The characteristic polynomial

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Establish algebraic criteria for determining exactly when a real number can occur as an eigenvalue of $A$ .

For an $n\times n$ matrix $A$ , the characteristic polynomial of $A$ is given by $p_A(t) := Det(A - tI)$

Note the matrix here is not strictly numerical. Precisely, $(A-tI)(i,j) = \begin {cases} A(i,j)\,\,\text {if } i\ne j\\A(i,i) - t\,\,\text {if }i=j\end {cases}$ . However, the determinant of such a matrix (using either one of the two equivalent definitions given above) is still well-defined. In Observation obs:eigen we noted that $\lambda$ is an eigenvalue of $A$ iff $A-\lambda I$ is singular. By the properties of the determinant listed above, we see that $A-\lambda I$ is singular iff its determinant is equal to zero. In other words,

$\lambda$ is an eigenvalue of $A$ iff $p_A(\lambda ) = 0$ ; that is, $\lambda$ is a root of the characteristic polynomial $p_A(t)$ of $A$ .

Consider the matrix $A = \begin {bmatrix} 2 & 2\\4 & -5\end {bmatrix}$ . Then $p_A(t) = Det\left (\begin {bmatrix} (2-t) & 2\\4 & (-5-t)\end {bmatrix}\right ) = (2-t)(-5-t) - 8 = t^2 + 3t - 18 = (t+6)(t-3)$ , indicating that there are two eigenvalues; $\lambda = -6$ and $\lambda = 3$ . Moreover, for each of these eigenvalues, we can find a basis for the corresponding eigenspace by computing a basis for the nullspace $N(A - \lambda Id)$ .

For $\lambda = -6$ , $E_{-6}(A) = N(A + 6Id) = N\left (\begin {bmatrix} 8 & 2\\4 & 1\end {bmatrix}\right )$ . Then $E_{-6}(A) = N(A + 6Id) = Span\left \{\begin {bmatrix} -1\\4\end {bmatrix}\right \}$ .
For $\lambda = 3$ , $E_{3}(A) = N(A - 3Id) = N\left (\begin {bmatrix} -1 & 2\\4 & -8\end {bmatrix}\right )$ . Then $E_{3}(A) = N(A - 3Id) = Span\left \{\begin {bmatrix} 2\\1\end {bmatrix}\right \}$ .

Let $A = \begin {bmatrix} 3 & 6\\2 & -1\end {bmatrix}$ .

Compute the characteristic polynomial $p_A(t)$ of $A$ .
For each eigenvalue of $A$ , find an eigenvector corresponding to that eigenvalue.

Consider the matrix $A = \begin {bmatrix} 2 & -4\\1 & 3\end {bmatrix}$ . Then $p_A(t) = Det\left (\begin {bmatrix} (2-t) & -4\\1 & (3-t)\end {bmatrix}\right ) = (2-t)(3-t) + 4 = t^2 - 5t + 10$ . In this case the quadratic formula shows that there are no real roots. There are, however, two complex roots which are conjugate. They can be computed using the quadratic formula: $\lambda = \frac {5 \pm \sqrt {25 - 40}}{2} = \frac 52 \pm \left (\frac {\sqrt {15}}{2}\right )i$ Because $A$ is a real matrix with complex eigenvalues, any eigenvectors will also be complex. Complex eigenvalues and eigenvectors will be discussed in more detail below.

Let $A = \begin {bmatrix} 3 & -2\\5 & 1\end {bmatrix}$ . Compute the characteristic polynomial of $A$ and find its roots (if necessary using the quadratic formula).

The following is an immediate consequence of the definition of $p_A(t)$ , and basic properties of polynomials.

If $A$ is an $n\times n$ matrix, then $p_A(t)$ is a polynomial of degree $n$ . Consequently, $A$ can have at most $n$ distinct eigenvalues.

Finally, suppose we are given a linear transformation $L:V\to V$ with $Dim(V) = n$ . We would like to define the characteristic polynomial of $L$ . The natural thing to do is to fix a basis $S$ of $V$ and set $p_L(t) := p_A(t),\qquad A = {}_{S}L_{S}$ However, in order for this definition to be valid, we will want to know that the resulting polynomial does not depend on the particular choice of basis. The following result provides that; it will be proven in the section below on similarity.

If $S,S'$ are two bases for $V$ , and $A = {}_{S}L_{S}$ , $A' = {}_{S'}L_{S'}$ , then $p_A(t) = p_{A'}(t)$ . Thus the expression for $p_L(t)$ given above is independent of the particular choice of basis for $V$ .