Eigenvalues of Diagonal and Triangular Matrices

Diagonal Matrices and Their Special Properties

For diagonal and triangular matrices, every eigenvalue can be $read directly from the main diagonal$ , with algebraic multiplicity. The matrix may have one of these forms:

The three forms below represent a diagonal, upper triangular, and lower triangular matrix $A \in \mathbb{K}^{n \times n}$ .

A = \begin{pmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{pmatrix}

For all three forms, the eigenvalues are the main diagonal entries:

\lambda_1 = a_{11}, \ldots, \lambda_n = a_{nn}

Why is this true? Since $\chi_A(t) = \det(A - t \cdot I) = (a_{11} - t) \cdots (a_{nn} - t)$ with roots $a_{11}, \ldots, a_{nn}$ .

No full determinant expansion or root-finding procedure is needed for a triangular matrix.

Upper and Lower Triangular Matrices

Triangular matrices have the same property as diagonal matrices. For both upper and lower triangular matrices, the eigenvalues are still the main diagonal entries.

This happens because when we calculate $\det(A - tI)$ , the entries above or below the main diagonal don't affect the determinant calculation. The triangular structure allows the determinant to be computed as the $product of diagonal entries$ .

For example, off-diagonal entries can change while the $same eigenvalues$ remain:

T=\begin{pmatrix}2&4&1\\0&2&3\\0&0&-1\end{pmatrix},\qquad \chi_T(t)=(2-t)^2(-1-t)

Thus 2 occurs twice and −1 once. This gives the eigenvalues, but does not by itself prove diagonalizability.

Computing Eigenvalues of Nontriangular Matrices

The diagonal rule is special to triangular matrices. Once a matrix is no longer triangular, the eigenvalues stay hidden in the characteristic polynomial, so the work moves from reading diagonal entries to $solving for roots$ .

Complex Eigenvalues

Consider the matrix:

A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = \det \begin{pmatrix} 1-t & -1 \\ 1 & 1-t \end{pmatrix}

t^2-2t+2=0\iff(t-1)^2=-1\iff t=1\pm i

$The roots of this characteristic polynomial$ are $\lambda_1 = 1 + i$ and $\lambda_2 = 1 - i$ .

Zero Eigenvalues

Consider the matrix:

A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = \det \begin{pmatrix} 1-t & -i \\ i & 1-t \end{pmatrix}

$The roots of this characteristic polynomial$ are $\lambda_1 = 2$ and $\lambda_2 = 0$ .

Characteristic Polynomial Factorization

When $A \in \mathbb{C}^{n \times n}$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ $counted with algebraic multiplicity$ , its characteristic polynomial factors as:

\chi_A(t) = (\lambda_1 - t) \cdots (\lambda_n - t)

Let $\sigma(A)$ denote the finite set of distinct eigenvalues of $A$ . Their algebraic multiplicities sum to $n$ :

\sum_{\lambda \in \sigma(A)} \mu_A(\lambda) = n

In a more compact form:

\chi_A(t) = \prod_{\lambda \in \sigma(A)} (\lambda - t)^{\mu_A(\lambda)}

This property holds naturally for complex eigenvalues of matrices with real entries. Eigenvalues can be real numbers or complex conjugate pairs.

Relationship Between Determinant and Trace

If the characteristic polynomial $\chi_A(t)$ splits into linear factors over $\mathbb{K}$ , the matrix $A$ has $n$ eigenvalues $\lambda_1, \ldots, \lambda_n$ there, counted with multiplicity. Their sum and product satisfy:

\det A = \prod_{i=1}^n \lambda_i

$The determinant is the product$ of all eigenvalues, and $the trace is the sum$ of all eigenvalues.

Verify with our previous examples:

For the following matrix with eigenvalues $\lambda_1 = 1 + i$ and $\lambda_2 = 1 - i$ :

A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}

We can check the eigenvalues using the determinant and trace:

\begin{aligned} \det A &= 1 \cdot 1 - 1 \cdot (-1) \\ &= 2 \\ &= \lambda_1 \cdot \lambda_2 \end{aligned}

For the following matrix with eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 0$ :

A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}

We can check the eigenvalues using the determinant and trace:

\begin{aligned} \det A &= 1 \cdot 1 - i \cdot (-i) \\ &= 0 \\ &= \lambda_1 \cdot \lambda_2 \end{aligned}

These identities are useful checks, but they do not determine every eigenvalue of a general matrix when $n > 2$ . One immediate consequence is:

\det A = 0 \quad \Longleftrightarrow \quad 0 \in \sigma(A)

So a square matrix is singular exactly when zero is one of its eigenvalues.

