Finding eigenspaces: Worked examples — Hinglish deep dive

4.5.30 · D3 · Maths › Linear Algebra (Full) › Finding eigenspaces

Yeh page Finding eigenspaces ka drill floor hai. Parent note ne recipe banai thi; yahan hum use har tarah ki matrix ke saath run karte hain jo tumhe mil sakti hai — clean roots, repeated roots, ek singular matrix, ek rotation jisme koi real eigenvector nahi, ek triangular matrix jise tum seedha dekh ke padh sakte ho, ek symmetric $3 \times 3$ , aur ek word problem. Har worked example ko scenario matrix ke cell ke saath tag kiya gaya hai jo use cover karta hai, toh end tak tumne har case hit kar liya hoga.

Shuru karne se pehle, kuch notation jis par hum rely karte hain, taaki koi cheez bina explanation ke use na ho:

Definition Size, dimension, rank, aur do multiplicities

Yeh paanch words hum neeche baar baar use karte hain, toh inhe abhi plain language mein samjho:

$n$ = matrix ki size: ek $n \times n$ matrix $n$ entries wale vectors par act karta hai. Ek $2 \times 2$ matrix ke liye $n = 2$ ; $3 \times 3$ example ke liye $n = 3$ .
$dim$ (dimension) of a space = us space mein independent directions ki sankhya. Ek line ka $dim = 1$ hota hai; ek flat plane ka $dim = 2$ ; ek single point (sirf $0$ ) ka $dim = 0$ .
rank of a matrix $M$ = row-reduction ke baad bachi hui independent rows ki sankhya — yaani kitni genuine equations $M v = 0$ actually impose hoti hain. Ek $2 \times 2$ matrix jo ek nonzero row tak reduce hoti hai uska rank $1$ hai; zero matrix ka rank $0$ hai.
algebraic multiplicity of $λ$ = kitni baar $(λ - woh value)$ characteristic polynomial $det (A - λ I)$ ke factor ke roop mein aata hai (dekho Characteristic polynomial). Double root $(λ - 3)^{2}$ ki algebraic multiplicity $2$ hai.
geometric multiplicity of $λ$ = uski eigenspace ki dimension, $dim E_{λ}$ — uske paas actually kitne independent eigenvectors hain. Rank-Nullity theorem se yeh is tarah connect hote hain: $dim E_{λ} = n - rank (A - λ I),$ yaani (variables ki sankhya) minus (independent equations ki sankhya) = (free directions ki sankhya).

Scenario matrix

Har eigenvalue problem in boxes mein se kisi ek mein fit hota hai. Neeche hamara kaam inhe sab fill karna hai.

Cell	Scenario	Kya galat ho sakta hai / surprising ho sakta hai	Example
A	Do distinct real eigenvalues, dono nonzero	routine, lekin $λ$ ka sign dekho	Ex 1
B	Ek eigenvalue zero ke barabar ( $A$ singular)	"zero eigenvalue nahi ho sakta" wala trap	Ex 2
C	Ek negative eigenvalue (ek flip)	eigenvector phir bhi real line hai, bas reversed	Ex 3
D	Repeated root, full (2D) eigenspace	geometric = algebraic multiplicity	Ex 4
E	Repeated root, deficient eigenspace	geometric < algebraic → not diagonalizable	Ex 5
F	Koi real eigenvalues nahi (pure rotation)	roots complex hain; koi real line survive nahi karti	Ex 6
G	Triangular matrix	eigenvalues seedha diagonal se padho	Ex 7
H	Symmetric $3 \times 3$	hamesha real, eigenspaces perpendicular	Ex 8
I	Word problem (population / limiting behaviour)	dominant eigenvalue ko physically interpret karo	Ex 9

Ex 1 — Cell A: do distinct nonzero real eigenvalues

Forecast: Compute karne se pehle guess karo — kya do eigenvalues bahut alag honge ya karib? (Row sums $5$ aur $5$ hain, ek hint ki $(1 1)$ shayad special ho.)

