Finding eigenspaces: Exercises — Maths deep dive

The whole game, in one sentence you already know from the parent:

Here $I$ is the identity matrix (1's on the diagonal, 0's elsewhere — the matrix that does nothing), and $Null (M)$ is the null space: all vectors $v$ with $M v = 0$ . If either phrase feels shaky, revisit Null space and solving homogeneous systems first.

LEVEL 1 — Recognition

Problem 1.1

You are told that for a matrix $A$ , the vector $v = (2 - 3)$ satisfies $A v = (8 - 12)$ . Is $v$ an eigenvector? If so, what is its eigenvalue?

Recall Solution 1.1

WHAT to check: an eigenvector satisfies $A v = λ v$ — the output must be the same arrow scaled, not turned. Compare output to input componentwise: $\frac{8}{2} = 4, \frac{- 12}{- 3} = 4.$ Both components scale by the same factor $4$ , so $A v = 4 v$ . Answer: Yes, $v$ is an eigenvector with eigenvalue $== λ = 4 ==$ .

Problem 1.2

For $A = (70 0 - 2)$ , write down the eigenvalues by inspection and one eigenvector for each.

Recall Solution 1.2

WHY inspection works: a diagonal matrix scales each coordinate axis independently — it never mixes them. So the standard axis arrows are already eigenvectors.

$A (10) = (70) = 7 (10)$ → $λ = 7$ , eigenvector $(10)$ .
$A (01) = (0 - 2) = - 2 (01)$ → $λ = - 2$ , eigenvector $(01)$ . Answer: eigenvalues $== 7 ==$ and $== - 2 ==$ along the two axes.

LEVEL 2 — Application

Problem 2.1

Find all eigenvalues and eigenspaces of $A = (4213) .$

Recall Solution 2.1

Step 1 — characteristic equation. Form $A - λ I = (4 - λ 2 1 3 - λ)$ (subtract $λ$ off the diagonal only). Its determinant is $det (A - λ I) = (4 - λ) (3 - λ) - (1) (2) = λ^{2} - 7 λ + 10.$ Set to $0$ : $λ^{2} - 7 λ + 10 = (λ - 2) (λ - 5) = 0$ , so $λ = 2, 5$ .

Step 2 — $E_{5}$ . $A - 5 I = (- 1 2 1 - 2)$ . Row 2 is $- 2 \times$ row 1, so it reduces to $(10 - 1 0)$ , meaning $v_{1} = v_{2}$ . Let $v_{2} = t$ : $E_{5} = span {(11)} .$ Step 3 — $E_{2}$ . $A - 2 I = (2211) \to (2010)$ , so $2 v_{1} + v_{2} = 0 \Rightarrow v_{2} = - 2 v_{1}$ : $E_{2} = span {(1 - 2)} .$ Check: $A (11) = (55) = 5 (11)$ ✓.

The picture below shows why those two lines are special: every arrow off the lines gets turned, the two eigen-lines only stretch.

Problem 2.2

Find the eigenvalues and eigenspaces of $A = (01 - 1 0) (a 90^{\circ} rotation).$

Recall Solution 2.2

Step 1. $det (- λ 1 - 1 - λ) = λ^{2} + 1 = 0$ . WHAT this means: $λ^{2} = - 1$ , so $λ = \pm i$ — no real eigenvalues. WHY, geometrically: this matrix rotates every real arrow by $9 0^{\circ}$ . No real arrow stays on its own line, so no real eigenvector can exist. The eigenvalues are complex. Answer: eigenvalues $== \pm i ==$ ; there are no real eigenspaces. (Over $C$ each eigenspace is a complex line, but over $R$ the eigenspace is ${0}$ only.)

LEVEL 3 — Analysis

Problem 3.1

Consider $A = 300030007 .$ Find every eigenvalue, its algebraic multiplicity (repeat count in the characteristic polynomial) and its geometric multiplicity ( $dim E_{λ}$ ). Comment on diagonalizability.

Recall Solution 3.1

Characteristic polynomial (diagonal ⇒ product of $(diag - λ)$ ): $det (A - λ I) = (3 - λ)^{2} (7 - λ) .$ So $λ = 3$ (algebraic multiplicity $2$ ) and $λ = 7$ (algebraic multiplicity $1$ ).

$E_{3}$ : $A - 3 I = 000000004$ , rank $1$ , so $dim E_{3} = 3 - 1 = 2$ . Basis: $100, 010$ . $E_{7}$ : $A - 7 I$ has rank $2$ , so $dim E_{7} = 3 - 2 = 1$ , basis $001$ .

Comment: geometric = algebraic for both eigenvalues ( $2 = 2$ and $1 = 1$ ), so $A$ is diagonalizable — indeed it's already diagonal.

Problem 3.2

For $A = (6026),$ find the eigenvalue, its algebraic and geometric multiplicities, and decide if $A$ is diagonalizable.

Recall Solution 3.2

Characteristic polynomial: upper-triangular, so $det (A - λ I) = (6 - λ)^{2}$ . One eigenvalue $λ = 6$ , algebraic multiplicity $2$ . $E_{6}$ : $A - 6 I = (0020)$ , rank $1$ . Equation $2 v_{2} = 0 \Rightarrow v_{2} = 0$ , $v_{1}$ free: $E_{6} = span {(10)}, dim E_{6} = 2 - 1 = 1.$ Analysis: geometric multiplicity $1 <$ algebraic $2$ . By Rank-Nullity theorem the null space is only 1-dimensional. So $A$ is not diagonalizable — the off-diagonal $2$ steals the second independent direction.

