Unit 9, lesson 46

Eigenvalues and Diagonalization Workshop

Mixed practice with defective, rotation and symmetric matrices, covering the choice of scalar field, the defective test, complex eigenpairs, orthogonal diagonalization, matrix powers and the hypothesis behind each method.

  • Video 46
  • Eigenvalues and Diagonalization Workshop

Three 2×22 \times 2 matrices can look equally harmless and still lead to three different verdicts. One has a repeated eigenvalue but too few eigenvectors, one turns every nonzero real vector off its own line, and one has perpendicular eigenvectors that make its powers and its quadratic form easy to read. This workshop practices deciding which scalar field and which hypotheses are in play before calculating, then running the method that applies and checking the result against the original matrix. Tuples such as (1,0)(1, 0) stand for column vectors.

Key terms

Choice of scalar field

Definition Let AA be a real n×nn \times n matrix and let FF be R\mathbb{R} or C\mathbb{C}. An eigenvalue of AA over FF is a number λ\lambda in FF with Av=λvA\mathbf{v} = \lambda\mathbf{v} for some nonzero v\mathbf{v} in FnF^n, and these are exactly the roots of χA(t)=det⁡(tI−A)\chi_A(t) = \det(tI - A) that lie in FF. Over C\mathbb{C} the characteristic polynomial always factors into nn linear factors, counted with multiplicity, while over R\mathbb{R} it may not. The matrix AA is diagonalizable over FF if A=PDP−1A = PDP^{-1} with PP invertible and DD diagonal, both with entries in FF, which happens exactly when FnF^n has a basis of eigenvectors of AA. Diagonalizable over R\mathbb{R} implies diagonalizable over C\mathbb{C}, but not conversely. Changing the field changes which scalars and vectors are allowed, never the entries of AA.

In plain words The same real matrix can be diagonalizable over C\mathbb{C} but not over R\mathbb{R}, so a claim that a matrix is diagonalizable should name the field. Fix the field first, then look for eigenvalues and eigenvectors in it.

Example The matrix

A=[0−10100002]A = \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 2 \end{bmatrix}

turns the xyxy-plane through a quarter turn and stretches the zz-axis by 22. Its characteristic polynomial is χA(t)=(t2+1)(t−2)\chi_A(t) = (t^2 + 1)(t - 2). Over R\mathbb{R} the only eigenvalue is 22, and A−2IA - 2I has rank 22, so its eigenspace is the line spanned by e3\mathbf{e}_3. One line cannot contain a basis of R3\mathbb{R}^3, so AA is not diagonalizable over R\mathbb{R}. Over C\mathbb{C} it has the three distinct eigenvalues ii, −i-i and 22, with eigenvectors (1,−i,0)(1, -i, 0), (1,i,0)(1, i, 0) and (0,0,1)(0, 0, 1), so it is diagonalizable over C\mathbb{C}.

Test for a defective matrix

Definition Let AA be a real n×nn \times n matrix and let λ\lambda be an eigenvalue of AA, real or complex. The algebraic multiplicity am⁡(λ)\operatorname{am}(\lambda) is the multiplicity of λ\lambda as a root of χA\chi_A, and the geometric multiplicity is

gm⁡(λ)=dim⁡Null⁡(A−λI)=n−rank⁡(A−λI).\begin{aligned} \operatorname{gm}(\lambda) &= \dim \operatorname{Null}(A - \lambda I) \\[0.4em] &= n - \operatorname{rank}(A - \lambda I). \end{aligned}

For a nonreal λ\lambda the null space and the rank are taken over C\mathbb{C}. These always satisfy 1≤gm⁡(λ)≤am⁡(λ)1 \le \operatorname{gm}(\lambda) \le \operatorname{am}(\lambda). The matrix is defective if gm⁡(λ)<am⁡(λ)\operatorname{gm}(\lambda) < \operatorname{am}(\lambda) for at least one of its eigenvalues, real or complex. To test a matrix, compute rank⁡(A−λI)\operatorname{rank}(A - \lambda I) for each repeated eigenvalue. A simple root has both multiplicities equal to 11 and needs no check. A defective matrix has fewer than nn independent eigenvectors, so it is diagonalizable neither over R\mathbb{R} nor over C\mathbb{C}. If every root of χA\chi_A lies in the field FF, then AA is diagonalizable over FF exactly when it is not defective. Without that condition a matrix can fail to be defective and still not be diagonalizable over FF, as the quarter turn over R\mathbb{R} shows.