$A - λ I$ banao. $A - λ I = (4 - λ 2 1 3 - λ) .$ Yeh step kyun? Eigenvectors sirf wahan exist karte hain jahan yeh matrix singular ho, toh pehle ise build karna zaroori hai.
Determinant zero set karo (dekho Determinants). $det = (4 - λ) (3 - λ) - (1) (2) = λ^{2} - 7 λ + 10 = 0.$ Yeh step kyun? Homogeneous system $M v = 0$ ka nonzero solution tab hoga jab $det M = 0$ . Singularity nahi, toh eigenvector nahi.
Quadratic solve karo. $λ^{2} - 7 λ + 10 = (λ - 2) (λ - 5) = 0 \Rightarrow λ = 2, 5.$ Yeh step kyun? Yeh do stretch factors hain; dono positive, toh dono genuine stretches hain (koi flip nahi).
Eigenspace for $λ = 5$ . $A - 5 I = (- 1 2 1 - 2) \to (10 - 1 0)$ , toh $v_{1} = v_{2}$ . $E_{5} = span {(11)} .$ Yeh step kyun? Row-reducing matrix equation ko ek plain equation $v_{1} - v_{2} = 0$ mein turn karta hai; free variable poori line trace karta hai.
Eigenspace for $λ = 2$ . $A - 2 I = (2211) \to (2010)$ , toh $2 v_{1} + v_{2} = 0$ , yaani $v_{2} = - 2 v_{1}$ . $E_{2} = span {(1 - 2)} .$ Yeh step kyun? Doosre eigenvalue ke liye same row-reduction idea; ek surviving equation $2 v_{1} + v_{2} = 0$ poori eigen-line define karta hai.

Verify: $A (1 1) = (5 5) = 5 (1 1)$ ✓ aur $A (- 2 1) = (- 4 2) = 2 (- 2 1)$ ✓.

Yeh figure kaise padhen. Axes ordinary $x$ – $y$ plane hain (standard Cartesian coordinates, origin centre mein). Blue line $E_{5}$ hai, direction $(1, 1)$ ; yellow line $E_{2}$ hai, direction $(1, - 2)$ . Grey arrow ek generic vector $v$ hai jo kisi bhi eigen-line par nahi hai, aur red arrow uska image $A v$ hai — notice karo yeh ek alag direction mein point karta hai (rotate ho gaya). Iska lesson: sirf woh arrows jo do coloured lines par hain unki direction $A$ ke under same rehti hai; baki sab nayi line par chale jaate hain.

Ex 2 — Cell B: ek eigenvalue zero ke barabar

Forecast: Bottom row top row ki half hai. Predict karo: kya yeh matrix invertible hai? Agar nahi, toh $det A$ — aur isliye ek eigenvalue — kya forced hoga?

Characteristic equation. $det (A - λ I) = (2 - λ)^{2} - 4 = λ^{2} - 4 λ = λ (λ - 4) = 0.$ Yeh step kyun? Standard: singularity condition. Notice karo constant term $det A = 4 - 4 = 0$ hai.
Roots padho. $λ = 0$ aur $λ = 4$ . Yeh step kyun? Root $λ = 0$ aana matlab $det A = 0$ : matrix kisi direction ko squash kar deta hai. Woh squashed direction hi eigenvalue $0$ wala eigenvector hai.
Eigenspace $E_{0} = Null (A)$ . $A v = 0$ solve karo: $(2142) \to (1020)$ , toh $v_{1} = - 2 v_{2}$ . $E_{0} = span {(- 2 1)} .$ Yeh step kyun? $A - 0 \cdot I = A$ , toh $0$ ka eigenspace literally $A$ ka null space hai.
Eigenspace $E_{4}$ . $A - 4 I = (- 2 1 4 - 2) \to (10 - 2 0)$ , toh $v_{1} = 2 v_{2}$ : $E_{4} = span {(1 2)}$ . Yeh step kyun? Nonzero eigenvalue ke liye same recipe; surviving equation $v_{1} - 2 v_{2} = 0$ $λ = 4$ ke liye eigen-line trace karta hai.

Verify: $A (1 - 2) = (0 0) = 0 \cdot (1 - 2)$ ✓ aur $A (1 2) = (4 8) = 4 (1 2)$ ✓.

Ex 3 — Cell C: ek negative eigenvalue (ek flip)

Forecast: Yeh matrix vector ki do entries swap karta hai: $(b a) \mapsto (a b)$ . Kaunsa arrow swap se reverse hota hai? Kaunsa same rehta hai?

