91Ó°ÊÓ

An eigenvalue, \(\lambda\), of a square matrix \(A\) is a scalar value such that there exists a non-zero vector \(x\) (called the eigenvector) which satisfies the following equation: \[Ax = \lambda x\] We need to show that for a triangular matrix, the eigenvalues are the diagonal elements.

Let's consider a square n x n matrix and let's call it \(A\). We will write the triangular matrix in its general form as: \[A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{pmatrix}\] Here, the matrix can be either upper or lower triangular - the proof will be the same for both cases.

Now, let's suppose that the matrix \(A\) has an eigenvalue \(\lambda\) and an eigenvector \(x\), such that \(Ax = \lambda x\). The eigenvector \(x\) can be represented as: \[ x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}\] Now, let's write down the equation for the triangular matrix as applied to the eigenvector \(x\): \[Ax = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix} = \begin{pmatrix} \lambda x_1 \\ \lambda x_2 \\ \vdots \\ \lambda x_n \end{pmatrix} = \lambda x \]

If we now look at the first row of this equation, we get: \(a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = \lambda x_1\) Now, by moving the term with the unknown \(\lambda\) to the other side, we get: \(a_{11}x_1 - \lambda x_1 = - a_{12}x_2 - \cdots - a_{1n}x_n\) Now, we see that \(x_1 \neq 0\) because eigenvector components cannot all be zero. Hence, we can factor out \(x_1\) from the left side: \(x_1(a_{11} - \lambda) = - a_{12}x_2 - \cdots - a_{1n}x_n\) Now, the right side of this equation is not equal to zero because eigenvectors are non-zero: \((a_{11} - \lambda) \neq 0\) Now, we can divide both sides by \(x_1\) and obtain the following equation: \(a_{11} - \lambda = - \frac{a_{12}}{x_1}x_2 - \cdots - \frac{a_{1n}}{x_1}x_n\) This equation represents a relationship between the eigenvalue \(\lambda\) and the matrix diagonal element \(a_{11}\). This relationship holds for all rows of the matrix \(A\). So, for each diagonal element, we get a similar equation: \(a_{ii} - \lambda \neq 0\) These equations must be true for any eigenvector component and diagonal matrix element. Therefore, we conclude that: \(\lambda = a_{11}, a_{22}, \cdots , a_{nn}\)

We have shown that for a triangular matrix, the eigenvalues are the diagonal elements of the matrix. This result holds for both upper and lower triangular matrices.