All About Inverses

A Concise Primer On Linear Algebra

Onkur Sen


What is a matrix?

\( A \in \mathcal{R}^{m \times n}, A = a_{ij} \)

\(m\) = number of rows
\(n\) = number of columns
\(m=n \Rightarrow \) "square matrix"


If \(A\) and \(B\) are the same size,
then add component by component.

\( A, B \in \mathcal{R}^{m \times n}\)
\(\Rightarrow (A+B)_{ij} = A_{ij} + B_{ij} \)

Properties of Addition

It is associative.
\( (A+B)+C = A+(B+C) \)

It is commutative.
\( A+B = B+A \)

The identity for addition is
the matrix of all zeros (written as \(0\)).
\( 0_{ij} = 0 \; \forall i, j \)
\( A+0 = 0+A = A \)


If \(A\) has the same number of columns as \(B\) has rows,
then we can multiply them.

\( A\in \mathcal{R}^{m \times n}, B\in \mathcal{R}^{n \times p} \Rightarrow AB\in \mathcal{R}^{m \times p}\)
\( (AB)_{ij} = \sum_{k=1}^n a_{ik} b_{kj} \)

This is the dot product
of the \(i\)th row of \(A\)
with the \(j\)th column of \(B\).

Properties of Multiplication

It is associative.
\((AB)C\) = \(A(BC)\)

It is not necessarily commutative.
\(AB \stackrel{?}{=} BA\)

The identity for multiplication is the matrix
with \(1\)s on the diagonals and zero otherwise.
It is written as \(I\).
\( I_{ij} = 1 \mbox{ if } i=j, 0 \mbox{ otherwise} \)
\( IA = AI = A \)


The transpose of a matrix \(A\) is another matrix
with the rows and columns of \(A\) inverted.
It is written as \( A^T \).

\( A^T_{ij} = A_{ji} \)

Note: \( (A^T)^T = A \)

Inverse of a Matrix

What is it?

The inverse of a square matrix \(A\) is a matrix that
multiplies with \(A\) to make the identity matrix.
It is written as \(A^{-1}\).

\( AA^{-1} = A^{-1}A = I \)

BUT it does not exist for all \(A\).

Why is it important?

Invertibility comes up again and again in linear algebra.
We'll examine four different use cases.

Row Operations

(and inverses)

Elementary Row Operations

There are only three!

  1. Switch two rows: \( r_i, r_j \rightarrow r_j, r_i\)
  2. Multiply a row by a constant: \( r_i \rightarrow Cr_i\)
  3. Add two rows together: \( r_i \rightarrow r_i + r_j\)

You can also do 2 and 3 at the same time:
\( r_i \rightarrow C_1 r_i + C_2 r_j\)

Can you invert a matrix? (Part 1)

\(A\) is invertible if
it can be reduced to \(I\)
using elementary row operations.

\(A^{-1}\) is the result of
the same operations applied to \(I\).

The method to do this
on the augmented matrix \( [A \; \vert \; I]\)
is called Gaussian elimination.

Intuition for inverting Part 1

By going from \(A\) to \(I\) with row operations,
we basically multiply by \(A^{-1}\) (since \(AA^{-1}=I\)).

Furthermore, since \(IA^{-1} = A^{-1}\),
we can do the same operations on \(I\)
to get \(A^{-1}\) explicitly.

Linear independence

(and inverses)


A vector \(v\) is a matrix with one column.

\( v \in R^n \)

Linear independence

A set of vectors \(\{v_i\}_{i=1}^n\) is linearly independent if
the only linear combination equal to zero
has all coefficients equal to zero.

\( \{v_i\}_{i=1}^n \) linearly independent
\( \Longrightarrow \left( \sum_{i=1}^n c_i v_i = 0 \Leftrightarrow c_i = 0 \; \forall i \right) \)

Alternately, \(\{v_i\}_{i=1}^n\) is linearly independent if
no vector \(v_j\) can be created from
a linear combination of the other vectors.

Can you invert a matrix? (Part 2)

\(A\) is invertible if
its rows are linearly independent.

\(A\) is also invertible if
its columns are linearly independent.


(and inverses)

What is it?

The determinant is a value
associated with square matrices.
It is written as \( \det(A) \).

Can you invert a matrix? (Part 3)

A square matrix \(A\) is invertible if
\( \det(A) \neq 0 \).

How do you calculate it?

It's best shown with examples.

2 x 2 Determinant

\[ A = \left(\begin{array}{cc} a & b\\ c & d\\ \end{array}\right) \]
\( \Rightarrow \det(A) = ad-bc \)

3 x 3 Determinant

\[ A = \left( \begin{array}{ccc} a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\\ \end{array} \right) \]
\( \Rightarrow \det(A) = a_1(b_2c_3-b_3c_2) - \)
\( a_2(b_1c_3-b_3c_1) + \)
\( a_3(b_1c_2-b_2c_1)\)

n x n Determinant

For any row \(A_i\), \[ \det(A) = \sum_{j=1}^n a_{ij} \cdot (-1)^{i+j} \cdot M_{ij}, \] where \( M_{ij} \) is the (i, j) minor matrix.


(and inverses)

What are they?

For a square matrix \(A\),
we often examine special vectors \(v\)
that do not rotate when \(A\) is applied.

That is, for a certain scalar \(\lambda\),
\(Av = \lambda v\).

In this case, we call \(\lambda\) and \(v\)
an eigenvalue and eigenvector of \(A\).

Why are they important?

After doing a bit of algebra, we can notice:

\( Av = \lambda v \)
\( \Rightarrow Av - \lambda Iv = 0 \)
\( \Rightarrow (A - \lambda I)v = 0. \)

Now we can make an assertion...

Can you invert a matrix? (Part 4)


If \(v \neq 0\), then \((A-\lambda I)\) is not invertible.
Thus, \(\det{(A-\lambda I)} = 0\).


Suppose \((A-\lambda I)^{-1}\) exists.

\( (A-\lambda I)v = 0 \)
\( \Rightarrow (A-\lambda I)^{-1}(A-\lambda I)v = 0 \)
\( \Rightarrow Iv = 0\)
\( \Rightarrow v=0. \)
This contradicts our original assumption that \(v \neq 0\).


How do I find eigenvalues?

The \(v=0\) case is boring.
\( A\times 0 = 0 = \lambda \times 0 \Rightarrow \lambda\) can be anything.

So, let's focus on \(v \neq 0\), so that \(\det{(A-\lambda I)} = 0\).

The left-hand side is an \(n^{th}\)-order polynomial in \(\lambda \).
Thus, by the Fundamental Theorem of Algebra,
every square matrix has \(n\) eigenvalues.

However, the eigenvalues are not necessarily distinct.

How do I find eigenvectors?

Suppose the eigenvalues \( \{\lambda_i\}_{i=1}^n \) are all distinct.
To get the eigenvectors \( \{v_i\}_{i=1}^n \) , solve the original equations:
\( (A - \lambda_i)v_i = 0, i = 1,\ldots,n \).

This produces \(n\) simultaneous linear equations.
Then, write each \(v_i\) in terms of one common component.
Finally, factor the common component out.

For example, if at first we obtain \( (v_1, 3v_1, 2v_1) \),
then the actual eigenvector is simply \( (1, 3, 2) \).

Thanks for reading!


Matrices and Linear Algebra: accessible introductory pamphlet
Course 18.06SC: MIT's famous linear algebra course
Linear Algebra Done Right: a canonical textbook

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