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Eigenvalues

The symbolic eigenvalues of a square matrix A or the symbolic eigenvalues and eigenvectors of A are computed, respectively, using the commands E = eig(A) and [V,E] = eig(A).

The variable-precision counterparts are E = eig(vpa(A)) and [V,E] = eig(vpa(A)).

The eigenvalues of A are the zeros of the characteristic polynomial of A, det(A-x*I), which is computed by charpoly(A).

The matrix H from the last section provides the first example:

H = sym([8/9 1/2 1/3; 1/2 1/3 1/4; 1/3 1/4 1/5])
H =
[ 8/9, 1/2, 1/3]
[ 1/2, 1/3, 1/4]
[ 1/3, 1/4, 1/5]

The matrix is singular, so one of its eigenvalues must be zero. The statement

[T,E] = eig(H)

produces the matrices T and E. The columns of T are the eigenvectors of H and the diagonal elements of E are the eigenvalues of H:

T =
[ 3/10, 218/285 - (4*12589^(1/2))/285, (4*12589^(1/2))/285 + 218/285]
[ -6/5,     292/285 - 12589^(1/2)/285,     12589^(1/2)/285 + 292/285]
[    1,                             1,                             1]

E =
[ 0,                       0,                       0]
[ 0, 32/45 - 12589^(1/2)/180,                       0]
[ 0,                       0, 12589^(1/2)/180 + 32/45]

It may be easier to understand the structure of the matrices of eigenvectors, T, and eigenvalues, E, if you convert T and E to decimal notation. To do so, proceed as follows. The commands

Td = double(T)
Ed = double(E)

return

Td =
0.3000   -0.8098    2.3397
-1.2000    0.6309    1.4182
1.0000    1.0000    1.0000

Ed =
0         0         0
0    0.0878         0
0         0    1.3344

The first eigenvalue is zero. The corresponding eigenvector (the first column of Td) is the same as the basis for the null space found in the last section. The other two eigenvalues are the result of applying the quadratic formula to ${x}^{2}-\frac{64}{45}x+\frac{253}{2160}$ which is the quadratic factor in factor(charpoly(H, x)):

syms x
g = factor(charpoly(H, x))/x
solve(g(3))
g =
[ 1/(2160*x), 1, (2160*x^2 - 3072*x + 253)/x]
ans =
32/45 - 12589^(1/2)/180
12589^(1/2)/180 + 32/45

Closed form symbolic expressions for the eigenvalues are possible only when the characteristic polynomial can be expressed as a product of rational polynomials of degree four or less. The Rosser matrix is a classic numerical analysis test matrix that illustrates this requirement. The statement

R = sym(rosser)

generates

R =
[  611,  196, -192,  407,   -8,  -52,  -49,   29]
[  196,  899,  113, -192,  -71,  -43,   -8,  -44]
[ -192,  113,  899,  196,   61,   49,    8,   52]
[  407, -192,  196,  611,    8,   44,   59,  -23]
[   -8,  -71,   61,    8,  411, -599,  208,  208]
[  -52,  -43,   49,   44, -599,  411,  208,  208]
[  -49,   -8,    8,   59,  208,  208,   99, -911]
[   29,  -44,   52,  -23,  208,  208, -911,   99]

The commands

p = charpoly(R, x);
factor(p)

produce

ans =

[ x, x - 1020, x^2 - 1040500, x^2 - 1020*x + 100, x - 1000, x - 1000]

The characteristic polynomial (of degree 8) factors nicely into the product of two linear terms and three quadratic terms. You can see immediately that four of the eigenvalues are 0, 1020, and a double root at 1000. The other four roots are obtained from the remaining quadratics. Use

eig(R)

to find all these values

ans =
0
1000
1000
1020
510 - 100*26^(1/2)
100*26^(1/2) + 510
-10*10405^(1/2)
10*10405^(1/2)

The Rosser matrix is not a typical example; it is rare for a full 8-by-8 matrix to have a characteristic polynomial that factors into such simple form. If you change the two “corner” elements of R from 29 to 30 with the commands

S = R;
S(1,8) = 30;
S(8,1) = 30;

and then try

p = charpoly(S, x)

you find

p =
x^8 - 4040*x^7 + 5079941*x^6 + 82706090*x^5...
- 5327831918568*x^4 + 4287832912719760*x^3...
- 1082699388411166000*x^2 + 51264008540948000*x...
+ 40250968213600000

You also find that factor(p) is p itself. That is, the characteristic polynomial cannot be factored over the rationals.

For this modified Rosser matrix

F = eig(S)

returns

F =
-1020.053214255892
-0.17053529728769
0.2180398054830161
999.9469178604428
1000.120698293384
1019.524355263202
1019.993550129163
1020.420188201505

Notice that these values are close to the eigenvalues of the original Rosser matrix.

It is also possible to try to compute eigenvalues of symbolic matrices, but closed form solutions are rare. The Givens transformation is generated as the matrix exponential of the elementary matrix

$A=\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right].$

Symbolic Math Toolbox™ commands

syms t
A = sym([0 1; -1 0]);
G = expm(t*A)

return

G =
[           exp(-t*1i)/2 + exp(t*1i)/2,
(exp(-t*1i)*1i)/2 - (exp(t*1i)*1i)/2]
[ - (exp(-t*1i)*1i)/2 + (exp(t*1i)*1i)/2,
exp(-t*1i)/2 + exp(t*1i)/2]

You can simplify this expression using simplify:

G = simplify(G)
G =
[  cos(t), sin(t)]
[ -sin(t), cos(t)]

Next, the command

g = eig(G)

produces

g =
cos(t) - sin(t)*1i
cos(t) + sin(t)*1i

You can rewrite g in terms of exponents:

g = rewrite(g, 'exp')
g =
exp(-t*1i)
exp(t*1i)

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