CHARACTERISTIC EQUATION

This is a special scalar equation associated with square matrices.

Example # 1: Find the characteristic equation and the eigenvalues of "A".

Find all scalars, l, such that: has a nontrivial solution. That matrix equation has nontrivial solutions only if the matrix is not invertible or equivalently its determinant is zero. This is the characteristic equation.

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Theorem: The determinant of a square n x n matrix "A" can be found from any echelon form, "U", obtained from "A" by row replacements and row interchanges ( without scaling ) using this formula: .

Example # 2: Find the determinant of "A" using the above formula and find the characteristic polynomial of "A" using a cofactor expansion.

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Theorem : Algebraic multiplicity of an eigenvalue is always greater than or equal to the dimension of the eigenspace corresponding to that eigenvalue.

Example # 3: Find "h" in the matrix below such that the eigenspace for l = 5 is 2-space.

The eigenspace is the null space of the matrix:

With l = 5 we get this result.

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Columns # 2 & # 4 are definitely pivot columns. Column # 3 is a pivot column if and only if . But that would result in a 1-dimensional eigenspace. Our problem calls for us to choose "h" such that our eigenspace corresponding to l = 5 be 2-dimensional. Equivalently, we can have no more than two pivot columns so that we get two free variables: if .

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Note that in our original "A" matrix the eigenvalue,

l = 5 has multiplicity of "2" which is greater than or equal to the dimension of its eigenspace which is "1" or "2" based on the value of "h".

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