Answer:
a) -4
b) 1
c) 1
Step-by-step explanation:
a) The matrix A is given by:
[tex]A=\left[\begin{array}{ccc}-3&0&1\\2&-4&2\\-3&-2&1\end{array}\right][/tex]
to find the eigenvalues of the matrix you use the following:
[tex]det(A-\lambda I)=0[/tex]
where lambda are the eigenvalues and I is the identity matrix. By replacing you obtain:
[tex]A-\lambda I=\left[\begin{array}{ccc}-3-\lambda&0&1\\2&-4-\lambda&2\\-3&-2&1-\lambda\end{array}\right][/tex]
and by taking the determinant:
[tex][(-3-\lambda)(-4-\lambda)(1-\lambda)+(0)(2)(-3)+(2)(-2)(1)]-[(1)(-4-\lambda)(-3)+(0)(2)(1-\lambda)+(2)(-2)(-3-\lambda)]=0\\\\-\lambda^3-6\lambda^2-12\lambda-16=0[/tex]
and the roots of this polynomial is:
[tex]\lambda_1=-4\\\\\lambda_2=-1+i\sqrt{3}\\\\\lambda_3=-1-i\sqrt{3}[/tex]
hence, the real eigenvalue of the matrix A is -4.
b) The multiplicity of the eigenvalue is 1.
c) The dimension of the eigenspace is 1 (because the multiplicity determines the dimension of the eigenspace)
