Eigenvalues & Eigenvectors - 2 | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

Example 4  is already singular (zero determinant). Find its λ’s and x’s.

When A is singular, λ = 0 is one of the eigenvalues. The equation Ax = 0x has solutions. They are the eigenvectors for λ = 0. But det (A - λI) = 0 is the way to find all λ’s and x’s. Always subtract λI from A:

Subtract λ from the diagonal to find 

Take the determinant “ad - bc” of this 2 by 2 matrix. From 1 - λ times 4 - λ, the “ad” part is λ² - 5λ + 4. The “bc” part, not containing λ, is 2 times 2.

Set this determinant λ² - 5λ to zero. One solution is λ = 0 (as expected, since A is singular). Factoring into λ times λ - 5, the other root is λ = 5:

det(A - λI) = λ² - 5λ = 0 yields die eigenvalues λ₁ = 0 and λ₂ = 5 .

Now find the eigenvectors. Solve (A - λI) x = 0 separately for λ₁ = 0 and λ₂ = 5:

The matrices A - 01 and A - 5I are singular (because 0 and 5 are eigenvalues). The eigenvectors (2, -1) and (1,2) are in the nullspaces: (A - λI)x = 0 is Ax = λx.

We need to emphasize: There is nothing exceptional about λ = 0. Like every other number, zero might be an eigenvalue and it might not. If A is singular, it is. The eigenvectors fill the nullspace: Ax = 0x = 0. If A is invertible, zero is not an eigenvalue. We shift A by a multiple of I to make it singular.

In the example, the shifted matrix A - 5I is singular and 5 is the other eigenvalue.

Summary To solve the eigenvalue problem for an n by n matrix, follow these steps:

1.  Compute the determinant of A - λI. With λ subtracted along the diagonal, this determinant starts with λⁿ or -λⁿ. It is a polynomial in λ of degree n.
2.  Find the roots of this polynomial, by solving det(A -λI) =0. The n roots are the n eigenvalues of A. They make A - λI singular.
3.  For each eigenvalue A, solve ( A - λI)x = 0 to find an eigenvector x.

A note on the eigenvectors of 2 by 2 matrices. When A - λI is singular, both rows are multiples of a vector (a, b). The eigenvector is any multiple of (b, -a). The example had λ = 0 and λ = 5

λ  = 0 : rows of A - 0I in the direction (1,2); eigenvector in the direction .(2, -1) 
λ = 5 : rows of A - 5I in the direction (-4, 2); eigenvector in the direction (2,4)

Previously we wrote that last eigenvector as (1, 2); Both (1, 2) and (2,4) are correct. There is a whole line of eigenvectors-any nonzero multiple of x is as good as x. MATLAB’s eig(A) divides by the length, to make the eigenvector into a unit vector.

We end with a warning. Some 2 by 2 matrices have only one line of eigenvectors.
This can only happen when two eigenvalues are equal. (On the other hand A = I has equal eigenvalues and plenty of eigenvectors.) Similarly some n by n matrices don’t have n independent eigenvectors. Without n eigenvectors, we don’t have a basis. We can’t write every v as a combination of eigenvectors. In the language of the next section, we can’t diagonalize a matrix without n independent eigenvectors.

Good News, Bad News

Bad news first: If you add a row of A to another row, or exchange rows, the eigenvalues usually change. Elimination does not preserve the λ’s. The triangular U has its eigenvalues sitting along the diagonal-they are the pivots. But they are not the eigenvalues of A!

Eigenvalues are changed when row 1 is added to row 2:

Good news second: The product λ₁ times λ₂ and the sum λ₁ + λ₂ can be found quickly from the matrix. For this A, the product is 0 times 7. That agrees with the determinant (which is 0). The sum of eigenvalues is 0 + 7. That agrees with the sum down the main diagonal (the trace is 1 + 6). These quick checks always work:

The product of the n eigenvalues equals the determinant. The sum of the n eigenvalues equals the sum of the n diagonal entries.

