

$\begin{array}{cc}& \det (𝐀)=\left|\begin{array}{ccccc}{a}_{11}& a_{12}& a_{13}& \dots & a_{1n}\\ a_{21}& a_{22}& a_{23}& \dots & a_{2n}\\ a_{31}& a_{32}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right|\\ & =a_{11}\left|\begin{array}{cccc}{a}_{22}& a_{23}& \dots & a_{2n}\\ a_{32}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n2}& a_{n3}& \dots & a_{nn}\end{array}\right|-a_{12}\left|\begin{array}{cccc}{a}_{21}& a_{23}& \dots & a_{2n}\\ a_{31}& a_{33}& \dots & a_{3n}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n3}& \dots & a_{nn}\end{array}\right|+\dots \pm a_{1n}\left|\begin{array}{cccc}{a}_{21}& a_{22}& \dots & a_{2(n-1)}\\ a_{31}& a_{32}& \dots & a_{3(n-1)}\\ ⋮& ⋮& \ddots & ⋮\\ a_{n1}& a_{n2}& \dots & a_{n(n-1)}\end{array}\right|\end{array}$

In short, the determinant of a matrix $𝐀$ has the value

$\det (𝐀)=\sum _{j=1}^n(-1{)}^{1+j}{a}_{1j}{M}_{1j}$

where $M_{ij}$ is the determinant of the submatrix of $𝐀$ formed by eliminating row $i$ and column $j$ from $𝐀$.

Some useful identities involving the determinant are given below.

• If $𝐀$ is a $n\times n$ matrix, then

$\det (𝐀)=\det ({𝐀}^T).$

• If $\lambda$ is a constant and $𝐀$ is a $n\times n$ matrix, then

$\det (\lambda 𝐀)={\lambda }^n\det (𝐀)⟹\det (-𝐀)=(-1{)}^n\det (𝐀).$

• If $𝐀$ and $𝐁$ are two $n\times n$ matrices, then

$\det (𝐀𝐁)=\det (𝐀)\det (𝐁).$

If you think you understand determinants, take the quiz.

Inverse of a matrix

Let $𝐀$ be a $n\times n$ matrix. The inverse of $𝐀$ is denoted by ${𝐀}^{-1}$ and is defined such that

$𝐀{𝐀}^{-1}=𝐈$

where $𝐈$ is the $n\times n$ identity matrix.

The inverse exists only if $\det (𝐀)\ne 0$. A singular matrix does not have an inverse.

An important identity involving the inverse is

$(𝐀𝐁{)}^{-1}={𝐁}^{-1}{𝐀}^{-1},$

since this leads to: $(𝐀𝐁{)}^{-1}(𝐀𝐁)=({𝐁}^{-1}{𝐀}^{-1})(𝐀𝐁)={𝐁}^{-1}{𝐀}^{-1}𝐀𝐁={𝐁}^{-1}({𝐀}^{-1}𝐀)𝐁={𝐁}^{-1}𝐈𝐁={𝐁}^{-1}𝐁=𝐈.$

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.

$\det (𝐀)=\frac{1}{\det ({𝐀}^{-1})}.$

• The determinant of a similarity transformation of a matrix

is equal to the original matrix.

$\det (𝐁𝐀{𝐁}^{-1})=\det (𝐀).$

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 :$𝐀=\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right],𝐯=\left[\begin{array}{c}6\\ -5\end{array}\right],𝐭=\left[\begin{array}{c}7\\ 4\end{array}\right].$

Which vector is an eigenvector for $𝐀$ ?

We have $𝐀𝐯=\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right]\left[\begin{array}{c}6\\ -5\end{array}\right]=\left[\begin{array}{c}-24\\ 20\end{array}\right]=-4\left[\begin{array}{c}6\\ -5\end{array}\right]$ , and $𝐀𝐭=\left[\begin{array}{cc}1& 6\\ 5& 2\end{array}\right]\left[\begin{array}{c}7\\ 4\end{array}\right]=\left[\begin{array}{c}31\\ 43\end{array}\right].$

Thus, $𝐯$ is an eigenvector.

• Is $𝐮=\left[\begin{array}{c}1\\ 4\end{array}\right]$ an eigenvector for $𝐀=\left[\begin{array}{cc}-3& -3\\ 1& 8\end{array}\right]$ ?

We have that since $𝐀𝐮=\left[\begin{array}{cc}-3& -3\\ 1& 8\end{array}\right]\left[\begin{array}{c}1\\ 4\end{array}\right]=\left[\begin{array}{c}-15\\ 33\end{array}\right]$ , $𝐮=\left[\begin{array}{c}1\\ 4\end{array}\right]$ is not an eigenvector for $𝐀=\left[\begin{array}{cc}-3& -3\\ 1& 8\end{array}\right].$

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