Characteristic equation. $det (- λ 1 1 - λ) = λ^{2} - 1 = 0 \Rightarrow λ = \pm 1.$ Yeh step kyun? Hum ek positive aur ek negative root expect karte hain kyunki swap kisi direction ko rakhta hai aur doosre ko flip karta hai.
$λ = 1$ . $A - I = (- 1 1 1 - 1) \to v_{1} = v_{2}$ : $E_{1} = span {(1 1)}$ . Yeh "same rehne wali" direction hai (swap of $(1 1)$ is $(1 1)$ ). Yeh step kyun? Row-reducing ek equation $v_{1} = v_{2}$ deta hai, jiski solution line woh fixed direction hai jo humne predict ki thi.
$λ = - 1$ . $A + I = (1111) \to v_{1} = - v_{2}$ : $E_{- 1} = span {(- 1 1)}$ . Yeh step kyun? $λ = - 1$ matlab arrow origin se flip ho jaata hai — same line, opposite direction. $(- 1 1)$ ko swap karne par $(1 - 1) = - (- 1 1)$ milta hai ✓.

Verify: $A (1 1) = (1 1) = 1 \cdot (1 1)$ ✓ aur $A (- 1 1) = (1 - 1) = - 1 \cdot (- 1 1)$ ✓.

Yeh figure kaise padhen. Standard $x$ – $y$ axes phir se. Blue line $E_{1}$ hai: direction $(1, 1)$ jise swap bilkul chhod deta hai. Red line $E_{- 1}$ hai: direction $(1, - 1)$ . Lower-right par red arrow eigenvector $(1, - 1)$ hai; upper-left par red arrow uska image $A (1, - 1) = (- 1, 1)$ hai — same line, lekin arrowhead opposite end par flip ho gaya hai. Lesson: ek negative eigenvalue vector ko uski apni line par rakhta hai lekin direction reverse kar deta hai.

Ex 4 — Cell D: repeated root, full 2D eigenspace

Forecast: $λ = 3$ ek double root hoga. Guess karo: kya eigenspace ek line hai ya poora plane?

Characteristic equation. $(3 - λ)^{2} = 0 \Rightarrow λ = 3$ , algebraic multiplicity $2$ . Yeh step kyun? $det (A - λ I) = 0$ set karna hamesha pehla move hai; yahan yeh humein ek single repeated root deta hai, toh next check karna hai ki uski eigenspace actually kitni badi hai.
$A - 3 I = (0000)$ banao. Yeh step kyun? Yeh jo bhi row-reduce hoga woh batayega kitne free variables (hence dimensions) eigenspace mein hain.
Har vector $0 \cdot v = 0$ solve karta hai. Dono variables free hain. $E_{3} = span {(10), (01)} = R^{2} .$ Yeh step kyun? Zero matrix ke saath satisfy karne ke liye koi equations nahi hain, toh dono coordinates free hain — null space, aur isliye eigenspace, poora plane hai.

Yahan size $n = 2$ hai, aur $rank (A - 3 I) = 0$ (zero matrix koi equations impose nahi karta), toh $dim E_{3} = n - rank (A - 3 I) = 2 - 0 = 2$ — geometric multiplicity (eigenspace ki dimension) algebraic multiplicity (double root) ke barabar hai. Toh $A$ diagonalizable hai (trivially, yeh already diagonal hai). Dekho Diagonalization.

Verify: $A (0 1) = 3 (0 1)$ aur $A (1 0) = 3 (1 0)$ ✓.

Ex 5 — Cell E: repeated root, deficient eigenspace

Forecast: Same double root $λ = 3$ jaise Ex 4 mein. Kya eigenspace phir se poora plane hoga? Step 3 se pehle predict karo.

Characteristic equation. $(3 - λ)^{2} = 0 \Rightarrow λ = 3$ , algebraic multiplicity $2$ — Ex 4 se identical. Yeh step kyun? Hum hamesha singularity condition se start karte hain; yahan interesting baat yeh hai ki polynomial Ex 4 se identical hai, phir bhi eigenspace alag nikalega.
$A - 3 I = (0010)$ banao. Yeh step kyun? Off-diagonal $1$ survive karta hai, aur yeh humara ek dimension kha jaayega.
Solve karo. Ek hi equation hai $v_{2} = 0$ ; $v_{1}$ free hai. $E_{3} = span {(10)} .$ Yeh step kyun? Surviving row $v_{2} = 0$ force karta hai, sirf ek free variable bachti hai — toh eigenspace ek single line hai, plane nahi.