The sum of the entries on the main diagonal is called the trace of A:

λ₁ + λ₂ + ... + λ_n = trace = a₁₁ + a₂₂ + ... + a_nn.                  (6)

Those checks are very useful. They are proved in Problems 16–17 and again in the next section. They don’t remove the pain of computing λ’s. But when the computation is wrong, they generally tell us so. To compute the correct λ’s, go back to det(A - λI) = 0.

The determinant test makes the product of the λ’s equal to the product of the pivots (assuming no row exchanges). But the sum of the λ’s is not the sum of the pivots-as the example showed. The individual λ’s have almost nothing to do with the pivots. In this new part of linear algebra, the key equation is really nonlinear: λ multiplies x.

Why do the eigenvalues of a triangular matrix lie on its diagonal?

Imaginary Eigenvalues

Example 5 The 90° rotation  has no real eigenvectors. Its eigenvalues are λ = i and λ = -i. Sum of λ’s = trace = 0. Product = determinant = 1.

After a rotation, no vector Qx stays in the same direction as x (except x = 0 which is useless). There cannot be an eigenvector, unless we go to imaginary numbers. Which we do.

To see how i can help, look at Q² which is -I. If Q is rotation through 90°, then Q² is rotation through 180°. Its eigenvalues are -1 and -1. (Certainly -Ix = -1x.) Squaring Q will square each λ, so we must have λ² = -1. The eigenvalues of the 90° rotation matrix Q are +i and - i, because i² = -1.
Those A’s come as usual from det(Q - λI) = 0. This equation gives λ² + 1 = 0. Its roots are i and -i. We meet the imaginary number i also in the eigenvectors:

Somehow these complex vectors x₁ = (1, i) and x₂ = (i, 1) keep their direction as they are rotated. Don’t ask me how. This example makes the all-important point that real matrices can easily have complex eigenvalues and eigenvectors. The particular eigenvalues i and-i also illustrate two special properties of Q:

1. Q is an orthogonal matrix so the absolute value of each λ is |λ| = 1.

2. Q is a skew-symmetric matrix so each λ is pure imaginary.

A symmetric matrix (A^T = A) can be compared to a real number. A skew-symmetric matrix (A^T = -A) can be compared to an imaginary number. An orthogonal matrix (A^T A = I) can be compared to a complex number with |λ| = 1. For the eigenvalues those are more than analogies-they are theorems to be proved in Section 6.4.
The eigenvectors for all these special matrices are perpendicular. Somehow (i, 1) and (1, i) are perpendicular (Chapter 10 explains the dot product of complex vectors).

Eigshow in MATLAB

There is a MATLAB demo (just type eigshow), displaying the eigenvalue problem for a 2 by 2 matrix. It starts with the unit vector x = (1, 0). The mouse makes this vector move around the unit circle. At the same time the screen shows Ax, in color and also moving.
Possibly Ax is ahead of x. Possibly Ax is behind x. Sometimes Ax is parallel to x.At that parallel moment, Ax = λx (at x₁ and x₂ in the second figure).

The eigenvalue λ is the length of Ax, when the unit eigenvector x lines up. The built-in choices for A illustrate three possibilities: 0; 1,or 2 directions where Ax crosses x.

0. There are no real eigenvectors. Ax stays behind or ahead of x. This means the eigenvalues and eigenvectors are complex, as they are for the rotation Q.

1. There is only one line of eigenvectors (unusual). The moving directions Ax and x touch but don’t cross over. This happens for the last 2 by 2 matrix below.

2. There are eigenvectors in two independent directions. This is typical! Ax crosses x at the first eigenvector x₁ , and it crosses back at the second eigenvector x₂ . Then Ax and x cross again at -x₁ and -x₂.

You can mentally follow x and Ax for these five matrices. Under the matrices I will count their real eigenvectors. Can you see where Ax lines up with x?

When A is singular (rank one), its column space is a line. The vector Ax goes up and down that line while x circles around. One eigenvector x is along the line. Another eigenvector appears when Ax₂ = 0. Zero is an eigenvalue of a singular matrix.