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I am using the textbook Linear Algebra done right by Sheldon Axler (fourth edition) to teach an undergraduate linear algebra course. Please find here a link to the book here.

Now Axler does not introduce the idea of determinants and characteristic polynomials while discussing eigenvalues and minimal polynomials. So after section 5B if one needs to solve the following question, how should one solve the question without using determinants?

The question is as follows: Find the eigenvalues and minimal polynomial of the given operator $T$ on $\mathbb{R}^3$, defined by $T(x,y,z)=(x+y, x+2y, 3z)$

Note: I am aware of this question, but here they are discussing only $2\times 2$ matrices. My operator above involves a $3\times 3$ matrix.

So basically I need to find values of $\lambda$ such that there exists $(x,y,z)\neq 0\in \mathbb{R}^3$, such that $T(x,y,z)=\lambda(x,y,z)$. i.e if the equation $(T-\lambda I)(x,y,z)=0$ has a non trivial solution in $\mathbb{R}^3$ i.e does the system $$\begin{bmatrix} 1-\lambda & 1 & 0 \\ 1 & 2-\lambda & 0\\ 0 & 0 & 3-\lambda \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$ have a non trivial solution.

Now for $(x,y,z)$ to be a non trivial solution , at least one of $x$ or $y$ or $z$ has to be non-zero.

Suppose I take $x\neq 0$ I get $\lambda$ to be the values $\frac{3}{2}-\frac{\sqrt{5}}{2}$ and $\frac{3}{2}+\frac{\sqrt{5}}{2}$. So I have got two eigenvalues out of maximum three. I see it is very evident that $\lambda=3$ is also an eigenvalue. I will get that if I assume $z\neq 0$.

My question is can I consider the two cases? i.e Case 1: Since $(x,y,z)\neq 0$ so I can assume $x\neq 0$ and Case 2: Since $(x,y,z)\neq 0$ so suppose $z\neq 0$.

Or is it, as I suppose I can consider only one case?

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  • $\begingroup$ This appears to be a problem from outside the book, is that correct? $\endgroup$ Commented Feb 6 at 19:42
  • $\begingroup$ Yes. We can also talk about Question 12 in exercise 5B. $\endgroup$
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    Commented Feb 6 at 19:49
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    $\begingroup$ Unfortunately, this seems more like a mathematics content question (even with the tool constraint) than a math-educator question. $\endgroup$ Commented Feb 6 at 20:27
  • $\begingroup$ I'm having difficulties to see the point of your example. This is a block diagonal matrix; its eigenvalues are the number $3$ and the eigenvalues of the $2\times 2$-matrix $\begin{bmatrix} 1&1\\1&2\end{bmatrix}$. $\endgroup$ Commented Feb 7 at 18:06
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    $\begingroup$ It seems to me that Steve Gubkin's answer matheducators.stackexchange.com/a/21200/51 to the 2x2 case is perfectly good for the 3x3 case as well. $\endgroup$ Commented Feb 8 at 2:16

3 Answers 3

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Let's assume I don't know anything about determinants and characteristic polynomials, but I understand that I'm interested in non-trivial solutions to $T(x,y,z)=\lambda(x,y,z)$ with the $T$ you provide. Then I proceed as follows.

I start with the system of equations \begin{array}{rl} x+y & =\lambda x\\ x+2y &=\lambda y \\ 3z &= \lambda z\end{array} Immediately I notice that if $z \neq 0$, then $\lambda =3$. This reduces the system to \begin{array}{rl} -2x+y &=0\\ x-y &=0 \end{array} which pretty quickly I see has $x=y=0$ (but $z \neq 0$, so not a trivial solution).

So now I return to the possibility that $z=0$. Then from the first equation I have $x+y=\lambda x$, so $$y=(\lambda-1)x.$$ Substituting into the second gives $$x+2(\lambda-1)x=\lambda(\lambda-1)x.$$

Now I notice that if $x \neq 0$, then I have $$1+2(\lambda-1)=\lambda(\lambda-1).$$ From which I solve $$\lambda=\frac{1}{2} \left(3\pm\sqrt{5}\right).$$

Lastly I check if $x=0$ then $y=0$ and this is the trivial solution.

The take-away, from a pedagogical point of view, is this: This method is annoying (and look how lucky we got with the first step with $z \neq 0$ being such an easy case). This is why we go off and develop more theory (determinants, etc.) to find a better way.

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    $\begingroup$ Yes. It is annoying to ignore deteminants! $\endgroup$ Commented Feb 7 at 0:56
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One can get a fair amount of information by using row reduction. Subtracting $1-\lambda$ times the second row from the first one and then exchanging the first two rows,

$$ \begin{vmatrix} 1-\lambda & 1 & 0 \\ 1 & 2-\lambda & 0\\ 0 & 0 & 3-\lambda \end{vmatrix} \xrightarrow{\phantom{xxxxxx}} \begin{bmatrix} 1 & 2-\lambda & 0\\ 0&1-(1-\lambda)(2-\lambda)&0\\ 0 & 0 & 3-\lambda \end{bmatrix} $$ Now the system is in echelon form. For the system to have a non-trivial solution, one of the diagonal entries needs to be zero. We are left with $\lambda=3$, and $0=1-(1-\lambda)(2-\lambda)=-\lambda^2+3\lambda-1$, which gives the roots $$ \lambda=\frac{3\pm\sqrt5}2. $$ As all eigenvalues are simple, the minimal polynomial agrees with the characteristic polynomial, $$ p(x)=(\lambda-3)(\lambda^2-3\lambda+1). $$

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The matrix form of $T$ is

\begin{bmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix}

At this point in Axler's book (linked in the OP), it should be obvious that $\lambda=3$ is an eigenvalue and that $T$ has an invariant subspace $W$ so that $T\vert_W$ has matrix \begin{bmatrix} 1&1\\ 1&2\\ \end{bmatrix}

Exercise 5B#11 page 151 says that for a 2x2 matrix \begin{bmatrix} a&c\\ b&d\\ \end{bmatrix} the minimal polynomial of a linear operator on a two-dimensional vector space with this matrix representation is (except for a degenerate case) $$z^2-(a+d)z+(ad-bc)$$.

Notice that the constant term of the quadratic is the determinant, and the linear term is the negative of the trace. In effect, this result (Exercise #11) gives us the green light to use $\det(A-\lambda I)$ to calculate eigenvalues of a 2x2 matrix, even if we are not formally allowed to utter the words "trace" and "determinant."

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    $\begingroup$ Axler's book and this whole line of thinking just makes me want to take determinants even harder. $\endgroup$ Commented Feb 11 at 12:10

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