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The matrix $(\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array})$ has characteristic polynomial $(x - 2) (x - 3)$ and two explicitly computed eigenspaces

Example

Over $R$ , let

$A = (\begin{array}{cc} 2 & 1 \\ 0 & 3 \end{array}) .$

Then $χ_{A} (x) = (x - 2) (x - 3)$ , $E_{2} (A) = span {(1, 0)}$ , and $E_{3} (A) = span {(1, 1)}$ .

Facts & Assumptions

Given: The displayed real matrix $A$ .

[L1]

The characteristic polynomial is $\det (x I - A)$ (For $A \in M_{n} (F)$ , the characteristic polynomial is $χ_{A} (x) = \det (x I_{n} - A)$ when $n \geq 1$ , with $χ_{A} (x) = 1$ for the unique $0 \times 0$ matrix).

[L2]

The eigenspace is $E_{λ} (A) = \ker (A - λ I)$ (Eigenvalues, eigenvectors, eigenspaces $E_{λ} (T) = \ker (T - λ I)$ , and the spectrum $σ_{F} (T)$ of an endomorphism).

Verification

technique · direct computation

1.1

By [L1], $χ_{A} (x) = \det (\begin{array}{cc} x - 2 & - 1 \\ 0 & x - 3 \end{array}) = (x - 2) (x - 3)$ .

L1algebra

1.2

The equation $(A - 2 I) (u, v) = (v, v) = (0, 0)$ gives $v = 0$ , hence $E_{2} (A) = {(u, 0) : u \in R} = span {(1, 0)}$ .

L2algebra

1.3

The equation $(A - 3 I) (u, v) = (- u + v, 0) = (0, 0)$ gives $v = u$ , hence $E_{3} (A) = span {(1, 1)}$ .

L2algebra

2.1

The two nonzero spanning vectors also directly exhibit both roots as eigenvalues.

step 1.1step 1.2step 1.3L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 27 results over 8 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

H. Pinkham, Linear Algebra, §12.1 (standard reference, not scraped)

The matrix (2103)(20​13​) has characteristic polynomial (x−2)(x−3)(x−2)(x−3) and two explicitly computed eigenspaces