Introduction to Matrices
YouTube lecture recording from October 2020
The following YouTube video was recorded for the 2020 iteration of the course.
The material is still very similar:
Simultaneous equations
Consider 2 simultaneous equations:
a 1 x + b 1 y = c 1 , ( 1 ) a 2 x + b 2 y = c 2 , ( 2 ) \begin{align*}
a_1x+b_1y &= c_1, \qquad \qquad (1)\\
a_2x+b_2y &= c_2, \qquad \qquad (2)
\end{align*} a 1 x + b 1 y a 2 x + b 2 y = c 1 , ( 1 ) = c 2 , ( 2 ) where the values x \;x\; x y \;y\; y a 1 , b 1 , a 2 , b 2 , c 1 \;a_1, \;b_1, \;a_2, \;b_2, \;c_1\; a 1 , b 1 , a 2 , b 2 , c 1 c 2 \;c_2\; c 2
( 1 ) × b 2 : b 2 a 1 x + b 2 b 1 y = b 2 c 1 , ( 3 ) ( 2 ) × b 1 : b 1 a 2 x + b 1 b 2 y = b 1 c 2 , ( 4 ) ( 3 ) − ( 4 ) : b 2 a 1 x − b 1 a 2 x = b 2 c 1 − b 1 c 2 . \begin{align*}
(1) \times b_2:~~~~~~~~~~~~~~~ b_2a_1x+b_2b_1y &= b_2c_1, \qquad \qquad (3)\\
(2) \times b_1:~~~~~~~~~~~~~~~ b_1a_2x+b_1b_2y &= b_1c_2, \qquad \qquad (4)\\
(3) - (4):~~~~~~~~~~~~~~~ b_2a_1x-b_1a_2x &= b_2c_1-b_1c_2.
\end{align*} ( 1 ) × b 2 : b 2 a 1 x + b 2 b 1 y ( 2 ) × b 1 : b 1 a 2 x + b 1 b 2 y ( 3 ) − ( 4 ) : b 2 a 1 x − b 1 a 2 x = b 2 c 1 , ( 3 ) = b 1 c 2 , ( 4 ) = b 2 c 1 − b 1 c 2 .
x = b 2 c 1 − b 1 c 2 b 2 a 1 − b 1 a 2 , \displaystyle x=\frac{b_2c_1-b_1c_2}{b_2a_1-b_1a_2}, x = b 2 a 1 − b 1 a 2 b 2 c 1 − b 1 c 2 ,
y = a 1 c 2 − a 2 c 1 a 1 b 2 − a 2 b 1 . \displaystyle y=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}. y = a 1 b 2 − a 2 b 1 a 1 c 2 − a 2 c 1 .
This works, provided that a 1 b 2 − a 2 b 1 ≠ 0. a_1b_2-a_2b_1\neq 0. a 1 b 2 − a 2 b 1 = 0.
While the algebraic manipulation is straightforward when solving two equations, it can get really messy when solving large systems.
What we want is a way to be able to easily manipulate linear systems, regardless of how big they are.
The matrix
Matrices are a structure that allow us to more easily manipulate linear systems.
Consider the original system
a 1 x + b 1 y = c 1 , a 2 x + b 2 y = c 2 . \begin{align*}
a_1x+b_1y &= c_1, \\
a_2x+b_2y &= c_2.
\end{align*} a 1 x + b 1 y a 2 x + b 2 y = c 1 , = c 2 . We rewrite this, in the form of a matrix as:
( a 1 b 1 a 2 b 2 ) ( x y ) = ( c 1 c 2 ) . \begin{equation*}
\left(\begin{matrix}a_1&b_1\\ a_2&b_2\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)
=\left(\begin{matrix}c_1\\ c_2 \end{matrix}\right).
\end{equation*} ( a 1 a 2 b 1 b 2 ) ( x y ) = ( c 1 c 2 ) . Think about how this form relates to the original linear system.
What is a matrix?
A matrix is an array of numbers such as:
( a b c d e f g h i ) \left(\begin{matrix}a&b&c\\ d&e&f\\ g&h&i\end{matrix}\right) a d g b e h c f i
3 × 3 3\times3 3 × 3 size of the matrix.
A 3 × 3 3\times3 3 × 3 square and have order (dimension) 3.
