A 2 × 2 real symmetric matrix A has eigenvalues 1 and 3. (2, -3) is an eigenvector corresponding to the eigenvalue λ = 1. (a) Find an eigenvector corresponding to the eigenvalue λ = 3. (b) Find A.
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Step 1: To find an eigenvector corresponding to the eigenvalue λ = 3, we need to solve the equation (A - 3I)v = 0, where A is the matrix and I is the identity matrix. Show more…
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The $2 \times 2$ real symmetric matrix $A$ has two distinct eigenvalues $\lambda_{1}$ and $\lambda_{2} .$ If $\mathbf{v}_{1}=(1,2)$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda_{1},$ determine an eigenvector corresponding to $\lambda_{2}$.
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The $3 \times 3$ real symmetric matrix $A$ has eigenvalues $\lambda_{1}$ and $\lambda_{2}$ (multiplicity 2 ). (a) If $\mathbf{v}_{1}=(1,-1,1)$ spans the eigenspace $E_{1},$ determine a basis for $E_{2}$ and hence find an orthogonal matrix $S,$ such that $S^{T} A S=\operatorname{diag}\left(\lambda_{1}, \lambda_{2}, \lambda_{2}\right)$ (b) Use your result from part (a) to find $A$.
Determine whether or not $\mathbf{x}$ is an eigenvector of $A .$ If it is, determine its associated eigenvalue. $$(\mathrm{a})A=\left[ \begin{array}{rr}{3} & {-1} \\ {2} & {0}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{1} \\ {2}\end{array}\right]$$ $$(\mathrm{b})A=\left[ \begin{array}{rrr}{-3} & {-1} & {5} \\ {-2} & {1} & {2} \\ {-2} & {-1} & {4}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\ {1} \\ {1}\end{array}\right]$$ $$(\mathrm{c})A=\left[ \begin{array}{rr}{1} & {-1} \\ {-2} & {0}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{1} \\ {0}\end{array}\right]$$ $$(\mathrm{d}) A=\left[ \begin{array}{rr}{1} & {2} \\ {-2} & {1}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\ {2 i}\end{array}\right]$$ $$(\mathrm{e}) A=\left[ \begin{array}{ll}{2+3 a} & {-2-2 a} \\ {3+3 a} & {-3-2 a}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\ {3}\end{array}\right]$$ $$(\mathrm{f}) A=\left[ \begin{array}{rrr}{-9} & {4} & {6} \\ {-6} & {3} & {4} \\ {-9} & {4} & {6}\end{array}\right] \quad \mathbf{x}=\left[ \begin{array}{l}{2} \\ {0} \\ {3}\end{array}\right]$$
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