Originally Posted by

**HallsofIvy** Then, yes, of course, that matrix will row-reduce to a matrix having the last row 0- The equation $\displaystyle (A- \lambda I)v= 0$ must, by definition of "eigenvalue" have a non-zero solution and since 0 obviously is a solution, cannot have a **unique** solution. In fact, the set of all solutions, the "eigenspace", is a subspace and so it must have an infinite number of solutions.

Doing exactly what you say reduces the matrix to

$\displaystyle \begin{bmatrix}-12.4171 & 5 & 1 \\ 0 & -9.40375 & 4.40267 \\ 0 & 0 & 0\end{bmatrix}$

which essentially means your equations reduce to -12.4171x+ 5y+ z= 0 and -9.40375y+ 4.40267z= 0.

From the second equation, z= (9.40375/4.40267)y= 2.13592y.

Putting that into the first equation we have -12.4171x+ 5y+ 2.13592y= -12.4171x+ 7.13592y= 0 so that x= (7.13592/12.4171)y= 0.57468y.

That is, any eigenvector is of the form (0.57468y, y, 2.13592y)= (0.57468, 1, 2.13592)y where y can be any non-zero number (1 is simplest of course).