Addition, subtraction, and scalar multiplication
We can add or subtract two matrices as long as they have the same size:
( 2 1 3 − 4 ) + ( 6 − 5 1 − 7 ) = ( 8 − 4 4 − 11 ) \left(\begin{matrix} 2&1 \\ 3&-4 \end{matrix}\right) +\left(\begin{matrix} 6&-5 \\ 1&-7 \end{matrix}\right)= \left(\begin{matrix} 8&-4 \\ 4&-11\end{matrix}\right) ( 2 3 1 − 4 ) + ( 6 1 − 5 − 7 ) = ( 8 4 − 4 − 11 ) We can multiply a matrix by a scalar, by multiplying every element:
5 × ( 2 1 3 − 4 ) = ( 10 5 15 − 20 ) 5\times\left(\begin{matrix} 2&1\\ 3&-4 \end{matrix}\right)=\left(\begin{matrix}10&5\\ 15&-20\end{matrix}\right) 5 × ( 2 3 1 − 4 ) = ( 10 15 5 − 20 ) Matrix multiplication
To multiply two matrices, we multiply each row in the first matrix by each column in the second one, and put the results into a new matrix.
A row and column are multiplied by summing up each element in the row, multiplied by the corresponding element in the column.
( 1 2 3 4 ) ( 5 6 7 8 ) = ( 1 × 5 + 2 × 7 1 × 6 + 2 × 8 3 × 5 + 4 × 7 3 × 6 + 4 × 8 ) = ( 19 22 43 46 ) \left(\begin{matrix} 1&2 \\ 3&4 \end{matrix}\right) \left(\begin{matrix} 5&6\\7&8\end{matrix}\right) = \left(\begin{matrix} 1 \times 5 + 2 \times 7 & 1 \times 6 + 2 \times 8 \\ 3 \times 5 + 4 \times 7 & 3 \times 6 + 4\times 8\end{matrix}\right) = \left(\begin{matrix} 19&22\\43&46\end{matrix}\right) ( 1 3 2 4 ) ( 5 7 6 8 ) = ( 1 × 5 + 2 × 7 3 × 5 + 4 × 7 1 × 6 + 2 × 8 3 × 6 + 4 × 8 ) = ( 19 43 22 46 ) ( 1 2 3 4 5 6 7 8 ) ( 1 2 3 4 5 6 7 8 9 10 11 12 ) = ( 70 80 90 158 184 210 ) \left(\begin{matrix} 1&2&3&4\\ 5&6&7&8 \end{matrix}\right) \left(\begin{matrix} 1&2&3\\ 4&5&6\\ 7&8&9\\ 10&11&12 \end{matrix}\right) = \left(\begin{matrix} 70&80&90\\ 158&184&210 \end{matrix}\right) ( 1 5 2 6 3 7 4 8 ) 1 4 7 10 2 5 8 11 3 6 9 12 = ( 70 158 80 184 90 210 ) ( a b c ) ( p q r s v w ) = ( a p + b r + c v a q + b s + c w ) . \left(\begin{matrix} a & b & c \end{matrix}\right)
\left(\begin{matrix} p & q \\ r & s \\ v & w \end{matrix}\right)
= \left(\begin{matrix} ap+br+cv & aq+bs+cw \end{matrix}\right). ( a b c ) p r v q s w = ( a p + b r + c v a q + b s + c w ) . If the number of columns in the first matrix doesn't match the number of rows in the second, they cannot be multiplied.
( 2 3 1 2 − 1 3 ) ( 1 0 − 1 − 4 ) = ? ? ? \left(\begin{matrix} 2 & 3 & 1 \\ 2 & -1 & 3\end{matrix}\right)\left(\begin{matrix} 1 & 0 \\ -1 & -4\end{matrix}\right) =\;?\;?\;? ( 2 2 3 − 1 1 3 ) ( 1 − 1 0 − 4 ) = ? ? ? Matrix multiplication is not commutative
This means that A × B A \times B A × B B × A B \times A B × A
( 3 x 2 m a t r i x ) × ( 2 x 2 m a t r i x ) = ( 3 x 2 m a t r i x ) (3 x 2 \rm{matrix}) \times (2 x 2 \rm{matrix}) = (3 x 2 \rm{matrix}) ( 3 x 2 matrix ) × ( 2x2 matrix ) = ( 3x2 matrix ) ( 2 x 2 m a t r i x ) × ( 3 x 2 m a t r i x ) = ? ? ? (2 x 2 \rm{matrix}) \times (3 x 2 \rm{matrix}) = ??? ( 2 x 2 matrix ) × ( 3x2 matrix ) = ???
