[Axler] Eigenvalues, Invariant Subspaces, Diagonal Matrix

Eigenvalues, Eigenvectors, and Invariant Subspaces (Part 2 & 3)

Based on Linear Algebra Done Right

This post covers the definition of the matrix of an operator, the properties of upper triangular matrices, and the concept of eigenspaces.

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1. The Matrix of an Operator

Previously, defining the matrix of a linear map from one vector space to another required two bases. However, for an operator (a linear map $T: V \to V$), we use a single basis.

Definition

Let $v_1, \dots, v_n$ be a basis of $V$. The matrix of the operator $T$ with respect to this basis is the $n \times n$ matrix determined by:

$$ T(v_k) = A_{1,k}v_1 + \dots + A_{n,k}v_n $$

The coefficients of this linear combination form the $k$-th column of the matrix.

• It is computed using a single basis of $V$.
• It is always a square matrix ($n \times n$).

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2. Upper Triangular Matrix

A matrix is called upper triangular if all entries below the diagonal are zero (i.e., $A_{j,k} = 0$ if $j > k$).

Example Calculation:
Define $T \in \mathcal{L}(\mathbf{R}^3)$ by $T(x, y, z) = (2x + y, 5y + 3z, 8z)$.
Using the standard basis $(1,0,0), (0,1,0), (0,0,1)$:

• $T(1,0,0) = (2,0,0) \implies$ Col 1: $[2, 0, 0]^T$
• $T(0,1,0) = (1,5,0) \implies$ Col 2: $[1, 5, 0]^T$
• $T(0,0,1) = (0,3,8) \implies$ Col 3: $[0, 3, 8]^T$

The resulting matrix $\mathcal{M}(T)$ is: $$ \begin{bmatrix} 2 & 1 & 0 \\ 0 & 5 & 3 \\ 0 & 0 & 8 \end{bmatrix} $$ Since all entries below the diagonal are zero, this is upper triangular.

Theorem: Conditions for Upper Triangular Matrix

Let $T \in \mathcal{L}(V)$ and let $v_1, \dots, v_n$ be a basis for $V$. The following are equivalent:

1. The matrix of $T$ with respect to this basis is upper triangular.
2. $T(v_j) \in \operatorname{span}(v_1, \dots, v_j)$ for each $j = 1, \dots, n$.
3. $\operatorname{span}(v_1, \dots, v_j)$ is invariant under $T$ for each $j = 1, \dots, n$.

Theorem: Existence of Upper-Triangular Form

If $V$ is a finite-dimensional complex vector space and $T \in \mathcal{L}(V)$, then there exists a basis of $V$ such that the matrix of $T$ is upper triangular.

*(Note: This requires a complex vector space because operators on real vector spaces may not have eigenvalues.)*

Connection to Eigenvalues: If the matrix of $T$ is upper triangular, the eigenvalues of $T$ are exactly the entries on the diagonal. (In the example above, eigenvalues are 2, 5, and 8.)

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3. Eigenspaces and Diagonal Matrices

Diagonal Matrix

A diagonal matrix is a square matrix where all entries off the diagonal are zero. (Every diagonal matrix is upper triangular.)

$$ \begin{bmatrix} 8 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} $$

If an operator has a diagonal matrix with respect to some basis, the diagonal entries are its eigenvalues.

Eigenspaces

For $\lambda \in F$, the eigenspace of $T$ corresponding to $\lambda$ is defined as:

$$ E(\lambda, T) = \operatorname{null}(T - \lambda I) $$

This subspace contains all eigenvectors corresponding to $\lambda$, plus the zero vector.

Example:
If $\mathcal{M}(T)$ with respect to $v_1, v_2, v_3$ is the diagonal matrix above (entries 8, 5, 5):

• $E(8, T) = \operatorname{span}(v_1)$
• $E(5, T) = \operatorname{span}(v_2, v_3)$

Theorem: Sum of Eigenspaces

Let $V$ be finite-dimensional and let $\lambda_1, \dots, \lambda_m$ be distinct eigenvalues of $T$. Then:

1. The sum of the eigenspaces is a direct sum: $$ E(\lambda_1, T) + \dots + E(\lambda_m, T) $$
2. $\dim(E(\lambda_1, T)) + \dots + \dim(E(\lambda_m, T)) \le \dim V$.

This theorem implies that eigenvectors corresponding to distinct eigenvalues are linearly independent.

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This concludes the summary of Part 2 & 3.