Table of Contents
Matrix Basics
This is an extra page to help explain some of the basics of linear algebra to those who are unfamiliar with it. It is highly recommended to watch 3blue1brown's Essence of Linear Algebra series, as it does a great job of providing a simple, visual intuition for all of the concepts it covers. If, on the other hand, you want a deeper understanding of linear algebra, you should use MIT's Linear Algebra course. The course is provided for free via MIT's OpenCourseWare.
A matrix is simply a 2D list of numbers. We can write a size $m \times n$ matrix as
$$ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ a_{31} & a_{32} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} $$When we say a matrix is size $m \times n$, we are saying that it has $m$ rows and $n$ columns. We write the position of an entry within a matrix the entry's subscript. The row number comes first and is followed by the column number. To represent the element at row $i$ and column $j$, we write $A_{ij}$. A matrix with only 1 column is known as a column vector, or just a vector. A matrix with only one row is known as a row vector.
$$ \vec{v} = \begin{bmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{bmatrix} \quad \text{(column vector)} $$$$ \vec{r} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{bmatrix} \quad \text{(row vector)} $$Matrix Operations
The transpose of a matrix, denoted with the subscript $\top$, is an operation which switches the row and column indices of the matrix, so $A_{ij}=A^\top_{ji}$. For example,
$$ A = \begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix}, \quad A^\top = \begin{bmatrix} a & c & e \\ b & d & f \end{bmatrix} $$In quantum computing, we will generally use the conjugate transpose, denoted with the subscript $\dag$, which transposes the matrix and applies the complex conjugate to every value of the matrix (flipping the sign of the imaginary part in a value). You will see its use as we progress, but for now, just know that it is the same as the transpose, just with the conjugate part stitched on.
If $A$ is a $m \times n$ matrix, and $B$ is a $n \times p$ matrix, the values of the matrix product $C=AB$, which is a $m \times p$ matrix, are defined as $C_{ij}=\sum_{k=1}^{n} A_{ik}B_{kj}$. The number of columns in $A$ must match the number of rows in $B$ for a matrix product to exist. For example,
$$ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 7 & 8 \\ 9 & 10 \\ 11 & 12 \end{bmatrix} = \begin{bmatrix} 58 & 64 \\ 139 & 154 \end{bmatrix} $$The inner product of two vectors $a$ and $b$ of size $n$ is the sum of the product of the corresponding entries in each vector.
$$ a^\dag b = a \cdot b = a_1b_1+\cdots+a_nb_n $$Think of the $C_{ij}$ entry in the matrix product as the inner product of the row vector $i$ in $A$ and the column vector $j$ in $B$, which is a visual interpretation of $C_{ij}=\sum_{k=1}^{n} A_{ik}B_{kj}$.
There are a few different ways to write the inner product. In quantum computing, we generally use bra-ket notation, where we would write this operation as $\langle a | b \rangle$. For more on bra-ket notation, see Bra-Ket Notation.
The outer product of two vectors $a$ and $b$ is the opposite operation, the product of all possible corresponding terms in each vector.
$$ ab^\dag = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix} \begin{bmatrix} b_1^* & b_2^* & b_3^* \end{bmatrix} = \begin{bmatrix} a_1b_1^* & a_1b_2^* & a_1b_3^* \\ a_2b_1^* & a_2b_2^* & a_2b_3^* \\ a_3b_1^* & a_3b_2^* & a_3b_3^* \end{bmatrix} $$In bra-ket notation, we would write this operation as $|a\rangle\langle b|$.
Basis
A set of vectors $\{v_1,\dots,v_n\}$ is a basis for some vector space (defined as every point we can reach by scaling the vectors), if every possible vector in that space1 we're in can be written as a linear combination of the vectors in the basis. The elements of a basis must be linearly independent, i.e. it cannot be possible to write one vector in the basis as a linear combination of the other vectors In other words,
$$ a = c_1v_1+\cdots+c_nv_n $$
where $a$ is any vector in the space we're in and $c_1,\dots,c_n$ are some constant values. Consequently, if we are working with vectors in $n$ dimensions, any basis must be of size $n$.
Eigenvalues and Eigenvectors
An eigenvalue and eigenvector are some value $\lambda$ and vector $\vec{v}$ such that
$$A\vec{v} = \lambda \vec{v}$$Conceptually, eigenvectors are vectors that don't change their direction in space when multiplied by a matrix. For example, any vector pointing along the direction of the Z-axis would be an eigenvector of the Z-rotation matrix (which rotates vectors around the Z-axis), since the Z-axis is its axis of rotation. In order to find an eigenvalue and its eigenvector(s), associated with some matrix $A$, we use the characteristic polynomial of $A$. It is given by the equation:
$$ \text{det}(A - \lambda I) = 0$$Where det is the determinant of the matrix.
