Finding eigenspaces

WHAT we are finding

Why is $E_\lambda$ a subspace? Because it's the null space of the matrix $(A-\lambda I)$, and every null space is a subspace (closed under addition and scalar multiplication, contains $\mathbf 0$). That's the key trick — finding an eigenspace = solving a homogeneous system.

HOW we derive the recipe (from scratch)

We want nonzero $\mathbf v$ with $A\mathbf v = \lambda \mathbf v$.

Step 1 — Move everything to one side. $A\mathbf v - \lambda \mathbf v = \mathbf 0.$ Why? To collect $\mathbf v$ and turn this into a single matrix times $\mathbf v$.

Step 2 — Factor out $\mathbf v$. We can't write $(A-\lambda)\mathbf v$ because $A$ is a matrix and $\lambda$ is a scalar — they live in different worlds. Insert the identity: $\lambda \mathbf v = \lambda I \mathbf v$. $(A - \lambda I)\mathbf v = \mathbf 0.$ Why? Now $A-\lambda I$ is a genuine matrix, so this is a clean homogeneous system.

Step 3 — Demand a nonzero solution. A homogeneous system $M\mathbf v = \mathbf 0$ has a nonzero solution iff $M$ is not invertible, i.e. $\boxed{\det(A - \lambda I) = 0.}$ Why? If $A-\lambda I$ were invertible, the only solution would be $\mathbf v = \mathbf 0$ — useless. So singularity is exactly the condition for eigenvectors to exist. This equation in $\lambda$ is the characteristic equation.

Step 4 — For each eigenvalue $\lambda$, solve $(A-\lambda I)\mathbf v = \mathbf 0$. The solution set (the null space) is the eigenspace. A basis of it = a set of independent eigenvectors.

WORKED EXAMPLE 1 — a clean $2\times2$

$A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}.$

Characteristic equation. $\det\begin{pmatrix}2-\lambda & 1 \\ 1 & 2-\lambda\end{pmatrix} = (2-\lambda)^2 - 1 = 0.$ Why this step? Eigenvectors exist only where the matrix is singular, so we set the determinant to $0$.

Expand: $(2-\lambda)^2 = 1 \Rightarrow 2-\lambda = \pm 1 \Rightarrow \lambda = 1,\ 3.$

Eigenspace for $\lambda = 3$. Form $A - 3I = \begin{pmatrix} -1 & 1 \\ 1 & -1\end{pmatrix}$. Row reduce → $\begin{pmatrix} 1 & -1 \\ 0 & 0\end{pmatrix}$, giving $v_1 = v_2$. Why? Free variable $v_2 = t$ ⇒ $\mathbf v = t\binom{1}{1}$. $E_3 = \operatorname{span}\left\{\begin{pmatrix}1\\1\end{pmatrix}\right\}.$

Eigenspace for $\lambda = 1$. $A - I = \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix}$ → $v_1 = -v_2$. $E_1 = \operatorname{span}\left\{\begin{pmatrix}1\\-1\end{pmatrix}\right\}.$

Check: $A\binom{1}{1} = \binom{3}{3} = 3\binom{1}{1}$ ✓.

WORKED EXAMPLE 2 — a repeated eigenvalue with a 2D eigenspace

$A = \begin{pmatrix} 5 & 0 \\ 0 & 5\end{pmatrix}.$ $\det(A-\lambda I) = (5-\lambda)^2 = 0 \Rightarrow \lambda = 5$ (algebraic multiplicity 2).

$A - 5I = \begin{pmatrix}0&0\\0&0\end{pmatrix}$. Every vector satisfies $(A-5I)\mathbf v=\mathbf 0$, so $E_5 = \operatorname{span}\left\{\begin{pmatrix}1\\0\end{pmatrix},\begin{pmatrix}0\\1\end{pmatrix}\right\} = \mathbb R^2.$ Why this matters: the eigenspace dimension (geometric multiplicity) here is $2$ — equal to the repeat count.

WORKED EXAMPLE 3 — repeat eigenvalue but a "deficient" eigenspace

$A = \begin{pmatrix} 5 & 1 \\ 0 & 5\end{pmatrix}.$ Again $\det(A-\lambda I) = (5-\lambda)^2$, so $\lambda = 5$ (multiplicity 2).

$A-5I = \begin{pmatrix}0&1\\0&0\end{pmatrix}$ → equation $v_2 = 0$, $v_1$ free. $E_5 = \operatorname{span}\left\{\begin{pmatrix}1\\0\end{pmatrix}\right\}, \quad \dim E_5 = 1.$ Why this is the lesson: algebraic multiplicity $2$ but geometric multiplicity $1$. Geometric ≤ algebraic always, and when it's strictly less the matrix is not diagonalizable.

Common mistakes

Active recall

Connections

• Characteristic polynomial — produces the eigenvalues you feed in here.
• Null space and solving homogeneous systems — the engine of step 4.
• Diagonalization — possible exactly when eigenspaces' dimensions sum to $n$.
• Determinants — the singularity test in step 3.
• Rank-Nullity theorem — gives $\dim E_\lambda = n - \operatorname{rank}(A-\lambda I)$.
• Symmetric matrices and spectral theorem — guarantees full, orthogonal eigenspaces.

Concept Map

Hinglish (regional understanding)