Eigenvalues and Eigenvectors

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Eigenvalues and Eigenvectors

The states a gate leaves alone (up to a scalar), why that scalar is called an eigenvalue, and where this shows up all over the algorithms track.

Suggested before this lesson

Matrices as Linear Operators

Most vectors, hit with a matrix, come out pointing in some new direction entirely. A special few don't: the matrix only stretches or flips them, without changing the direction they point in. Those special vectors, and the amount they get scaled by, turn out to matter enormously for quantum computing, all the way up to Quantum Phase Estimation and Shor's algorithm later in this course.

The Definition

For a matrix $A$ , a nonzero vector $v$ is an eigenvector of $A$ if

$Av = \lambda v$

for some scalar $\lambda$ , called the corresponding eigenvalue. $A$ doesn't rotate $v$ into some other direction; it just multiplies it by $\lambda$ .

Finding Them: the Characteristic Equation

Rearranging $Av = \lambda v$ gives $(A - \lambda I)v = 0$ . A nonzero $v$ satisfying this only exists when $A - \lambda I$ is singular (non-invertible), which for a $2\times2$ matrix $A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$ means its determinant is zero:

$\det(A - \lambda I) = (a-\lambda)(d-\lambda) - bc = 0$

This is the characteristic equation; its roots are $A$ 's eigenvalues.

Worked Example: the $Z$ Gate

$Z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$ . The characteristic equation is $(1-\lambda)(-1-\lambda) = 0$ , giving $\lambda = 1$ or $\lambda = -1$ .

For $\lambda = 1$ : solving $(Z - I)v = 0$ gives $v = \begin{pmatrix}1\\0\end{pmatrix} = |0\rangle$ . Check: $Z|0\rangle = |0\rangle = 1\cdot|0\rangle$ . ✓
For $\lambda = -1$ : solving gives $v = \begin{pmatrix}0\\1\end{pmatrix} = |1\rangle$ . Check: $Z|1\rangle = -|1\rangle$ (you computed exactly this in the last lesson's quiz). ✓

$Z$ 's eigenvectors are just $|0\rangle$ and $|1\rangle$ , with eigenvalues $+1$ and $-1$ : unsurprising, since $Z$ was already built to leave the computational basis states alone (up to sign).

Worked Example: the $X$ Gate

$X = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$ has the same characteristic equation, $-(1-\lambda^2)\cdot(-1)$ simplifies to $\lambda^2 = 1$ , so again $\lambda = \pm1$ , but this time the eigenvectors aren't $|0\rangle$ / $|1\rangle$ ( $X$ visibly swaps those, it doesn't leave them alone). Solving $(X-I)v=0$ and $(X+I)v=0$ gives:

$|+\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix} \text{ for } \lambda=+1, \qquad |-\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix} \text{ for } \lambda=-1$

That second line, $X|-\rangle = -|-\rangle$ , is exactly the fact the phase-kickback lesson leans on entirely, stated there without the "eigenvalue" name. Now you have the formal machinery behind it: $|-\rangle$ is an eigenvector of $X$ with eigenvalue $-1$ , full stop, and that's why applying $X$ to it only ever multiplies it by $-1$ instead of moving it somewhere else.

Eigenbasis

$|0\rangle,|1\rangle$ form a complete basis of eigenvectors for $Z$ (its eigenbasis); $|+\rangle,|-\rangle$ form one for $X$ . Every $2\times2$ matrix relevant to this course has an eigenbasis like this: a basis of eigenvectors that the matrix acts on by pure scaling, no rotation. Later, Quantum Phase Estimation (in the algorithms track) is built entirely around a single question: given an eigenvector of some unitary $U$ , what is its eigenvalue? Since a unitary's eigenvalues always have the form $e^{i\theta}$ (the Unitary Matrices lesson right after this one covers why), "find the eigenvalue" really means "find the phase $\theta$ ," which is where that lesson's name comes from.

Hunt for an eigendirection

Sweep $v$ around the circle for $X$ or $Z$ and watch for the exact angles where $Mv$ snaps onto the same line as $v$ , that's an eigenvector, and the badge shows its eigenvalue $\lambda$ .

Vector transform

$Angle of v$

20°

$v$

(0.94, 0.34)

$M v$

(0.94, -0.34)

$∥ M v ∥$

1.000

M v

is not parallel to

v

Sweep the angle until the dashed input arrow $v$ and the solid output arrow $Mv$ point along the same line (or exactly opposite): that's an eigendirection, and the badge shows its eigenvalue $\lambda$ .

Check your understanding

v = \begin{pmatrix}1\\2\end{pmatrix}

an eigenvector of

A = \begin{pmatrix}2 & 0\\0 & 3\end{pmatrix}

Yes, with eigenvalue 2.

Yes, with eigenvalue 3.

No:

Av = (2, 6)^T

, which isn't any scalar multiple of

(1, 2)^T

No, because

A

isn't a valid quantum gate.