However, even when sizes match, the product is usually not the same.
The identity matrix
I I I
A I = I A = A A I = I A = A A I = I A = A
for a square matrix A A A
The 2x2 identity matrix is:
I 2 = ( 1 0 0 1 ) . I_2 = \left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right). I 2 = ( 1 0 0 1 ) .
The 3x3 identity matrix is:
I 3 = ( 1 0 0 0 1 0 0 0 1 ) . I_3 = \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right). I 3 = 1 0 0 0 1 0 0 0 1 .
and so on for higher dimensions.
Inverting matrices
The determinant
A = ( p q r s ) \displaystyle A = \left(\begin{matrix} p & q \\ r & s\end{matrix}\right) A = ( p r q s )
then the determinant of A is:
∣ A ∣ = p s − q r |A| = ps-qr ∣ A ∣ = p s − q r
That is, (top left × \times × × \times ×
If ∣ A ∣ = 0 |A| = 0 ∣ A ∣ = 0 singular (and it has no inverse).
Inverting 2x2 matrices
If A B = I AB = I A B = I
A is called the inverse of B, and vice versa. I.e.
A = B − 1 , B = A − 1 \displaystyle A = B^{-1}, B = A^{-1} A = B − 1 , B = A − 1
A = ( p q r s ) \displaystyle A = \left(\begin{matrix} p & q \\ r & s\end{matrix}\right) A = ( p r q s )
A − 1 = 1 p s − q r ( s − q − r p ) \displaystyle A^{-1} = \frac{1}{ps-qr} \left(\begin{matrix} s & -q \\ -r & p\end{matrix}\right) A − 1 = p s − q r 1 ( s − r − q p )
Example of inverting a 2x2 matrix
Let us take a matrix A = ( 2 − 3 − 2 4 ) . \displaystyle A=\left(\begin{matrix}2&-3\\ -2&4\end{matrix}\right). A = ( 2 − 2 − 3 4 ) .
First, calculate its determinant:
∣ A ∣ = ( 2 × 4 ) − ( − 3 × − 2 ) = 8 − 6 = 2. \displaystyle |A|=(2\times 4)-(-3\times-2)=8-6=2. ∣ A ∣ = ( 2 × 4 ) − ( − 3 × − 2 ) = 8 − 6 = 2.
A − 1 = 1 2 ( 4 3 2 2 ) \displaystyle A^{-1}={1\over 2}\left(\begin{matrix}4&3\\ 2&2\end{matrix}\right) A − 1 = 2 1 ( 4 2 3 2 )
As a check, calculate A − 1 A A^{-1}A A − 1 A
A − 1 A = 1 2 ( 4 3 2 2 ) ( 2 − 3 − 2 4 ) \displaystyle A^{-1}A= \frac{1}{2}\left(\begin{matrix}4&3\\ 2&2\end{matrix}\right)\left(\begin{matrix}2&-3\\ -2&4\end{matrix}\right) A − 1 A = 2 1 ( 4 2 3 2 ) ( 2 − 2 − 3 4 ) = 1 2 ( 2 0 0 2 ) \displaystyle = \frac{1}{2}\left(\begin{matrix}2&0\\ 0&2\end{matrix}\right) = 2 1 ( 2 0 0 2 ) = ( 1 0 0 1 ) \displaystyle = \left(\begin{matrix}1&0\\ 0&1\end{matrix}\right) = ( 1 0 0 1 ) = I 2 . \displaystyle =I_2. = I 2 .
