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 $\mathbf {A} ={\begin{bmatrix}-3&-3\\1&8\end{bmatrix}}~.$

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