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In general, the eigenvalues of a diagonal matrix are given by the entries on the diagonal. Verify this for 2 × 2 and 3 × 3 matrices.

User Losttime
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Eigenvalues:

Eigenvalues of a matrix A are all the values of π for the following equation:

d = det (πI-A) = 0

For a 2x2 diagonal matrix:


Let A =\left[\begin{array}{ccc}a1&0\\0&a2\\\end{array}\right] \\and\\I=\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]


\pi I-A=\left[\begin{array}{ccc}\pi-a1&0\\0&\pi-a2\\\end{array}\right] \\

Now,

d = det (πI-A)

π1-a1=0 and π2-a2=0

a1=π1 and a2=π2

Hence proved that the eigenvalues of a diagonal matrix are given by the entries on the diagonal.

For a 3x3 diagonal matrix:


Let A =\left[\begin{array}{ccc}a1&0&0\\0&a2&0\\0&0&a3\end{array}\right] \\and\\I=\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]


\pi I-A=\left[\begin{array}{ccc}\pi-a1&0&0\\0&\pi-a2\\0&0&\pi-a3\end{array}\right] \\

Now,

d = det (πI-A)

π1-a1=0 ; π2-a2=0 ; π3-a3=0

a1=π1 ; a2=π2 ; a3=π3

Hence proved that the eigenvalues of a diagonal matrix are given by the entries on the diagonal.

User Charlie Page
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