Diagonal Matrices and Their Special Properties

For diagonal and triangular matrices, every eigenvalue can be $read directly from the main diagonal$ , with algebraic multiplicity. The matrix may have one of these forms:

The three forms below represent a diagonal, upper triangular, and lower triangular matrix $A \in \mathbb{K}^{n \times n}$ .

A = \begin{pmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{pmatrix}

For all three forms, the eigenvalues are the main diagonal entries:

\lambda_1 = a_{11}, \ldots, \lambda_n = a_{nn}

Why is this true? Since $\chi_A(t) = \det(A - t \cdot I) = (a_{11} - t) \cdots (a_{nn} - t)$ with roots $a_{11}, \ldots, a_{nn}$ .

No full determinant expansion or root-finding procedure is needed for a triangular matrix.

Upper and Lower Triangular Matrices

Triangular matrices have the same property as diagonal matrices. For both upper and lower triangular matrices, the eigenvalues are still the main diagonal entries.

For example, off-diagonal entries can change while the $same eigenvalues$ remain:

T=\begin{pmatrix}2&4&1\\0&2&3\\0&0&-1\end{pmatrix},\qquad \chi_T(t)=(2-t)^2(-1-t)

Thus 2 occurs twice and −1 once. This gives the eigenvalues, but does not by itself prove diagonalizability.

Computing Eigenvalues of Nontriangular Matrices

Complex Eigenvalues

Consider the matrix:

A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = \det \begin{pmatrix} 1-t & -1 \\ 1 & 1-t \end{pmatrix}

t^2-2t+2=0\iff(t-1)^2=-1\iff t=1\pm i

$The roots of this characteristic polynomial$ are $\lambda_1 = 1 + i$ and $\lambda_2 = 1 - i$ .

Zero Eigenvalues

Consider the matrix:

A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}

Its characteristic polynomial is:

\chi_A(t) = \det \begin{pmatrix} 1-t & -i \\ i & 1-t \end{pmatrix}

$The roots of this characteristic polynomial$ are $\lambda_1 = 2$ and $\lambda_2 = 0$ .

Characteristic Polynomial Factorization

When $A \in \mathbb{C}^{n \times n}$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ $counted with algebraic multiplicity$ , its characteristic polynomial factors as:

\chi_A(t) = (\lambda_1 - t) \cdots (\lambda_n - t)

Let $\sigma(A)$ denote the finite set of distinct eigenvalues of $A$ . Their algebraic multiplicities sum to $n$ :

\sum_{\lambda \in \sigma(A)} \mu_A(\lambda) = n

In a more compact form:

\chi_A(t) = \prod_{\lambda \in \sigma(A)} (\lambda - t)^{\mu_A(\lambda)}

This property holds naturally for complex eigenvalues of matrices with real entries. Eigenvalues can be real numbers or complex conjugate pairs.

Relationship Between Determinant and Trace

\det A = \prod_{i=1}^n \lambda_i

$The determinant is the product$ of all eigenvalues, and $the trace is the sum$ of all eigenvalues.

Verify with our previous examples:

For the following matrix with eigenvalues $\lambda_1 = 1 + i$ and $\lambda_2 = 1 - i$ :

A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}

We can check the eigenvalues using the determinant and trace:

\begin{aligned} \det A &= 1 \cdot 1 - 1 \cdot (-1) \\ &= 2 \\ &= \lambda_1 \cdot \lambda_2 \end{aligned}

For the following matrix with eigenvalues $\lambda_1 = 2$ and $\lambda_2 = 0$ :

A = \begin{pmatrix} 1 & -i \\ i & 1 \end{pmatrix}

We can check the eigenvalues using the determinant and trace:

\begin{aligned} \det A &= 1 \cdot 1 - i \cdot (-i) \\ &= 0 \\ &= \lambda_1 \cdot \lambda_2 \end{aligned}

These identities are useful checks, but they do not determine every eigenvalue of a general matrix when $n > 2$ . One immediate consequence is:

\det A = 0 \quad \Longleftrightarrow \quad 0 \in \sigma(A)

So a square matrix is singular exactly when zero is one of its eigenvalues.