Hello, let's note A the matrix, we need to find [tex]\lambda[/tex] such that A[tex]\lambda[/tex]=[tex]\lambda[/tex] I, where I is the identity matrix, so the determinant is 0, giving us the characteristic equation as
[tex]\left|\begin{array}{cc}1-\lambda&3\\-3&1-\lambda\end{array}\right|\\\\=(1-\lambda)^2+9\\\\=\lambda^2-2\lambda+10\\\\=0[/tex]
We just need to solve this equation using the discriminant.
[tex]\Delta=b^2-4ac=2^2-40=-36=(6i)^2[/tex]
And then the eigenvalues are.
[tex]\lambda_1=\dfrac{2-6i}{2}=\boxed{1-3i}\\\\\lambda_2=\boxed{1+3i}[/tex]
To find the basis, we have to solve the system of equations.
[tex]A\lambda_1-\lambda_1 I=\left[\begin{array}{cc}3i&3\\-3&3i\end{array}\right] \\\\=3\left[\begin{array}{cc}i&1\\-1&i\end{array}\right] \\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}ai+b=0\\-a+bi=0\end{cases}\\\\\text{(1,-i) is a base of this space, as i-i=0 and -1-}i^2\text{=-1+1=0.}[/tex]
[tex]A\lambda_2-\lambda_2 I=\left[\begin{array}{cc}-3i&3\\-3&-3i\end{array}\right] \\\\=3\left[\begin{array}{cc}-i&1\\-1&-i\end{array}\right]\\\\\text{For a vector (a,b), we need to find a and b such that.}\\\\\begin{cases}-ai+b=0\\-a-bi=0\end{cases}\\\\\text{(1,i) is a base of this space as -i+i=0 and -1-i*i=0.}[/tex]
Thank you