Check your understanding

What does it mean for

|+\rangle

and

|-\rangle

to form an 'eigenbasis' of

X

X

can only be applied to states written in that basis.

Every state can be written as a combination of

|+\rangle

and

|-\rangle

, and

X

acts on each of those two by pure scaling (

\pm1

), not by mixing them into new directions.

|+\rangle

and

|-\rangle

are the only two states

X

can ever produce.

X

and

|+\rangle

|-\rangle

commute, which is unrelated to eigenvectors.

Finding Them: the Characteristic Equation

Rearranging

Av = \lambda v

gives

(A - \lambda I)v = 0

. A nonzero

v

satisfying this only exists when

A - \lambda I

is singular (non-invertible), which for a

2\times2

matrix

A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}

means its determinant is zero:

\det(A - \lambda I) = (a-\lambda)(d-\lambda) - bc = 0

This is the characteristic equation; its roots are

A

's eigenvalues.

Worked Example: the

Z

Gate

Z = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}

. The characteristic equation is

(1-\lambda)(-1-\lambda) = 0

, giving

\lambda = 1

\lambda = -1

For

\lambda = 1

: solving

(Z - I)v = 0

gives

v = \begin{pmatrix}1\\0\end{pmatrix} = |0\rangle

. Check:

Z|0\rangle = |0\rangle = 1\cdot|0\rangle

. ✓

For

\lambda = -1

: solving gives

v = \begin{pmatrix}0\\1\end{pmatrix} = |1\rangle

. Check:

Z|1\rangle = -|1\rangle

(you computed exactly this in the last lesson's quiz). ✓

Z

's eigenvectors are just

|0\rangle

and

|1\rangle

, with eigenvalues

+1

and

-1

: unsurprising, since

Z

was already built to leave the computational basis states alone (up to sign).

Worked Example: the

X

Gate

X = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}

has the same characteristic equation,

-(1-\lambda^2)\cdot(-1)

simplifies to

\lambda^2 = 1

, so again

\lambda = \pm1

, but this time the eigenvectors aren't

|0\rangle

|1\rangle

(

X

visibly swaps those, it doesn't leave them alone). Solving

(X-I)v=0

and

(X+I)v=0

gives:

|+\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\1\end{pmatrix} \text{ for } \lambda=+1, \qquad |-\rangle = \tfrac{1}{\sqrt2}\begin{pmatrix}1\\-1\end{pmatrix} \text{ for } \lambda=-1

That second line,

X|-\rangle = -|-\rangle

, is exactly the fact the phase-kickback lesson leans on entirely, stated there without the "eigenvalue" name. Now you have the formal machinery behind it:

|-\rangle

is an eigenvector of

X

with eigenvalue

-1

, full stop, and that's why applying

X

to it only ever multiplies it by

-1

instead of moving it somewhere else.

Eigenbasis

|0\rangle,|1\rangle

form a complete basis of eigenvectors for

Z

(its eigenbasis);

|+\rangle,|-\rangle

form one for

X

. Every

2\times2

matrix relevant to this course has an eigenbasis like this: a basis of eigenvectors that the matrix acts on by pure scaling, no rotation. Later, Quantum Phase Estimation (in the algorithms track) is built entirely around a single question: given an eigenvector of some unitary

U

, what is its eigenvalue? Since a unitary's eigenvalues always have the form

e^{i\theta}

(the Unitary Matrices lesson right after this one covers why), "find the eigenvalue" really means "find the phase

\theta

," which is where that lesson's name comes from.

Hunt for an eigendirection

Sweep $v$ around the circle for $X$ or $Z$ and watch for the exact angles where $Mv$ snaps onto the same line as $v$ , that's an eigenvector, and the badge shows its eigenvalue $\lambda$ .

Vector transform

$Angle of v$

20°

$v$

(0.94, 0.34)

$M v$

(0.94, -0.34)

$∥ M v ∥$

1.000

M v

is not parallel to

v

Check your understanding

v = \begin{pmatrix}1\\2\end{pmatrix}

an eigenvector of

A = \begin{pmatrix}2 & 0\\0 & 3\end{pmatrix}

Yes, with eigenvalue 2.

Yes, with eigenvalue 3.

No:

Av = (2, 6)^T

, which isn't any scalar multiple of

(1, 2)^T

No, because

A

isn't a valid quantum gate.

Check your understanding

What does it mean for

|+\rangle

and

|-\rangle

to form an 'eigenbasis' of

X

X

can only be applied to states written in that basis.

Every state can be written as a combination of

|+\rangle

and

|-\rangle

, and

X

acts on each of those two by pure scaling (

\pm1

), not by mixing them into new directions.

|+\rangle

and

|-\rangle

are the only two states

X

can ever produce.

X

and

|+\rangle

|-\rangle

commute, which is unrelated to eigenvectors.