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Recitation 8B

Problem 1: Is x=[3−21]\mathbf{x}=\begin{bmatrix}3\\-2\\1\end{bmatrix} an eigenvector of A=[−4332−3−2−10−2]A=\begin{bmatrix}-4&3&3\\2&-3&-2\\-1&0&-2\end{bmatrix}?

Solution: Because Ax=[−1510−5]=(−5)xA\mathbf{x}=\begin{bmatrix}-15\\10\\-5\end{bmatrix}=(-5)\mathbf{x}, it is an eigenvector.

Problem 2: Is λ=4\lambda=4 an eigenvalue of A=[30−1231−345]A=\begin{bmatrix}3&0&-1\\2&3&1\\-3&4&5\end{bmatrix}.

Solution: Because A−λI=[−10−12−11−341]A-\lambda I=\begin{bmatrix}-1&0&-1\\2&-1&1\\-3&4&1\end{bmatrix} is not invertible, it is an eigenvalue.

Problem 3: Find a basis for the eigenspace corresponding to λ=−5\lambda=-5 of A=[−4112−3233−2]A=\begin{bmatrix}-4&1&1\\2&-3&2\\3&3&-2\end{bmatrix}.

Solution: It is enough to find a basis for the solution set of (A−λI)x=0(A-\lambda I)\mathbf{x}=\mathbf{0}. By row reduction, we obtain [111222333]∼[111000000].\begin{bmatrix}1&1&1\\2&2&2\\3&3&3\end{bmatrix}\sim \begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}. Therefore the parametric vector form of the solutions is [x1x2x3]=[−110]x2+[−101]x3\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=\begin{bmatrix}-1\\1\\0\end{bmatrix}x_2+\begin{bmatrix}-1\\0\\1\end{bmatrix}x_3. In other words, a basis can be [−110],[−101].\begin{bmatrix}-1\\1\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\1\end{bmatrix}.

Problem 4: If A2A^2 is the zero matrix, then the only eigenvalue of AA is $$.

Solution: Suppose λ\lambda is an eigenvalue of AA and x≠0\mathbf{x}\neq \mathbf{0} that satisfies Ax=λxA\mathbf{x}=\lambda\mathbf{x}. Now we have 0=A2x=A(Ax)=A(λx)=λ(Ax)=λ(λx)=λ2x\mathbf{0}=A^2\mathbf{x}=A(A\mathbf{x})=A(\lambda\mathbf{x})=\lambda(A\mathbf{x})=\lambda(\lambda\mathbf{x})=\lambda^2\mathbf{x}. Since x≠0\mathbf{x}\neq 0, λ2=0\lambda^2=0, which implies λ=0\lambda=0.

Problem 5: Find the characteristic equation of [40−104−1102]\begin{bmatrix}4&0&-1\\0&4&-1\\1&0&2\end{bmatrix}.

Solution: The characteristic polynomial is det(λI−A)\mathrm{det}\left(\lambda I - A\right), which is equal to ∣λ−4010λ−41−10λ−2∣=(λ−4)∣λ−410λ−2∣+∣0λ−4−10∣=(λ−4)(λ−4)(λ−2)+(λ−4)=(λ−4)(λ−3)2.\begin{aligned}\begin{vmatrix}\lambda-4 & 0 & 1 \\ 0 & \lambda-4 & 1 \\ -1 & 0 & \lambda-2\end{vmatrix} &=(\lambda-4)\begin{vmatrix}\lambda-4 & 1 \\ 0 & \lambda - 2\end{vmatrix}+\begin{vmatrix}0 & \lambda-4 \\ -1 & 0\end{vmatrix}\\&=(\lambda-4)(\lambda-4)(\lambda-2)+(\lambda-4)\\&=(\lambda-4)(\lambda-3)^2.\end{aligned} Therefore, the characteristic equation is (λ−4)(λ−3)2=0(\lambda-4)(\lambda-3)^2=0.