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Given $\lambda$ is -2 and after calculating $A-\lambda I$, $$\left(\begin{array}{rrr} 3 & 3 & 3\\ -3 & -3 & -3\\ 3 & 3 & 3 \end{array}\right).$$

The reduced row echelon form is: $$\left(\begin{array}{ccc} 1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right).$$

The solution is actually: $$\left(\begin{array}{aaa} -1\\ 1\\ 0 \end{array}\right).$$ and

$$\left(\begin{array}{bbb} -1\\ 0\\ 1 \end{array}\right).$$

Can someone explain how the book arrived at this?

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The greek letter $\lambda$'s name is "lambda". – Arturo Magidin Apr 18 '12 at 19:22
I copied down the wrong matrix! Forgive me. – Bob Apr 18 '12 at 19:28

Your reduced row echelon form is $$\left(\begin{array}{ccc} 1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{array}\right).$$ You have one leading variable (the first term), and two free variables. The system is equivalent to the single equation $$x+y+z=0.$$ That means that you have two degrees of freedom, $y$ and $z$. You can express $x$ in terms of $y$ and $z$ and $x=-y-z$. So the solutions are given by: $$\begin{align*} x &= -s - t\\ y & = s\\ z &= t \end{align*}\qquad\qquad s,t\in\mathbb{R}.$$ You get a basis for the space of solutions by taking the parameters (in this case, $s$ and $t$), and putting one of them equal to $1$ and the rest to $0$, one at a time. Setting $s=1$ and $t=0$, we get $x=-1$, $y=1$, $z=0$, leading to the vector $(-1,1,0)$; setting $s=0$ and $t=1$ we get $x=-1$, $y=0$, $z=1$, leading to the vector $(-1,0,1)$.

Note: What you claim is "the solution" is just one possible basis for the eigenspace. The eigenspace is equivalent to the nullspace of $A-\lambda I$; which is equivalent to the nullspace of the reduced row echelon form of $A-\lambda I$. So this question really comes down to:

Do you know how to find a basis for the nullspace of a matrix $A$?


Do you know how to express the set of solutions to a homogeneous system of equations in vector form?

Above, we end up with the single-equation "system" given by $$x+y+z=0.$$ Do you know how to write down/describe the set of all solutions to this equation in terms of vectors?

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I don't understand why are you just setting y = s and z=t or is this just how eigenvalues work? Shouldn't y = -x-z? Sorry if this is a dumb question. – Bob Apr 18 '12 at 19:37
@Bob: That part has nothing to do with eigenvalues. We solve for the leading variables (in this case $x$) in terms of the free variables (in this case, $y$ and $z$). We don't simultaneously try to solve for $x$ in terms of $y$ and $z$ and for $y$ in terms of $x$ and $z$. The point is to isolate the things that are free on one side, and the things that will be forced once the free choices are made. – Arturo Magidin Apr 18 '12 at 19:41
@Bob: (cont) So while it is true that it is also the case that $y=-x-z$, we don't want to say it at the same time as we are saying $x=-y-z$; that just gives us circularity: "Well, $x$ depends on $y$ and $z$, and $y$ depends on $x$ and $z$, and $z$ depends on $x$ and $y$..." and you just go around and around in circles. So $y$ and $z$ are free; they can be anything. Once we decide what $y$ and $z$ will be, this gives us the corresponding value of $x$. I set them equal to $s$ and $t$ because I want to keep $x$, $y$, and $z$ reserved for the names of the variables/components, (cont) – Arturo Magidin Apr 18 '12 at 19:43
@Bob:, and so I use two new letters to make it clear that $y$ and $z$ can be chosen to be any arbitrary numbers. I picked $s$ and $t$, but you can pick any names you want (or none at all, though I think that would invite confusion, making it seem that you haven't solved the system, just rewritten the equation). It just so happens that the solutions to this system give you the eigenvectors of the matrix, but we are trying to describe the solutions without regard to the fact that they correspond to eigenvectors. This is how you write solutions to systems of linear equations parametrically. – Arturo Magidin Apr 18 '12 at 19:44

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