The transpose of a Matrix
A T A^T A T A A A
Swap elements across the leading diagonal so that A i j T = A j i A^T_{ij}= A_{ji} A ij T = A ji
A = ( 2 1 2 1 4 6 1 − 1 2 ) \displaystyle A=\left(\begin{matrix}2&1&2\\ 1&4&6\\ 1&-1&2\end{matrix}\right) A = 2 1 1 1 4 − 1 2 6 2 A T = ( 2 1 1 1 4 − 1 2 6 2 ) \displaystyle A^T=\left(\begin{matrix}2&1&1\\ 1&4&-1\\ 2&6&2\end{matrix}\right) A T = 2 1 2 1 4 6 1 − 1 2
Solving a linear system using matrices
To solve a matrix system A x = b \displaystyle A {\bf x} = {\bf b} A x = b x {\bf x} x
If it's of order 2 then use the formula to write A − 1 A^{-1} A − 1 x = A − 1 b {\bf x} = A^{-1}{\bf b} x = A − 1 b
If it's larger ( 3 × 3 ) (3\times3) ( 3 × 3 ) A − 1 A^{-1} A − 1
Use an analytical method (Gaussian elimination) to find the inverse (not in this course).
Use a numerical scheme to find an approximation to x {\bf x} x
Solve using linear algebra software (e.g. in Python, which we will see shortly).
Example of solving a 2x2 linear system
A − 1 A x = A − 1 b \displaystyle A^{-1}A{\bf x}=A^{-1}{\bf b} A − 1 A x = A − 1 b
x = A − 1 b \displaystyle {\bf x}=A^{-1}{\bf b} x = A − 1 b
x + 5 y = 11 , − x + 5 y = 9 \begin{align*}
x+5y &= 11, \\
-x+5y &= 9
\end{align*} x + 5 y − x + 5 y = 11 , = 9 In matrix form, this gives:
( 1 5 − 1 5 ) ( x y ) = ( 11 9 ) \displaystyle \left(\begin{matrix}1 &5\\ -1&5\end{matrix}\right) \left(\begin{matrix}x\\ y\end{matrix}\right) = \left(\begin{matrix}11\\ 9\end{matrix}\right) ( 1 − 1 5 5 ) ( x y ) = ( 11 9 )
A − 1 = 1 10 ( 5 − 5 1 1 ) \displaystyle A^{-1}= \frac{1}{10} \left(\begin{matrix}5 &-5\\ 1&1 \end{matrix}\right) A − 1 = 10 1 ( 5 1 − 5 1 )
( x y ) = 1 10 ( 5 − 5 1 1 ) ( 11 9 ) = 1 10 ( 10 20 ) \displaystyle \left(\begin{matrix}x\\ y\end{matrix}\right) = \frac{1}{10}\left(\begin{matrix}5 &-5\\ 1&1\end{matrix}\right)\left(\begin{matrix}11\\ 9\end{matrix}\right) =\frac{1}{10} \left(\begin{matrix}10\\ 20\end{matrix}\right) ( x y ) = 10 1 ( 5 1 − 5 1 ) ( 11 9 ) = 10 1 ( 10 20 ) = ( 1 2 ) \displaystyle =\left(\begin{matrix}1\\ 2\end{matrix}\right) = ( 1 2 )
And x = 1 , y = 2 x=1, y=2 x = 1 , y = 2
This process seems like more effort than its worth for small systems.
But it allows for a much more systematic approach when dealing with large systems.
As the size of the matrix grows, this process can be easily performed with Python (or other tools).
Example. Solving a 4x4 system in Python
x + 5 y + 3 z − w = 5 , x − 2 y + z + 4 w = 2 , − 3 x + y − z + 2 w = − 5 , x + y + z = 0. \begin{align*}
x + 5y + 3z - w &= 5, \\
x - 2y + z + 4w &= 2, \\
-3x + y - z + 2w &= -5, \\
x + y + z &= 0.
\end{align*} x + 5 y + 3 z − w x − 2 y + z + 4 w − 3 x + y − z + 2 w x + y + z = 5 , = 2 , = − 5 , = 0. In matrix form, this gives:
( 1 5 3 − 1 1 − 2 1 4 − 3 1 − 1 2 1 1 1 0 ) ( x y z w ) = ( 5 2 − 5 0 ) . \displaystyle \left(\begin{matrix} 1 & 5 & 3 & -1 \\ 1 & -2 & 1 & 4 \\ -3 & 1 & -1 & 2\\ 1 & 1 & 1 & 0 \end{matrix}\right) \left(\begin{matrix} x \\ y \\ z \\ w\end{matrix}\right) = \left(\begin{matrix} 5 \\ 2 \\ -5 \\ 0\end{matrix}\right). 1 1 − 3 1 5 − 2 1 1 3 1 − 1 1 − 1 4 2 0 x y z w = 5 2 − 5 0 .