In plain words Count independent eigenvectors, not roots. A repeated root is a signal to check the eigenspace, not a verdict.

Example The matrices

J=[2102],2I=[2002]\begin{gathered} J = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}, \\[0.4em] 2I = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \end{gathered}

both have the characteristic polynomial (t−2)2(t - 2)^2, so am⁡(2)=2\operatorname{am}(2) = 2 for each. The matrix J−2IJ - 2I has rank 11, its only equation is y=0y = 0, and gm⁡(2)=1\operatorname{gm}(2) = 1. So JJ is defective, and all its eigenvectors lie on the line spanned by (1,0)(1, 0). Complex scalars change nothing, because the equation y=0y = 0 stays the same. For 2I2I the matrix 2I−2I2I - 2I is zero, so gm⁡(2)=2\operatorname{gm}(2) = 2 and 2I2I is already diagonal. Replacing JJ by diag⁡(2,2)\operatorname{diag}(2, 2) would predict J2=4IJ^2 = 4I, but in fact

J2=[4404].J^2 = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}.

Complex eigenpairs of a rotation

Definition Let aa and bb be real numbers with b≠0b \ne 0, and let

C=[a−bba].C = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}.

With r=a2+b2r = \sqrt{a^2 + b^2}, the matrix CC rotates the plane through the angle θ\theta with cos⁡θ=a/r\cos\theta = a/r and sin⁡θ=b/r\sin\theta = b/r and scales by rr. Its characteristic polynomial χC(t)=(t−a)2+b2\chi_C(t) = (t - a)^2 + b^2 has no real root, so CC has no real eigenvalue and no real eigenvector. Over C\mathbb{C} it has the conjugate eigenvalues a±bia \pm bi, with conjugate eigenvectors,

C[1−i]=(a+bi)[1−i],C[1i]=(a−bi)[1i].\begin{aligned} C\begin{bmatrix} 1 \\ -i \end{bmatrix} &= (a + bi)\begin{bmatrix} 1 \\ -i \end{bmatrix}, \\[0.4em] C\begin{bmatrix} 1 \\ i \end{bmatrix} &= (a - bi)\begin{bmatrix} 1 \\ i \end{bmatrix}. \end{aligned}

The eigenvector matrix with columns (1,−i)(1, -i) and (1,i)(1, i) has determinant 2i≠02i \ne 0, so CC is diagonalizable over C\mathbb{C}. Both eigenvalues have modulus rr, the scale factor.

In plain words A rotation that is not a multiple of a half turn moves every real line off itself, so there is no real eigenvector. Complex coordinates still provide two eigenvectors, and they come as a conjugate pair. They are columns of complex numbers, not arrows in the real plane.

Example The matrix

R=[0−220]R = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}

has a=0a = 0 and b=2b = 2, so it turns every vector through a quarter turn and doubles its length. Its characteristic polynomial is t2+4t^2 + 4, with roots ±2i\pm 2i. For 2i2i the check is R(1,−i)=(2i,2)=2i(1,−i)R(1, -i) = (2i, 2) = 2i(1, -i), using 2i⋅(−i)=22i \cdot (-i) = 2. Real consequences can be checked with real arithmetic. Since (±2i)2=−4(\pm 2i)^2 = -4, the diagonalization gives R2=−4IR^2 = -4I, and indeed two quarter turns make a half turn while lengths are multiplied by 44, so R2(1,0)=(−4,0)R^2(1, 0) = (-4, 0).

Real orthogonal diagonalization

Definition A real n×nn \times n matrix AA has a real orthogonal diagonalization

A=QDQ⊤,A = QDQ^{\top},

with QQ real orthogonal and DD real diagonal, if and only if AA is symmetric. This is the real spectral theorem together with the fact that QDQ⊤QDQ^{\top} is always symmetric. The method is to check A⊤=AA^{\top} = A, find the eigenvalues, which are real, find an orthonormal basis of each eigenspace, with Gram-Schmidt inside any eigenspace of dimension 22 or more, and place these vectors in QQ with their eigenvalues in matching order in DD. Because the columns of QQ are orthonormal, Q−1=Q⊤Q^{-1} = Q^{\top}, so no inverse has to be computed.

