Find eigenvalues of $T$

Question

Let the linear transformation $S \in \mathbb{C}$ be defined as

$$S(x_1, x_2) = (5x_1 - 3x_2, -6x_1 + 2x_2) = \lambda(x_1, x_2)$$

Then I get the following system of linear equations

$$5x_1 - 3x_2 = \lambda x_1 \\ -6x_1 + 2x_2 = \lambda x_2$$

I tried to substitute one equation into the other and find $\lambda$, but it just seemed to make the equation more difficult. Not sure what to do here, just that there’s a general idea to solve for $\lambda$

Answer 1

$$\left\{\begin{matrix} 5x_1 - 3x_2 = \lambda x_1 \\ -6x_1 + 2x_2 = \lambda x_2 \end{matrix}\right.$$

Then, doing some algebraic manipulations in the second line, we have that $$-6x_1 = -2x_2+\lambda x_2$$ $$x_1=\frac{(2-\lambda)}{6}x_2$$

Replacing the value of $x_1$ in the first line,

$$5\frac{(2-\lambda)}{6}x_2-3x_2=\lambda\frac{(2-\lambda)}{6}x_2$$ $$x_2\left(\frac{(5-\lambda)(2-\lambda)}{6}-3\right)=0$$ $$x_2\left((5-\lambda)(2-\lambda)-18\right)=0$$ $$x_2\left((\lambda+1)(\lambda-8)\right)=0$$

Can you conclude from this point?

Answer 2

It might be easier to do this with the standard method of finding the characteristic polynomial of the transformation. For this, note that the matrix that represents this transformation is given by:

$T=\begin{bmatrix} 5&-3\\-6&2\end{bmatrix}$

You thus get:

$$\det \left( \begin{bmatrix} 5-\lambda&-3\\-6&2-\lambda\end{bmatrix} \right)=0$$ $$(5-\lambda)(2-\lambda)-18=0$$ $$\lambda^2-7\lambda-8=0$$ $$\lambda=-1, 8$$

Answer 3

If you want to solve this without determinants, let's first simplify the equations. Note that if $x_1 = 0$ then the first equation becomes $-3x_2 = 0$, which forces $x_2 = 0$. Since $(x_1, x_2) = (0,0)$ is not a valid eigenvector, we must have $x_1 \neq 0$.

Now note that if $(x_1, x_2)$ is an eigenvector, then so is $c(x_1, x_2)$ for any nonzero scalar $c$. Thus, since $x_1 \neq 0$, we can scale the eigenvector so that $x_1 = 1$. If we do this, the system of equations becomes $$5 - 3x_2 = \lambda$$ $$-6 + 2x_2 = \lambda x_2$$ Solving the first equation for $x_2$ gives us $x_2 = (5-\lambda)/3$. Substituting this into the second equation gives us $$-6 + \frac{2}{3}(5 - \lambda) = \frac{1}{3}\lambda(5 - \lambda)$$ which is a quadratic equation in $\lambda$. Solving this gives you the eigenvalues.

All of the above is a rather convoluted workaround to deal with Axler's phobia of determinants. :-)

Answer 4

Adding the two equations, we have $-1(x_1+x_2)=\lambda(x_1+x_2)$, so either $\lambda=-1$ or $x_1=-x_2$. Substituting the first back into the original equations reduces them to $2x_1=x_2$, which gives you one eigenvalue/eigenvector pair. Substituting the second leads to the equation $(\lambda-8)x_2=0$. We know from $x_1=-x_2$ that $x_2\ne0$, so we must have $\lambda=8$ for your second eigenvalue/eigenvector pair.

This is similar to what I might do when working with the coefficient matrix directly: for artificial exercises like this one, it’s often fruitful to try a few simple linear combinations of the columns before plunging into a general method. In this case, subtracting the second column from the first produces $8(1,-1)^T$, which gives you one eigenvalue/eigenvector pair and the remaining eigenvalue is obtained “for free” from the trace of the matrix. Alternatively, since the transpose of a matrix has the same eigenvalues, it’s easy to spot $-1$ as the common column sum of the matrix (making the vector of all $1$s a left eigenvector) and as before the remaining eigenvalue can be extracted from the trace.