Numerically, using NumPy
Copy # In python, we use numpy arrays to store the needed matrices
# the procedure linalg.solve, solves the system Ax = b
# We could also calculate the inverse of A (linalg.inv), and then multiply.
# But this is faster
import numpy as np
A = np . array ( [ [ 1 , 5 , 3 , - 1 ] , [ 1 , - 2 , 1 , 4 ] , [ - 3 , 1 , - 1 , 2 ] , [ 1 , 1 , 1 , 0 ] ] )
b = np . array ( [ 5 , 2 , - 5 , 0 ] )
x = np . linalg . solve ( A , b )
print ( x )
print ( np . matmul ( A , x ) )
Copy [-5.94444444 -5.11111111 11.05555556 -3.33333333]
[ 5.0000000e+00 2.0000000e+00 -5.0000000e+00 -8.8817842e-16]
Symbolically, using SymPy
Copy import sympy as sp
A = sp . Matrix ( [ [ 1 , 5 , 3 , - 1 ] , [ 1 , - 2 , 1 , 4 ] , [ - 3 , 1 , - 1 , 2 ] , [ 1 , 1 , 1 , 0 ] ] )
A . inv ( ) * sp . Matrix ( [ 5 , 2 , - 5 , 0 ] )
[ − 107 18 − 46 9 199 18 − 10 3 ] \displaystyle \left[\begin{matrix}- \frac{107}{18}\\- \frac{46}{9}\\\frac{199}{18}\\- \frac{10}{3}\end{matrix}\right] − 18 107 − 9 46 18 199 − 3 10
Introductory problems
Introductory problems 1 Write the following system of equations:
x + y + z = 7 2 x − y + z = 2 x − 2 y + 2 z = 5 \begin{aligned}
x + y + z &= 7\\
2x - y + z &= 2\\
x - 2y + 2z &= 5
\end{aligned} x + y + z 2 x − y + z x − 2 y + 2 z = 7 = 2 = 5
A x = b \displaystyle A\mathbf{x} = \mathbf{b} A x = b
y B = c \displaystyle \mathbf{y}B = \mathbf{c} y B = c
Where A A A B B B 3 × 3 3\times3 3 × 3 x \mathbf{x} x y \mathbf{y} y b \mathbf{b} b c \mathbf{c} c
Check that your answers make sense by expanding your expressions to ensure you get back to the original equations.
Introductory problems 2
A = ( 2 1 3 4 ) \displaystyle A = \begin{pmatrix} 2 & 1 \\ 3 & 4 \end{pmatrix} A = ( 2 3 1 4 ) B = ( 1 4 7 2 ) \displaystyle B = \begin{pmatrix} 1 & 4 \\ 7 & 2 \end{pmatrix} B = ( 1 7 4 2 ) C = ( 3 − 1 − 5 2 ) \displaystyle C = \begin{pmatrix} 3 & -1 \\ -5 & 2 \end{pmatrix} C = ( 3 − 5 − 1 2 ) D = ( 1 3 ) \displaystyle D = \begin{pmatrix} 1 \\ 3 \end{pmatrix} D = ( 1 3 ) E = ( 2 − 1 ) \displaystyle E = \begin{pmatrix} 2 & -1 \end{pmatrix} E = ( 2 − 1 )
Write down a 21 a_{21} a 21 b 12 b_{12} b 12 c 22 c_{22} c 22
Calculate, where possible, or explain why the product is not defined:
Do A and B commute? Do A and C commute?