In plain words Symmetry guarantees a basis of perpendicular eigenvectors, and perpendicular unit eigenvectors make the inverse free. Without symmetry a real diagonalization may still exist, but it uses a slanted eigenbasis and a genuine inverse.

Example For

S=[3113],S = \begin{bmatrix} 3 & 1 \\ 1 & 3 \end{bmatrix},

the characteristic polynomial is (t−3)2−1=(t−4)(t−2)(t - 3)^2 - 1 = (t - 4)(t - 2). The matrix S−4IS - 4I gives the equation y=xy = x and the eigenvector (1,1)(1, 1), and S−2IS - 2I gives y=−xy = -x and the eigenvector (1,−1)(1, -1). Their dot product is 00, and normalizing gives q1=12(1,1)\mathbf{q}_1 = \frac{1}{\sqrt{2}}(1, 1) and q2=12(1,−1)\mathbf{q}_2 = \frac{1}{\sqrt{2}}(1, -1). With Q=[q1 q2]Q = [\mathbf{q}_1 \ \mathbf{q}_2] and D=diag⁡(4,2)D = \operatorname{diag}(4, 2), the spectral decomposition checks the result,

4q1q1⊤=[2222],2q2q2⊤=[1−1−11],\begin{aligned} 4\mathbf{q}_1\mathbf{q}_1^{\top} &= \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}, \\[0.4em] 2\mathbf{q}_2\mathbf{q}_2^{\top} &= \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}, \end{aligned}

and the sum of these two matrices is SS.

Matrix power from eigenvalues

Definition Suppose AA has an eigenbasis v1,…,vn\mathbf{v}_1, \dots, \mathbf{v}_n with Avi=λiviA\mathbf{v}_i = \lambda_i\mathbf{v}_i, so that A=PDP−1A = PDP^{-1} with P=[v1 ⋯ vn]P = [\mathbf{v}_1 \ \cdots \ \mathbf{v}_n] and D=diag⁡(λ1,…,λn)D = \operatorname{diag}(\lambda_1, \dots, \lambda_n). For every integer k≥0k \ge 0, Ak=PDkP−1A^k = PD^kP^{-1} with Dk=diag⁡(λ1k,…,λnk)D^k = \operatorname{diag}(\lambda_1^k, \dots, \lambda_n^k). To apply AkA^k to a vector, write x=c1v1+⋯+cnvn\mathbf{x} = c_1\mathbf{v}_1 + \cdots + c_n\mathbf{v}_n, where c=P−1x\mathbf{c} = P^{-1}\mathbf{x}, and then

Akx=c1λ1kv1+⋯+cnλnkvn.A^k\mathbf{x} = c_1\lambda_1^k\mathbf{v}_1 + \cdots + c_n\lambda_n^k\mathbf{v}_n.

With a real orthonormal eigenbasis q1,…,qn\mathbf{q}_1, \dots, \mathbf{q}_n the coefficients are dot products, ci=qi⋅xc_i = \mathbf{q}_i \cdot \mathbf{x}. The method needs an eigenbasis, so it does not apply to a defective matrix, and every coefficient must be kept, not only the one with the largest eigenvalue.

In plain words Split the vector along eigenvectors, multiply each piece by its eigenvalue raised to the kkth power, and add the pieces back up.

Example For SS above and e1=(1,0)\mathbf{e}_1 = (1, 0), the coefficients are q1⋅e1=q2⋅e1=1/2\mathbf{q}_1 \cdot \mathbf{e}_1 = \mathbf{q}_2 \cdot \mathbf{e}_1 = 1/\sqrt{2}, so e1=12(1,1)+12(1,−1)\mathbf{e}_1 = \frac{1}{2}(1, 1) + \frac{1}{2}(1, -1). Then

Ske1=4k2[11]+2k2[1−1]=12[4k+2k4k−2k].\begin{aligned} S^k\mathbf{e}_1 &= \frac{4^k}{2}\begin{bmatrix} 1 \\ 1 \end{bmatrix} + \frac{2^k}{2}\begin{bmatrix} 1 \\ -1 \end{bmatrix} \\[0.4em] &= \frac{1}{2}\begin{bmatrix} 4^k + 2^k \\ 4^k - 2^k \end{bmatrix}. \end{aligned}

For k=0k = 0 this returns (1,0)(1, 0), and for k=2k = 2 it gives (10,6)(10, 6). Direct multiplication agrees, since Se1=(3,1)S\mathbf{e}_1 = (3, 1) and S(3,1)=(10,6)S(3, 1) = (10, 6).