Below is an example of how to find the eigenvalues and eigenvectors of the matrix
$$ A = \begin{pmatrix}4 & 2\\ 1 & 3\end{pmatrix} $$The characteristic polynomial is
$$ \begin{align*} \det(A - \lambda I) &= \det\begin{pmatrix}4-\lambda & 2\\ 1 & 3-\lambda\end{pmatrix} \\ &= (4-\lambda)(3-\lambda) - 2 \cdot 1 \\ &= \lambda^2 - 7\lambda + 10. \end{align*} $$Solving
$$ \lambda^2 - 7\lambda + 10 = 0 $$gives the integer roots
$$ \lambda = 5, \: \lambda = 2. $$Now, let's solve for the first eigenvector, $(A - 5I) \vec{v} = 0$.
$$ \begin{align*} (A - 5I) &= \begin{pmatrix}-1 & 2\\ 1 & -2\end{pmatrix} \\ \begin{pmatrix}-1 & 2\\ 1 & -2\end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} &= \begin{pmatrix}0\\ 0\end{pmatrix} \\ -x + 2y &= 0 \\ x &= 2y. \end{align*} $$Choosing $y = 1$ gives:
$$ \mathbf v_5 = \begin{pmatrix}2\\1\end{pmatrix}. $$Now, let's solve for the other eigenvector, $(A - 2I) \vec{v} = 0$.
$$ \begin{align*} (A - 2I) &= \begin{pmatrix}2 & 2\\ 1 & 1\end{pmatrix} \\ \begin{pmatrix}2 & 2\\ 1 & 1\end{pmatrix} \begin{pmatrix}x\\ y\end{pmatrix} &= \begin{pmatrix}0\\ 0\end{pmatrix} \\ x + y &= 0 \\ y &= -x. \end{align*} $$Choosing $x = 1$ gives:
$$ \mathbf v_2 = \begin{pmatrix}1\\-1\end{pmatrix}. $$The matrix
$$ A = \begin{pmatrix}4 & 2\\ 1 & 3\end{pmatrix} $$has integer eigenvalues 5 and 2, with corresponding integer eigenvectors
$$ \begin{pmatrix}2\\1\end{pmatrix} \text{ and } \begin{pmatrix}1\\-1\end{pmatrix} $$This is arguably the most important concept in linear algebra, both inside and outside of quantum computing. Some examples of their applications include Schrödinger's equation and molecular orbitals.
We will be looking at a quantum algorithm to estimate the eigenvalues of a matrix later on in the lessons.
Normalization
Another important concept is normalization. A vector is normalized if its magnitude (also called its norm, denoted as $||v||$) is equal to 1. In the world of linear algebra, we call this the unit vector (like a unit circle, where the radius is 1). For a vector $\vec{v}$ with components $v_1, v_2, \dots, v_n$, the norm is given by the Pythagorean theorem:
$$ \|v\| = \sqrt{|v_1|^2 + |v_2|^2 + \cdots + |v_n|^2} $$To normalize a vector, divide each component by its norm:
$$ \hat{v} = \frac{\vec{v}}{\|v\|} $$where $\hat{v}$ is the normalized vector.
Below is an example of a normalized vector calculation:
$$ \begin{align*} \vec{v} &= \begin{bmatrix} 3 \\ 4 \end{bmatrix} \\ \quad \|v\| &= \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \\ \hat{v} &= \frac{1}{5}\begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 3/5 \\ 4/5 \end{bmatrix} \end{align*} $$In quantum computing, when you "measure" a particle that physically exists (whose probability is not zero), you are bound to measure something. That is, the sum of the probabilities of all possible outcomes must be equal to 1 or 100%. Mathematically, this means that all statevectors must be normalized.
Throughout these pages, you will see expressions normalized to satisfy this rule. An example is $|+\rangle$ where:
$$ |+\rangle = \frac{|0\rangle+|1\rangle}{\sqrt{2}} $$Don't worry what $|+\rangle$ or other related symbols mean just yet; it will be covered as we progress, but notice the denominator. Since there is a $\sqrt2$ in the denominator, this means that the probability of measuring $|+\rangle$ as $|0\rangle$ or $|1\rangle$ has been normalized to 1.
Ultimately, the best way to become more familiar with linear algebra is just to practice it. There are plenty of videos and resources online to help you master the topics. A very strong understanding of linear algebra will make a lot of quantum computing concepts easier to understand.
1. In the world of linear algebra, a space is a set of vectors that can be reached by scaling the vectors in the basis.