LEVEL 4 — Synthesis

Problem 4.1

Find the eigenvalues and eigenspaces of the symmetric matrix $A = 2 - 1 0 - 1 2 - 1 0 - 1 2 .$

Recall Solution 4.1

Step 1 — characteristic polynomial. Expanding $det (A - λ I)$ along the first row: $det (A - λ I) = (2 - λ) [(2 - λ)^{2} - 1] - (- 1) [- (2 - λ)] .$ $= (2 - λ) [(2 - λ)^{2} - 1] - (2 - λ) = (2 - λ) [(2 - λ)^{2} - 2] .$ Set $= 0$ : either $2 - λ = 0 \Rightarrow λ = 2$ , or $(2 - λ)^{2} = 2 \Rightarrow λ = 2 \pm 2$ . Eigenvalues: $λ = 2, 2 + 2, 2 - 2$ . Three distinct real values — expected, since symmetric matrices always have real eigenvalues.

$E_{2}$ : $A - 2 I = 0 - 1 0 - 1 0 - 1 0 - 1 0$ . Rows give $v_{2} = 0$ and $v_{1} + v_{3} = 0$ . Free $v_{3} = t$ : $E_{2} = span ⎩ ⎨ ⎧ 10 - 1 ⎭ ⎬ ⎫ .$ $E_{2 + 2}$ : solve $(A - (2 + 2) I) v = 0$ . Row 1: $- 2 v_{1} - v_{2} = 0 \Rightarrow v_{2} = - 2 v_{1}$ . Row 3: $- v_{2} - 2 v_{3} = 0 \Rightarrow v_{3} = - v_{2} / 2 = v_{1}$ . Take $v_{1} = 1$ : $E_{2 + 2} = span ⎩ ⎨ ⎧ 1 - 2 1 ⎭ ⎬ ⎫ .$ $E_{2 - 2}$ : same algebra with $+ 2$ : $v_{2} = 2 v_{1}$ , $v_{3} = v_{1}$ : $E_{2 - 2} = span ⎩ ⎨ ⎧ 121 ⎭ ⎬ ⎫ .$ Spectral bonus: the three eigenvectors are mutually orthogonal (check $10 - 1 \cdot 1 - 2 1 = 1 + 0 - 1 = 0$ ✓), exactly as the spectral theorem promises for symmetric $A$ .

LEVEL 5 — Mastery

Problem 5.1

Design a matrix. Build a $2 \times 2$ real matrix $A$ (not diagonal) whose eigenvalues are $λ = 1$ with eigenvector $(11)$ and $λ = 4$ with eigenvector $(1 - 1)$ .

Recall Solution 5.1

WHAT the tool is — diagonalization run backwards. If eigenvectors are the columns of $P$ and eigenvalues sit on a diagonal $D$ , then $A = P D P^{- 1}$ reconstructs $A$ . WHY this works: by construction $A P = P D$ column-by-column says $A (col i) = λ_{i} (col i)$ — exactly the eigen-equation. $P = (11 1 - 1), D = (1004), P^{- 1} = \frac{1}{- 2} (- 1 - 1 - 1 1) = \frac{1}{2} (11 1 - 1) .$ Multiply: $P D = (11 4 - 4), A = P D P^{- 1} = \frac{1}{2} (11 4 - 4) (11 1 - 1) = \frac{1}{2} (5 - 3 - 3 5) .$ $A = (2.5 - 1.5 - 1.5 2.5) .$ Verify: $A (11) = (11)$ (= $1 \cdot$ ) ✓, and $A (1 - 1) = (4 - 4)$ (= $4 \cdot$ ) ✓.

Problem 5.2

Constrain the free parameter. For which value(s) of the real number $c$ does $A = (20 c 2)$ have a 2-dimensional eigenspace? Explain using the Characteristic polynomial.

Recall Solution 5.2

The characteristic polynomial is $(2 - λ)^{2}$ for every $c$ (triangular), so $λ = 2$ with algebraic multiplicity $2$ always. The eigenspace is $Null (A - 2 I) = Null (00 c 0)$ .

If $c \neq = 0$ : rank $= 1$ , so $dim E_{2} = 2 - 1 = 1$ — a line only.
If $c = 0$ : the matrix is the zero matrix, rank $= 0$ , so $dim E_{2} = 2 - 0 = 2 = R^{2}$ . Answer: only $== c = 0 ==$ gives a 2-dimensional eigenspace (and only then is $A$ diagonalizable). The off-diagonal entry is exactly what collapses the eigenspace.

Wrap-up recall

Recall One-line summary of the whole ladder

Eigenvalues come from $det (A - λ I) = 0$ ; eigenspaces are $Null (A - λ I)$ ; dimension is $n - rank$ ; the repeat count only bounds that dimension; and you can run the whole thing backwards with $A = P D P^{- 1}$ .

Connections

Finding eigenspaces — the parent recipe these exercises drill.
Characteristic polynomial — where every $λ$ in these problems is born.
Null space and solving homogeneous systems — the machinery behind every eigenspace here.
Determinants — used in every characteristic equation.
Rank-Nullity theorem — the $dim E_{λ} = n - rank$ formula.
Diagonalization — L3 and L5 test when it succeeds and how to reverse it.
Symmetric matrices and spectral theorem — L4's orthogonal eigenvectors.