The EigenvaluesTutor(M) command by default opens a Maplet window which allows you to work interactively through solving for the eigenvalues of M. Options provide other ways to show the step-by-step solutions, as described below.

The EigenvaluesTutor(M) command presents the techniques used in finding the eigenvalues of the square matrix $M$ by:

Creating the matrix M - lambda*Id where Id is an identity matrix with dimensions equal to that of M

Taking the determinant of M - lambda*Id

Finding the roots of the resulting characteristic polynomial

The Matrix M must be square and of dimension 4 at most.

Floating-point numbers in M are converted to rationals before computation begins.

If the symbolic expression representing an eigenvalue grows too large, then the value displayed in the Maplet application window is a floating-point approximation to it (obtained by applying evalf).  The underlying computations continue to be performed using exact arithmetic, however.

The EigenvaluesTutor(M) command returns the eigenvalues as a column Vector.

The following options can be used to control how the problem is displayed and what output is returned, giving the ability to generate step-by-step solutions directly without going through the Maplet tutor interface:

output = steps,canvas,script,record,list,print,printf,typeset,link (default: maplet)

displaystyle= columns,compact,linear,brief (default: linear)

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{LinearAlgebra}\right]\right)\:$

$M≔⟨⟨1\,2\,0⟩\left|⟨2\,3\,2⟩\right|⟨0\,2\,1⟩⟩$

$M≔\left[\begin{array}{ccc}1& 2& 0\\ 2& 3& 2\\ 0& 2& 1\end{array}\right]$

$\mathrm{EigenvaluesTutor}\left(M\right)$

$\mathrm{EigenvaluesTutor}\left(M\,\mathrm{output}=\mathrm{steps}\right)$

$\begin{array}{lll}& & \text{Compute the Eigenvalues}\\ & & \left[\begin{array}{ccc}1& 2& 0\\ 2& 3& 2\\ 0& 2& 1\end{array}\right]\\ \text{•}& & \text{Calculate A=M-t*Id}\\ & & \left[\begin{array}{ccc}1-t& 2& 0\\ 2& 3-t& 2\\ 0& 2& 1-t\end{array}\right]\\ \text{▫}& & \text{Find the determinant; this is also called the characteristic polynomial of M.}\\ & \text{◦}& \text{Use cofactor expansion on the}3\text{by}3\text{matrix}\\ & & \left(1-t\right)\cdot \left[\begin{array}{cc}3-t& 2\\ 2& 1-t\end{array}\right]+\left(\mathrm{-1}\right)\cdot 2\cdot \left[\begin{array}{cc}2& 2\\ 0& 1-t\end{array}\right]+1\cdot 0\cdot \left[\begin{array}{cc}2& 3-t\\ 0& 2\end{array}\right]\\ & \text{◦}& \text{Find the determinant of the 2 by 2 matrices by multiplying the diagonals}\\ & & \left(1-t\right)\cdot \left(\left(3-t\right)\cdot \left(1-t\right)-2\cdot 2\right)+\left(\mathrm{-1}\right)\cdot 2\cdot \left(2\cdot \left(1-t\right)-0\cdot 2\right)+1\cdot 0\cdot \left(2\cdot 2-0\cdot \left(3-t\right)\right)\\ & \text{◦}& \text{Evaluate inside the brackets}\\ & & \left(1-t\right)\cdot \left(\left(3-t\right)\left(1-t\right)-4\right)+\left(\mathrm{-1}\right)\cdot 2\cdot \left(2-2t\right)+1\cdot 0\cdot 4\\ & \text{◦}& \text{Multiply}\\ & & \left(1-t\right)\left(\left(3-t\right)\left(1-t\right)-4\right)+\left(-4+4t\right)+0\\ & \text{◦}& \text{Evaluate}\\ & & \left(1-t\right)\left(\left(3-t\right)\left(1-t\right)-4\right)-4+4t\\ & & \text{Find the determinant; this is also called the characteristic polynomial of M.}\\ & & -t^3+5t^2+t-5\\ \text{•}& & \text{Solve; the eigenvalues are the roots of the characteristic polynomial.}\\ & & \left[\begin{array}{c}5\\ 1\\ \mathrm{-1}\end{array}\right]\end{array}$