Hypothesis of a method

Definition The hypothesis of a method is the condition under which its conclusion is guaranteed, and it is checked before the method is used. The methods of this unit need the following.

  • Diagonalization A=PDP−1A = PDP^{-1} over a field FF needs nn linearly independent eigenvectors in FnF^n.
  • Real orthogonal diagonalization A=QDQ⊤A = QDQ^{\top} needs a real symmetric matrix.
  • The eigenvalue test for definiteness needs the symmetric matrix of the form, which for x⊤Mx\mathbf{x}^{\top}M\mathbf{x} is 12(M+M⊤)\frac{1}{2}(M + M^{\top}).
  • The power formula Ak=PDkP−1A^k = PD^kP^{-1} needs an eigenbasis.

When a hypothesis fails, the conclusion can fail too, and the failure points to what to do instead, such as working over C\mathbb{C}, computing powers directly or replacing a matrix by its symmetric part.

In plain words Before calculating, name the theorem you plan to use and check its conditions. Being square does not make a matrix eligible for every method.

Example The matrix

N=[1401]N = \begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix}

has the single eigenvalue 11, which is positive, but its form x⊤Nx=x2+4xy+y2\mathbf{x}^{\top}N\mathbf{x} = x^2 + 4xy + y^2 takes the value 1−4+1=−21 - 4 + 1 = -2 at (1,−1)(1, -1). So the eigenvalues of NN do not decide the sign of its form, because the eigenvalue test needs a symmetric matrix and NN is not symmetric. The symmetric matrix of this form has rows (1,2)(1, 2) and (2,1)(2, 1), with eigenvalues 33 and −1-1, so the form is indefinite. Similarly, the matrix with rows (1,1)(1, 1) and (0,2)(0, 2) is diagonalizable, with eigenvectors (1,0)(1, 0) and (1,1)(1, 1) for the eigenvalues 11 and 22, but it has no orthonormal eigenbasis. Each of its eigenspaces is a line, and the two lines are not perpendicular.

Check yourself

  1. Is the matrix AA with rows (4,1)(4, 1) and (−1,2)(-1, 2) diagonalizable over R\mathbb{R}? Over C\mathbb{C}?
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The characteristic polynomial is

χA(t)=(t−4)(t−2)+1=t2−6t+9=(t−3)2,\begin{aligned} \chi_A(t) &= (t - 4)(t - 2) + 1 \\[0.4em] &= t^2 - 6t + 9 \\[0.4em] &= (t - 3)^2, \end{aligned}

so 33 is the only eigenvalue, with am⁡(3)=2\operatorname{am}(3) = 2. The matrix A−3IA - 3I has rows (1,1)(1, 1) and (−1,−1)(-1, -1) and rank 11, so gm⁡(3)=1\operatorname{gm}(3) = 1, and the eigenspace is spanned by (1,−1)(1, -1). Check that A(1,−1)=(3,−3)A(1, -1) = (3, -3). Since gm⁡(3)<am⁡(3)\operatorname{gm}(3) < \operatorname{am}(3), the matrix is defective and is diagonalizable neither over R\mathbb{R} nor over C\mathbb{C}. Complex scalars do not help, because the only eigenvalue is already real and A−3IA - 3I has rank 11 over either field. The matrix is not symmetric either, so the spectral theorem gives no guarantee here.

  1. Find the complex eigenvalues and eigenvectors of the matrix BB with rows (3,−4)(3, -4) and (4,3)(4, 3). Explain why BB has no real eigenvector, and use the eigenvalues to find B2B^2.
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The matrix has the rotation form with a=3a = 3 and b=4b = 4. Its characteristic polynomial is (t−3)2+16=t2−6t+25(t - 3)^2 + 16 = t^2 - 6t + 25, whose discriminant 36−100=−6436 - 100 = -64 is negative, so there is no real eigenvalue and therefore no real eigenvector. Over C\mathbb{C} the eigenvalues are 3±4i3 \pm 4i, with eigenvector (1,−i)(1, -i) for 3+4i3 + 4i and (1,i)(1, i) for 3−4i3 - 4i. Check the first, B(1,−i)=(3+4i,4−3i)B(1, -i) = (3 + 4i, 4 - 3i), and (3+4i)(−i)=4−3i(3 + 4i)(-i) = 4 - 3i. Both eigenvalues have modulus 55, and BB rotates through the angle θ\theta with cos⁡θ=35\cos\theta = \frac{3}{5} and sin⁡θ=45\sin\theta = \frac{4}{5} and scales by 55.

