- #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?
- 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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