(1 point) The matrix A=⎡⎣⎢−4−4−40−8−4084⎤⎦⎥A=[−400−4−88−4−44] has two real eigenvalues, one of multiplicity 11 and one of multiplicity 22. Find the eigenvalues and a basis of each eigenspace. λ1λ1 = equation editorEquation Editor has multiplicity 11, with a basis of equation editorEquation Editor . λ2λ2 = equation editorEquation Editor has multiplicity 22, with a basis of equation editorEquation Editor .

Answer:We have the matrix [tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right][/tex]To find the eigenvalues of A we need find the zeros of the polynomial characteristic [tex]p(\lambda)=det(A-\lambda I_3)[/tex]Then[tex]p(\lambda)=det(\left[\begin{array}{ccc}-4-\lambda&-4&-4\\0&-8-\lambda&-4\\0&8&4-\lambda\end{array}\right] )\\=(-4-\lambda)det(\left[\begin{array}{cc}-8-\lambda&-4\\8&4-\lambda\end{array}\right] )\\=(-4-\lambda)((-8-\lambda)(4-\lambda)+32)\\=-\lambda^3-8\lambda^2-16\lambda[/tex]Now, we fin the zeros of [tex]p(\lambda)[/tex].[tex]p(\lambda)=-\lambda^3-8\lambda^2-16\lambda=0\\\lambda(-\lambda^2-8\lambda-16)=0\\\lambda_{1}=0\; o \; \lambda_{2,3}=\frac{8\pm\sqrt{8^2-4(-1)(-16)}}{-2}=\frac{8}{-2}=-4[/tex]Then, the eigenvalues of A are [tex]\lambda_{1}=0[/tex] of multiplicity 1 and [tex]\lambda{2}=-4[/tex] of multiplicity 2.Let's find the eigenspaces of A. For [tex]\lambda_{1}=0[/tex]: [tex]E_0 = Null(A- 0I_3)=Null(A)[/tex].Then, we use row operations to find the echelon form of the matrix[tex]A=\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&8&4\end{array}\right]\rightarrow\left[\begin{array}{ccc}-4&-4&-4\\0&-8&-4\\0&0&0\end{array}\right][/tex]We use backward substitution and we obtain1. [tex]-8y-4z=0\\y=\frac{-1}{2}z[/tex]2.[tex]-4x-4y-4z=0\\-4x-4(\frac{-1}{2}z)-4z=0\\x=\frac{-1}{2}z[/tex]Therefore,[tex]E_0=\{(x,y,z): (x,y,z)=(-\frac{1}{2}t,-\frac{1}{2}t,t)\}=gen((-\frac{1}{2},-\frac{1}{2},1))[/tex]For [tex]\lambda_{2}=-4[/tex]: [tex]E_{-4} = Null(A- (-4)I_3)=Null(A+4I_3)[/tex].Then, we use row operations to find the echelon form of the matrix[tex]A+4I_3=\left[\begin{array}{ccc}0&-4&-4\\0&-4&-4\\0&8&8\end{array}\right] \rightarrow\left[\begin{array}{ccc}0&-4&-4\\0&0&0\\0&0&0\end{array}\right][/tex]We use backward substitution and we obtain1. [tex]-4y-4z=0\\y=-z[/tex]Then,[tex]E_{-4}=\{(x,y,z): (x,y,z)=(x,z,z)\}=gen((1,0,0),(0,1,1))[/tex]