Does there exist a 2x2 non-singular matrix with only one 1d eigenspace?

[zenterix]

Homework Statement

How do we find a matrix $A$ whose only eigenvectors are multiples of $(1,4)$ and that is invertible?

Relevant Equations

Let's try to find a $2\times 2$ matrix. Suppose the matrix $A$ is given by

$$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$

Since we have only a single 1d eigenspace for this matrix, there can only be one eigenvalue, $\lambda$.

We know this eigenvalue cannot be 0 since the matrix is non-singular.

Before going through calculations/reasoning, let me summarize what my questions will be

- In order to obtain the desired matrix, I impose five constraints on $a,b,c,d,$ and $\lambda$.

- These five constraints are four equations and an inequality. I am not sure how to work with the inequality.

- I can leave the inequality out and solve the system of four equations for five unknowns.

- By guessing values together with the solution to these four equations, I can eventually reach the desired matrix.

- However, in this process of guessing it became clear that one constraint is definitely missing: the equations I have guarantee solutions (ie, matrices) with a sole eigenvalue; but they don't guarantee an eigenspace with only a single dimension.

- I would like to know how to add this constraint.

Now let me go through my reasoning.

$$\begin{vmatrix} a-\lambda & b\\ c & d-\lambda \end{vmatrix}=0\tag{2}$$

$$\lambda^2-\lambda(a+d)+(ad-bc)=0$$

which has discriminant

$$\Delta=(a+d)^2-4(ad-bc)=0\tag{3}$$

We equate $\Delta$ to zero because we there to be a single $\lambda$.

Thus

$$\lambda=\frac{a+d}{2}\tag{4}$$

We have three further constraints on the variables.

First, $A$ is non-singular if

$$ad-bc\neq 0\tag{5}$$

In addition, $A(1,4)=\lambda (1,4)$ so

$$a+4b=\lambda\tag{6}$$

$$c+4d=4\lambda\tag{7}$$

At this point, equations (3), (4), (5), (6), and (7) are five constraints and we have the five unknowns $a,b,c,d,$ and $\lambda$.

One of the constraints is an inequality, however.

Suppose I leave out the inequality and just solve (3), (4), (6), and (7). Using Maple, I get the solution

$$a$$

$$b=\frac{d-a}{8}$$

$$c=2(a-d)$$

$$d$$

$$\lambda=\frac{a+d}{2}$$

If I let $a=d=1$ then it turns out that $b=c=0$. Thus, matrix is the identity matrix.

The problem is that the eigenspace for the sole eigenvalue of $1$ is 2-dimensional not 1-dimensional as desired.

By guessing values, I was able to obtain a matrix with all the desired constraints.

Let $a=2$, $d=3$. Then $b=\frac{1}{8}$ and $c=-2$.

Thus, the matrix is

$$\begin{bmatrix} 2& \frac{1}{8}\\-2 &3\end{bmatrix}$$

We have $ad-bc=\frac{25}{4}\neq 0$ so the matrix is invertible.

The sole eigenvalue is $\frac{5}{2}$ and the eigenspace is the span of $(1,4)$.

My questions are

1) how do I guarantee that my solution (ie, the matrix $A$) will have one eigenspace only, and this eigenspace is one-dimensional? What additional constraint do I need in the system of equations?

2) how do I work with a constraint that is an inequality?

 

[nuuskur]

The eigenspace corresponding to $\lambda$ is the solution space of $A-\lambda E=0$ (i.e $\mathrm{Ker}(A-\lambda E))$. What you are currently after is that
<br /> \left(\begin{array}{cc}a-\lambda &amp;b \\ c&amp;d-\lambda \end{array}\right)<br />
is of rank one i.e that the determinant is zero and at least one of the entries in the above is nonzero. Then you further require that
<br /> \left(\begin{array}{cc} a&amp;b\\c&amp;d \end{array}\right)\left(\begin{array}{c}1\\4 \end{array}\right) = \lambda \left(\begin{array}{c}1\\4 \end{array}\right)<br />
There is no inequality restriction at play here. You correctly deduce that $\lambda = \frac{a+d}{2}$ because there can only be one eigenvalue. So we have
<br /> \begin{cases}(a-\lambda)(d-\lambda) - bc = 0 \\ a+4b = \lambda \\ c+4d = 4\lambda \\ a+d = 2\lambda \end{cases}<br />
We have four conditions and five variables. Since you didn't restrict the eigenvalue we'll pick one. Since $A$ is nonsingular, we must pick $\lambda \neq 0$. So let's pick $\lambda = 1$ for instance. Thus, we simplify to
<br /> \begin{cases}<br /> 1-(a+d)+(ad-bc) = 0 \\<br /> a+4b=1 \\<br /> c+4d = 4\\<br /> a+d=2<br /> \end{cases}<br />
We can temporarily ignore the first condition, because it's not linear. The other three conditions give us
<br /> \left(\begin{array}{cccc|c} 1&amp;4&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;1&amp;4&amp;4 \\ 1&amp;0&amp;0&amp;1&amp;2 \end{array}\right)<br />
The solutions for this are
<br /> \left(\begin{array}{c}a\\4b\\c\\d \end{array}\right) = \left(\begin{array}{c}2-s \\ -1+s \\ 4-4s \\ s \end{array}\right),\quad s\in F.<br />
So all that's left is to plug it in the nonlinear condition and solve the resulting quadratic and that will determine what $A$ could be if we assumed that $\lambda =1$. It could also be that the quadratic has no solutions in which case try another $\lambda$.
edit: It turns out plugging in the solution to the nonlinear condition yields $0=0$, so pick any $s$ and fire away.

Instead of fixing the eigenvalue, we could also fix one of the entries of $A$ such that $A-\lambda E$ is of rank one.

---

Picking $s=0$ in the above gives us
<br /> A=\left(\begin{array}{cc} 2 &amp; -\frac{1}{4} \\ 4 &amp; 0\end{array}\right),<br />
which is clearly nonsingular with characteristic polynomial $(\lambda-1)^2$ as expected and the system
<br /> A-E = \left(\begin{array}{cc}1 &amp; -\frac{1}{4} \\ 4 &amp; -1\end{array}\right)<br />
is clearly of rank one whose solutions are generated by $(1,4)$.

 

[pasmith]

Do we not have that the Jordan normal form of A is \begin{pmatrix} \lambda &amp; 1 \\ 0 &amp; \lambda \end{pmatrix}? The first basis vector is (1,4)^T and the second can be any linearly independent vector - say (0,1)^T. Then <br /> \begin{split}<br /> A \begin{pmatrix} 1 \\ 4 \end{pmatrix} &amp;= \lambda \begin{pmatrix} 1 \\ 4 \end{pmatrix} \\<br /> A \begin{pmatrix} 0 \\ 1 \end{pmatrix} &amp;= \lambda \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix} \end{split} and \begin{split}<br /> A\begin{pmatrix} x \\ y \end{pmatrix} &amp;=<br /> A\left(x\begin{pmatrix} 1 \\ 4 \end{pmatrix} + (y - 4x) \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right)\\<br /> &amp;= x\lambda \begin{pmatrix} 1 \\ 4 \end{pmatrix} + (y - 4x) \left( \lambda \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix} \right) \end{split} giving <br /> A = \begin{pmatrix} \lambda - 4 &amp; 1 \\ -16 &amp; \lambda + 4 \end{pmatrix}.