Ab dimension count karo. Yahan $n = 2$ hai aur matrix $(0010)$ mein ek independent nonzero row hai, toh $rank (A - 3 I) = 1$ . Isliye $dim E_{3} = n - rank (A - 3 I) = 2 - 1 = 1$ (definitions box se Rank-Nullity theorem relation use karke). Toh algebraic multiplicity $2$ hai lekin geometric multiplicity $1$ hai, aur kyunki $1 < 2$ , yeh matrix not diagonalizable hai.

Verify: $A (0 1) = (0 3) = 3 (0 1)$ ✓. Aur $λ = 3$ ke liye genuinely koi doosra independent eigenvector nahi hai.

Recall Same root, alag fate

Ex 4 aur Ex 5 dono ka same characteristic polynomial $(λ - 3)^{2}$ hai. Repeat count sirf eigenspace dimension ko upar se bound karta hai — actual dimension $A - λ I$ ke rank se aati hai. Off-diagonal 1 ek dimension kyun kill karta hai? ::: Yeh $rank (A - 3 I)$ ko $0$ se $1$ tak raise karta hai, toh $dim E_{3} = 2 - 1 = 1$ .

Ex 6 — Cell F: koi real eigenvalues nahi (ek rotation)

Forecast: Pure rotation har arrow ko turn karta hai. Predict karo: kitne arrows apni khud ki line par same reh sakte hain? Iska real eigenvalues ke liye kya matlab hoga?

Characteristic equation. $det (- λ 1 - 1 - λ) = λ^{2} + 1 = 0.$ Yeh step kyun? Same recipe — lekin ab constant ka sign dekho.
Solve karo. $λ^{2} = - 1$ . Koi real number nahi hai jiska square $- 1$ ho; roots $λ = \pm i$ (imaginary) hain. Yeh step kyun? Ek real $λ$ ek real fixed line demand karta. $9 0^{\circ}$ rotation koi real direction fix nahi karta, toh algebra ko real answer refuse karna hi padega — aur karta hai.
Conclusion. Real numbers par, $A$ ke paas koi eigenvalues aur koi eigenspaces nahi hain. (Complex numbers par $λ = \pm i$ hain, lekin koi real invariant line nahi.) Yeh step kyun? Hum real-vs-complex distinction report karte hain kyunki is cell ka pura point ek aisi matrix hai jisme koi real eigenvector dhundhne ko nahi hai.

Verify: Koi bhi nonzero real $v = (b a)$ lo; $A v = (a - b)$ . Iske liye $λ (b a)$ ke barabar hone ke liye hamein $- b = λa$ aur $a = λb$ chahiye, jo $a = - λ^{2} a$ deta hai, yaani $λ^{2} = - 1$ — real $λ$ ke liye impossible jab tak $a = b = 0$ na ho. ✓ Toh koi real eigenvector exist nahi karta.

Yeh figure kaise padhen. Standard $x$ – $y$ axes, origin centred. Blue arrows circle ke around har direction mein point karte aath sample unit vectors $v$ hain. Har green arrow us vector ka image $A v$ hai — quarter-turn rotation. Kisi bhi blue arrow ko uske green partner se compare karo: green wala $9 0^{\circ}$ counter-clockwise swing ho gaya hai, toh har baar alag line par land karta hai. Lesson: kyunki koi bhi arrow apni original line par end nahi hota, koi real eigenvector nahi hai — yeh complex eigenvalues ka visual signature hai.

Ex 7 — Cell G: triangular matrix (diagonal se seedha padho)

Forecast: Lower-left corner mein sab zeros hain. Determinant compute kiye bina eigenvalues guess karo.