Does ( A B ) C = A ( B C ) (AB)C = A(BC) ( A B ) C = A ( BC )
Does A C + B C = ( A + B ) C AC + BC = (A+B)C A C + BC = ( A + B ) C
Main problems
Main problems 1 If A = 1 2 ( 1 1 1 1 ) \displaystyle A = \frac{1}{2}\left(\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array}\right) A = 2 1 ( 1 1 1 1 ) A 2 A^2 A 2 A 3 A^3 A 3
Main problems 2
A = ( 2 1 3 3 − 2 1 − 1 0 1 ) ; B = ( 0 − 1 1 − 5 2 − 1 3 0 2 ) \displaystyle A = \begin{pmatrix} 2 & 1 & 3 \\ 3 & -2 & 1 \\ -1 & 0 & 1 \end{pmatrix}; \qquad B = \begin{pmatrix} 0 & -1 & 1 \\ -5 & 2 & -1 \\ 3 & 0 & 2 \end{pmatrix} A = 2 3 − 1 1 − 2 0 3 1 1 ; B = 0 − 5 3 − 1 2 0 1 − 1 2
Main problems 3 Find the determinant, ∣ A ∣ |A| ∣ A ∣
A = ( 1 2 1 6 ) \displaystyle A = \begin{pmatrix} 1 & 2 \\ 1 & 6 \end{pmatrix} A = ( 1 1 2 6 )
A = ( 1 − 1 2 − 4 ) \displaystyle A = \begin{pmatrix} 1 &-1 \\ 2 &-4 \end{pmatrix} A = ( 1 2 − 1 − 4 )
Main problems 4 Find the inverse, A − 1 A^{-1} A − 1
A = ( 2 5 − 1 4 ) A = \begin{pmatrix} 2 & 5 \\-1 & 4 \end{pmatrix} A = ( 2 − 1 5 4 )
A = ( − 3 2 − 1 7 ) A = \begin{pmatrix} -3 & 2 \\-1 & 7 \end{pmatrix} A = ( − 3 − 1 2 7 )
Main problems 5 Use Python's numpy.linalg.solve to solve the following systems of equations:
x + 2 y − 3 z = 9 , 2 x − y + z = 0 , 4 x − y + z = 4. \begin{aligned} x + 2y - 3z &= 9,\\ 2x - y + z &= 0,\\ 4x - y + z &= 4. \end{aligned} x + 2 y − 3 z 2 x − y + z 4 x − y + z = 9 , = 0 , = 4.
x + 5 y + 3 z = 17 , 5 x + y − 2 z = 4 , x + 2 y + z = 7. \begin{aligned}x + 5y + 3z &= 17,\\ 5x + y - 2z &= 4,\\ x + 2y + z &= 7. \end{aligned} x + 5 y + 3 z 5 x + y − 2 z x + 2 y + z = 17 , = 4 , = 7.
2 y + z = − 8 , x − 2 y − 3 z = 0 , − x + y + 2 z = 3. \begin{aligned} 2y + z &= -8,\\ x - 2y - 3z &= 0,\\ -x + y + 2z &= 3. \end{aligned} 2 y + z x − 2 y − 3 z − x + y + 2 z = − 8 , = 0 , = 3.
Copy # hint
import numpy as np
A = np . array ( [ [ 1 , 2 , - 3 ] , [ 2 , - 1 , 1 ] , [ 4 , - 1 , 1 ] ] )
b = np . array ( [ 9 , 0 , 4 ] )
x = np . linalg . solve ( A , b )
print ( x )
Main problems 6
X = ( 1 2 3 4 ) ; Y = ( 5 6 7 8 ) \displaystyle X = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}; \qquad Y = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix} X = ( 1 3 2 4 ) ; Y = ( 5 7 6 8 )
Extension problems
Extension problems 1
( 3 1 1 3 ) ( 1 2 1 2 ) = λ 1 ( 1 2 1 2 ) \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{pmatrix} = \lambda_1 \begin{pmatrix} \frac{1}{\sqrt 2} \\ \frac{1}{\sqrt 2} \end{pmatrix} ( 3 1 1 3 ) ( 2 1 2 1 ) = λ 1 ( 2 1 2 1 ) ( 3 1 1 3 ) ( 1 2 − 1 2 ) = λ 2 ( 1 2 − 1 2 ) \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt 2} \\ \frac{-1}{\sqrt 2} \end{pmatrix} = \lambda_2 \begin{pmatrix} \frac{1}{\sqrt 2} \\ \frac{-1}{\sqrt 2} \end{pmatrix} ( 3 1 1 3 ) ( 2 1 2 − 1 ) = λ 2 ( 2 1 2 − 1 )
where λ 1 \lambda_1 λ 1 λ 2 \lambda_2 λ 2
