Algebra and Geometry | Part IA, 2001

Assume that xp\mathbf{x}_{p} is a particular solution to the equation Ax=bA \mathbf{x}=\mathbf{b} with x,bR3\mathbf{x}, \mathbf{b} \in \mathbb{R}^{3} and a real 3×33 \times 3 matrix AA. Explain why the general solution to Ax=bA \mathbf{x}=\mathbf{b} is given by x=xp+h\mathbf{x}=\mathbf{x}_{p}+\mathbf{h} where h\mathbf{h} is any vector such that Ah=0A \mathbf{h}=\mathbf{0}.

Now assume that AA is a real symmetric 3×33 \times 3 matrix with three different eigenvalues λ1,λ2\lambda_{1}, \lambda_{2} and λ3\lambda_{3}. Show that eigenvectors of AA with respect to different eigenvalues are orthogonal. Let xk\mathbf{x}_{k} be a normalised eigenvector of AA with respect to the eigenvalue λk\lambda_{k}, k=1,2,3k=1,2,3. Show that the linear system

(AλkI)x=b\left(A-\lambda_{k} I\right) \mathbf{x}=\mathbf{b}

where II denotes the 3×33 \times 3 unit matrix, is solvable if and only if xkb=0\mathbf{x}_{k} \cdot \mathbf{b}=0. Show that the general solution is given by

x=ikb.xiλiλkxi+βxk,βR\mathbf{x}=\sum_{i \neq k} \frac{\mathbf{b} . \mathbf{x}_{i}}{\lambda_{i}-\lambda_{k}} \mathbf{x}_{i}+\beta \mathbf{x}_{k}, \quad \beta \in \mathbb{R}

[Hint: consider the components of x\mathbf{x} and b\mathbf{b} with respect to a basis of eigenvectors of AA.]

Consider the matrix AA and the vector b\mathbf{b}

A=(122+163122+163133122+163122+163133133133233),b=(2+32+323)A=\left(\begin{array}{ccc} -\frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & \frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & -\frac{1}{3} \sqrt{3} \\ \frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & -\frac{1}{2} \sqrt{2}+\frac{1}{6} \sqrt{3} & -\frac{1}{3} \sqrt{3} \\ -\frac{1}{3} \sqrt{3} & -\frac{1}{3} \sqrt{3} & \frac{2}{3} \sqrt{3} \end{array}\right), \quad \mathbf{b}=\left(\begin{array}{c} \sqrt{2}+\sqrt{3} \\ -\sqrt{2}+\sqrt{3} \\ -2 \sqrt{3} \end{array}\right)

Verify that 13(1,1,1)T\frac{1}{\sqrt{3}}(1,1,1)^{T} and 12(1,1,0)T\frac{1}{\sqrt{2}}(1,-1,0)^{T} are eigenvectors of AA. Show that Ax=bA \mathbf{x}=\mathbf{b} is solvable and find its general solution.

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