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

In summary, a 2x2 non-singular matrix cannot have only one one-dimensional eigenspace. Such a matrix must have two distinct eigenvalues, leading to two one-dimensional eigenspaces. If it had only one eigenspace, it would imply the presence of a repeated eigenvalue, resulting in a singular matrix, which contradicts the non-singularity condition. Therefore, the existence of a 2x2 non-singular matrix with only one one-dimensional eigenspace is impossible.
  • #1
zenterix
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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?
 
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  • #2
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
[tex]
\left(\begin{array}{cc}a-\lambda &b \\ c&d-\lambda \end{array}\right)
[/tex]
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
[tex]
\left(\begin{array}{cc} a&b\\c&d \end{array}\right)\left(\begin{array}{c}1\\4 \end{array}\right) = \lambda \left(\begin{array}{c}1\\4 \end{array}\right)
[/tex]
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
[tex]
\begin{cases}(a-\lambda)(d-\lambda) - bc = 0 \\ a+4b = \lambda \\ c+4d = 4\lambda \\ a+d = 2\lambda \end{cases}
[/tex]
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
[tex]
\begin{cases}
1-(a+d)+(ad-bc) = 0 \\
a+4b=1 \\
c+4d = 4\\
a+d=2
\end{cases}
[/tex]
We can temporarily ignore the first condition, because it's not linear. The other three conditions give us
[tex]
\left(\begin{array}{cccc|c} 1&4&0&0&1 \\ 0&0&1&4&4 \\ 1&0&0&1&2 \end{array}\right)
[/tex]
The solutions for this are
[tex]
\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.
[/tex]
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
[tex]
A=\left(\begin{array}{cc} 2 & -\frac{1}{4} \\ 4 & 0\end{array}\right),
[/tex]
which is clearly nonsingular with characteristic polynomial ##(\lambda-1)^2## as expected and the system
[tex]
A-E = \left(\begin{array}{cc}1 & -\frac{1}{4} \\ 4 & -1\end{array}\right)
[/tex]
is clearly of rank one whose solutions are generated by ##(1,4)##.
 
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  • #3
Do we not have that the Jordan normal form of [itex]A[/itex] is [itex]\begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}[/itex]? The first basis vector is [itex](1,4)^T[/itex] and the second can be any linearly independent vector - say [itex](0,1)^T[/itex]. Then [tex]
\begin{split}
A \begin{pmatrix} 1 \\ 4 \end{pmatrix} &= \lambda \begin{pmatrix} 1 \\ 4 \end{pmatrix} \\
A \begin{pmatrix} 0 \\ 1 \end{pmatrix} &= \lambda \begin{pmatrix} 0 \\ 1 \end{pmatrix} + \begin{pmatrix} 1 \\ 4 \end{pmatrix} \end{split}[/tex] and [tex]\begin{split}
A\begin{pmatrix} x \\ y \end{pmatrix} &=
A\left(x\begin{pmatrix} 1 \\ 4 \end{pmatrix} + (y - 4x) \begin{pmatrix} 0 \\ 1 \end{pmatrix} \right)\\
&= 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} [/tex] giving [tex]
A = \begin{pmatrix} \lambda - 4 & 1 \\ -16 & \lambda + 4 \end{pmatrix}.[/tex]
 
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FAQ: Does there exist a 2x2 non-singular matrix with only one 1d eigenspace?

What is a 2x2 non-singular matrix?

A 2x2 non-singular matrix is a square matrix with two rows and two columns that has a non-zero determinant. This means that the matrix is invertible, meaning there exists another matrix that, when multiplied with the original, yields the identity matrix.

What is an eigenspace?

An eigenspace of a matrix is the set of all eigenvectors corresponding to a particular eigenvalue, along with the zero vector. For a given eigenvalue, the eigenspace consists of all vectors that, when the matrix is applied to them, result in a scalar multiple of the original vector.

Can a 2x2 non-singular matrix have only one 1-dimensional eigenspace?

Yes, a 2x2 non-singular matrix can have only one 1-dimensional eigenspace. This typically occurs when the matrix has a repeated eigenvalue, meaning it has a single eigenvalue with algebraic multiplicity of 2, but the geometric multiplicity (dimension of the eigenspace) is 1.

What is an example of a 2x2 non-singular matrix with only one 1-dimensional eigenspace?

An example of such a matrix is:\[ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \]This matrix has a single eigenvalue of 1 with an eigenspace spanned by the vector \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\).

Why does a 2x2 non-singular matrix with a repeated eigenvalue have only one 1-dimensional eigenspace?

When a 2x2 non-singular matrix has a repeated eigenvalue but only one 1-dimensional eigenspace, it means that the matrix is not diagonalizable. The eigenspace corresponding to the repeated eigenvalue does not have enough linearly independent eigenvectors to form a basis for the entire space, leading to a single eigenspace of dimension 1.

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