Let PP have the columns (1,−i)(1, -i) and (1,i)(1, i) and let D=diag⁡(3+4i,3−4i)D = \operatorname{diag}(3 + 4i, 3 - 4i), so that B=PDP−1B = PDP^{-1}. The matrix B2=PD2P−1B^2 = PD^2P^{-1} has the same eigenvectors with the eigenvalues (3±4i)2=−7±24i(3 \pm 4i)^2 = -7 \pm 24i. The rotation form with a=−7a = -7 and b=24b = 24 has exactly these eigenpairs, and a matrix is determined by its action on a basis, so

B2=[−7−2424−7].B^2 = \begin{bmatrix} -7 & -24 \\ 24 & -7 \end{bmatrix}.

Direct multiplication agrees. Both diagonal entries are 9−16=−79 - 16 = -7, the entry above the diagonal is −12−12=−24-12 - 12 = -24, and the entry below it is 12+12=2412 + 12 = 24.

  1. Let TT be the matrix with rows (1,2)(1, 2) and (2,−2)(2, -2). Orthogonally diagonalize TT, use the result to find T3e1T^3\mathbf{e}_1, and classify the quadratic form x⊤Tx\mathbf{x}^{\top}T\mathbf{x}.
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The matrix is symmetric, so the real spectral theorem applies. Its characteristic polynomial is (t−1)(t+2)−4=t2+t−6=(t−2)(t+3)(t - 1)(t + 2) - 4 = t^2 + t - 6 = (t - 2)(t + 3). For λ=2\lambda = 2, the system (T−2I)x=0(T - 2I)\mathbf{x} = \mathbf{0} reduces to −x+2y=0-x + 2y = 0, giving (2,1)(2, 1). For λ=−3\lambda = -3, it reduces to 4x+2y=04x + 2y = 0, giving (1,−2)(1, -2). These are orthogonal with length 5\sqrt{5}, so

Q=15[211−2],D=[200−3].\begin{gathered} Q = \frac{1}{\sqrt{5}}\begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix}, \\[0.4em] D = \begin{bmatrix} 2 & 0 \\ 0 & -3 \end{bmatrix}. \end{gathered}

The coefficients of e1\mathbf{e}_1 are q1⋅e1=2/5\mathbf{q}_1 \cdot \mathbf{e}_1 = 2/\sqrt{5} and q2⋅e1=1/5\mathbf{q}_2 \cdot \mathbf{e}_1 = 1/\sqrt{5}, so e1=25(2,1)+15(1,−2)\mathbf{e}_1 = \frac{2}{5}(2, 1) + \frac{1}{5}(1, -2). Then

T3e1=2⋅85[21]+(−3)35[1−2]=15[32−2716+54]=[114].\begin{aligned} T^3\mathbf{e}_1 &= \frac{2 \cdot 8}{5}\begin{bmatrix} 2 \\ 1 \end{bmatrix} \\[0.4em] &\quad + \frac{(-3)^3}{5}\begin{bmatrix} 1 \\ -2 \end{bmatrix} \\[0.4em] &= \frac{1}{5}\begin{bmatrix} 32 - 27 \\ 16 + 54 \end{bmatrix} = \begin{bmatrix} 1 \\ 14 \end{bmatrix}. \end{aligned}

Direct multiplication agrees, since Te1=(1,2)T\mathbf{e}_1 = (1, 2), T(1,2)=(5,−2)T(1, 2) = (5, -2) and T(5,−2)=(1,14)T(5, -2) = (1, 14).

The form is x2+4xy−2y2x^2 + 4xy - 2y^2. In the coordinates z1=(2x+y)/5z_1 = (2x + y)/\sqrt{5} and z2=(x−2y)/5z_2 = (x - 2y)/\sqrt{5} it becomes 2z12−3z222z_1^2 - 3z_2^2. One eigenvalue is positive and one negative, so the form is indefinite, with witnesses q(1,0)=1q(1, 0) = 1 and q(0,1)=−2q(0, 1) = -2.

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