Triangular shortcut use karo. Ek (upper- ya lower-) triangular matrix ke liye, $det (A - λ I)$ $A - λ I$ ki diagonal entries ka product hai: $det (A - λ I) = (2 - λ) (5 - λ) (1 - λ) .$ Yeh step kyun? Ek triangular matrix ka determinant bas uski diagonal entries ka product hota hai (diagonal ke neeche zeros ki wajah se saare off-diagonal cofactor terms vanish ho jaate hain — dekho Determinants). Toh hame poora characteristic polynomial already factored form mein milta hai, bina koi expansion work kiye.
Eigenvalues: $λ = 2, 5, 1.$ Yeh step kyun? Har factor $(d - λ)$ tab zero hoga jab $λ$ us diagonal entry $d$ ke barabar ho, toh diagonal entries hi eigenvalues hain.
Eigenvector for $λ = 2$ . $A - 2 I = 000730 - 3 4 - 1$ . Rows 2 aur 3 force karte hain $v_{3} = 0$ phir $v_{2} = 0$ ; $v_{1}$ free. $E_{2} = span ⎩ ⎨ ⎧ 100 ⎭ ⎬ ⎫ .$ Yeh step kyun? Hum phir bhi homogeneous system solve karte hain — shortcut ne sirf $λ$ 's dhundhe, vectors nahi.

Verify: $A 100 = 200 = 2100$ ✓, aur product $(2 - λ) (5 - λ) (1 - λ)$ true characteristic polynomial tak expand hota hai (neeche check kiya).

Ex 8 — Cell H: symmetric $3 \times 3$ (perpendicular eigenspaces)

Forecast: $A$ apne transpose ke barabar hai (diagonal ke across mirror). Symmetric matrices and spectral theorem ke through solve karne se pehle do facts predict karo: kya eigenvalues real honge? kya eigenvectors perpendicular honge?

Characteristic polynomial (pehle column ke along expand karo kyunki yeh block isolate karta hai): $det (A - λ I) = (2 - λ) [(3 - λ)^{2} - 1] .$ Yeh step kyun? Top-left $2$ decoupled hai (uski row aur column aur jagah zero hain), toh problem ek $1 \times 1$ aur ek $2 \times 2$ piece mein split hoti hai — full $3 \times 3$ expansion se kahin aasan.
Solve karo. $(3 - λ)^{2} - 1 = 0 \Rightarrow 3 - λ = \pm 1 \Rightarrow λ = 2, 4$ . Front factor $λ = 2$ ke saath: $λ = 2 (do baar), 4.$ Yeh step kyun? Har factor ke roots padhne se eigenvalues milte hain; $λ = 2$ ke do alag sources hume batate hain ki yeh double root hai (algebraic multiplicity $2$ ).
$E_{4}$ . $A - 4 I = - 2 00 0 - 1 1 01 - 1$ : force karta hai $v_{1} = 0$ , $v_{2} = v_{3}$ . $E_{4} = span ⎩ ⎨ ⎧ 011 ⎭ ⎬ ⎫ .$ Yeh step kyun? Row-reducing $v_{1} = 0$ aur $v_{2} - v_{3} = 0$ deta hai, ek free variable bachti hai — ek single eigen-line.
$E_{2}$ . $A - 2 I = 000011011$ : ek equation $v_{2} + v_{3} = 0$ ; $v_{1}$ free, $v_{3}$ free. $E_{2} = span ⎩ ⎨ ⎧ 100, 01 - 1 ⎭ ⎬ ⎫, dim E_{2} = 2.$ Yeh step kyun? Yahan $n = 3$ hai aur $A - 2 I$ mein sirf ek independent nonzero row hai, toh $rank (A - 2 I) = 1$ aur $dim E_{2} = 3 - 1 = 2$ . Symmetric matrices hamesha diagonalizable hote hain, toh double root $λ = 2$ ko zaroor full 2-dimensional eigenspace deni chahiye — aur ek surviving equation $v_{2} + v_{3} = 0$ exactly do free variables chodti hai, jo ise confirm karta hai: geometric multiplicity $=$ algebraic multiplicity $= 2$ .

Verify: Pehle eigen-equations check karo: $A 011 = 044 = 4011$ ✓, $A 100 = 200 = 2100$ ✓, aur $A 01 - 1 = 02 - 2 = 2 01 - 1$ ✓. Ab spectral-theorem ka promise confirm karo ki alag eigenspaces ke eigenvectors perpendicular hote hain (dot product $= 0$ ): $011 \cdot 100 = 0, 011 \cdot 01 - 1 = 0, 100 \cdot 01 - 1 = 0.$ Teeno dot products vanish karte hain ✓ — exactly jaisa Symmetric matrices and spectral theorem guarantee karta hai: ek real symmetric matrix ke hamesha real eigenvalues aur mutually perpendicular eigenvectors ka full set hota hai.

