Determine the entries $x$ and $y$ in a matrix so that its only eigenvalue is $1$.

Question

I am doing some self-study in preparation for an exam, and in this problem I am given the following matrix in $R^{3×3}$:

$\begin{pmatrix} 1&0&1\\ 0&1&-1\\ 0&x&y\\ \end{pmatrix}$

I need to determine the set of x and y values so that the only eigenvalue of this matrix is one.

At first glance, we have x = 0 and y = 1, since that would make this into a unipotent matrix and I know that the only eigenvalue of a unipotent matrix equals 1.

Now consider the characteristic polynomial of the matrix, which I have determined to be

$p(\lambda) := -\lambda^3 + (y + 2)\lambda^2 - (x + 2x + 1)\lambda + (x+y)$.

We see that $a_2 = (y + 2)$ and $a_0 = (x+y)$, and that $a_2$ is the sum of eigenvalues and that $a_0$ is the product of eigenvalues, both of which must be one. Solving for x and y gives us

x = 2 and y = -1.

For the third method, consider the companion matrix of the characteristic polynomial.

Transforming $p(\lambda) := -\lambda^3 + (y + 2)\lambda^2 - (x + 2x + 1)\lambda + (x+y)$ to the monic polynomial $p(\lambda) := \lambda^3 - (y + 2)\lambda^2 + (x + 2x + 1)\lambda - (x+y)$ and then determining the companion matrix, we get:

C(p) :=

$\begin{pmatrix} 0&0&x+y\\ 1&0&-(x+2x+1)\\ 0&1&y+2\\ \end{pmatrix}$

If the sum of all columns equals 1, then one of the eigenvalues of this matrix equals 1. Summing the third column gives us a value of 1 for all x and y. Therefore, knowing that the eigenvalues of C(p) are the roots of $p(\lambda)$, we see that 1 is an eigenvalue of the original matrix for all x and y.

To see that it is the only real eigenvalue, I factor $(\lambda - 1)$ out of the original monic polynomial and solve for the remaining quadratic coefficients. I get:

$p(\lambda) := (\lambda - 1)(\lambda^2 - (y + 1)\lambda + (x+y))$

Solving for the quadratic coefficients gives us:

$\lambda_{2,3} = \frac{(y+1) ± \sqrt{(y^2 - 2y -4x + 1)}}{2}$

So if we find x and y in the real numbers so that the term under the radical is less than zero, we have found x and y where 1 is the only real eigenvalue.

From this point I am stuck.

Answer 1

In your second method, the sum of the eigenvalues shouldn't be $1$ but actually $3$, since we want $1$ to be an eigenvalue with multiplicity $3$ so sum of the eigenvalues is $1+1+1 = 3$.

Regarding the third method, the eigenvalues could be complex numbers but what we want is to demand that $\lambda_{2,3} = 1$. So we set

$$ \begin{align} y^2-2y-4x+1 &= 0 \\ y+1 &= 2. \end{align} $$

So we get only the solution $x=0, y=1$.

Answer 2

The characteristic polynomial of the matrix is: $$\begin{align*}p_A(\lambda)&=\det(A-\lambda I)\\&=\det \begin{bmatrix} 1-\lambda & 0 & 1 \\ 0 & 1-\lambda & -1 \\ 0 & x & y-\lambda \end{bmatrix}\\&=(1-\lambda)\det\begin{bmatrix} 1-\lambda & -1\\ x & y-\lambda \end{bmatrix}\\&=(1-\lambda)\left[(1-\lambda)(y-\lambda)+x\right]\\&=(1-\lambda)(\lambda^2-(y+1)\lambda+(x+y))\end{align*}$$ So we have to determine $x$ and $y$ such that the quadratic polynomial $\lambda^2-(y+1)\lambda+(x+y)$ has only the root $1$. It's roots are: $$\begin{align*}\lambda_{1,2}&=\frac{y+1\pm\sqrt{(y+1)^2-4(x+y)}}{2}\\&=\frac{y+1\pm\sqrt{y^2+2y+1-4x-4y}}{2}\\&=\frac{y+1\pm\sqrt{(y-1)^2-4x}}{2}\end{align*}$$ We have to impose that: $$\begin{cases}(y-1)^2-4x=0\\y+1=2\end{cases}\:\Longrightarrow\:\begin{cases}x=0\\y=1\end{cases}$$ That's the only solution.

Answer 3

ploosu2 already discussed what is wrong with your second attempt.

I suggest a simpler method for your third attempt. You want

$$ q(\lambda) = \lambda^2-(y+1)\lambda + (x+y) $$ to only have $1$ as a root. Since the leading coefficient of this polynomial is $1$ this means that you must have $$ q(\lambda) = (\lambda-1)^2 $$ for all $\lambda$. Expanding the expression $(\lambda -1)^2 = \lambda^2 -2\lambda + 1$ and equating the coefficients of both polynomials leads to the system of equations $$ \begin{align} x + y &= 1 \\ y + 1 &= 2. \end{align} $$ The solution is $x = 0, y = 1$.