Ex 9 — Cell I: word problem (limiting behaviour)

Forecast: Lambe time baad split change hona band ho jaata hai. "Change hona band" kaunse eigenvalue se correspond karta hai? Step 2 se pehle guess karo.

Steady state = eigenvector for $λ = 1$ . Ek distribution jo aur move nahi karta $A v = v = 1 \cdot v$ satisfy karta hai. Yeh step kyun? "Saal dar saal unchanged" precisely eigenvalue- $1$ condition hai — steady state wahi eigenvector hai jo hum chahte hain.
Confirm karo ki $λ = 1$ ek eigenvalue hai. $A$ ke column sums $1$ hain (yeh ek stochastic matrix hai), jo $λ = 1$ force karta hai. Check: $det (A - I) = det (- 0.2 0.2 0.3 - 0.3) = (- 0.2) (- 0.3) - (0.3) (0.2) = 0.06 - 0.06 = 0$ ✓ (singular, toh $1$ actually ek eigenvalue hai). Yeh step kyun? Hume verify karna hoga ki $λ = 1$ sach mein ek eigenvalue hai uska eigenvector solve karne se pehle; zero determinant confirm karta hai ki $A - I$ singular hai.
$E_{1}$ solve karo. $A - I = (- 0.2 0.2 0.3 - 0.3)$ : equation $- 0.2 v_{1} + 0.3 v_{2} = 0 \Rightarrow v_{1} = \frac{3}{2} v_{2}$ . $E_{1} = span {(32)} .$ Yeh step kyun? $A - I$ ko row-reduce karne par ek relation $v_{1} = \frac{3}{2} v_{2}$ milta hai, jiski direction $(3, 2)$ unchanging split hai.
Percentages mein interpret karo. Scale karo taaki entries sum to $1$ hoon: split $= (\frac{3}{5}, \frac{2}{5}) = (60%, 40%)$ . Yeh step kyun? Populations proportions hain, toh hum eigenvector ko normalise karte hain taaki woh $1$ tak sum kare aur percentages padh sakein.

Verify: $A (2 3) = (0.2 \cdot 3 + 0.7 \cdot 2 0.8 \cdot 3 + 0.3 \cdot 2) = (0.6 + 1.4 2.4 + 0.6) = (2 3)$ ✓ — split $60%/40%$ hamesha ke liye khud ko reproduce karta hai. (Doosra eigenvalue $λ = 0.5$ hai, jo decay ho jaata hai, toh koi bhi starting split $60/40$ ki taraf funnel karta hai.)

Recall

Kaunse cell mein koi real eigenspace nahi hai, aur kyun? ::: Cell F (rotation): $λ^{2} + 1 = 0$ ke koi real roots nahi hain kyunki rotation se koi real line fix nahi hoti. Triangular matrix ke eigenvalues instantly kaise milte hain? ::: Diagonal se seedha padho — $det (A - λ I)$ diagonal entries ka product hai. Ek stochastic (column-sum-1) matrix mein kaunsa eigenvalue hamesha aata hai? ::: $λ = 1$ ; uska eigenvector (normalised) steady-state distribution hota hai. Do matrices ka polynomial $(λ - 3)^{2}$ share karte hain — unki eigenspaces dimension mein kyun differ kar sakti hain? ::: $dim E_{3} = n - rank (A - 3 I)$ ; rank $0$ ho sakta hai (full 2D space) ya $1$ (deficient line). Algebraic aur geometric multiplicity mein kya fark hai? ::: Algebraic = kitni baar $λ$ characteristic polynomial ka root hai; geometric = $dim E_{λ}$ . Hamesha $1 \leq$ geometric $\leq$ algebraic.

Connections

Finding eigenspaces — yeh parent recipe hai jise yeh examples drill karte hain.
Characteristic polynomial — upar har "characteristic equation" step issi se aata hai.
Null space and solving homogeneous systems — har $E_{λ}$ ek null space hai jise hum row-reduce karte hain.
Determinants — $det (A - λ I) = 0$ singularity test.
Rank-Nullity theorem — $dim E_{λ} = n - rank (A - λ I)$ deta hai (Ex 5).
Diagonalization — Ex 4 vs Ex 5 exactly diagonalizable / not test hai.
Symmetric matrices and spectral theorem — Ex 8 ke real, perpendicular eigenspaces.