6.7E: Exercises for Section 6.7

Exercise \(\PageIndex{1}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). \[A=\left[\begin{array}{rrr}11&-1&-4 \\ -1&11&-4\\-4&-4&14\end{array}\right]\nonumber\] Hint: Two eigenvalues are \(12\) and \(18\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\1\\1\end{array}\right]\right\}\leftrightarrow 6,\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}-1\\1\\0\end{array}\right]\right\}\leftrightarrow 12,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}-1\\-1\\2\end{array}\right]\right\}\leftrightarrow 18\nonumber\]

Exercise \(\PageIndex{2}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). \[A=\left[\begin{array}{rrr}4&1&-2\\1&4&-2\\-2&-2&7\end{array}\right]\nonumber\] Hint: One eigenvalue is \(3\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}-1\\1\\0\end{array}\right],\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\1\\1\end{array}\right]\right\}\leftrightarrow 3,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}-1\\-1\\2\end{array}\right]\right\}\leftrightarrow 9\nonumber\]

Exercise \(\PageIndex{3}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{rrr}-1&1&1\\1&-1&1\\1&1&-1\end{array}\right]\nonumber\] Hint: One eigenvalue is \(-2\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2} \\ 0\end{array}\right],\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ -\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow -2\nonumber\] \[\left[\begin{array}{c}\sqrt{3}/3&-\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&0&\frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]^T\left[\begin{array}{rrr}-1&1&1\\1&-1&1\\1&1&-1\end{array}\right]\nonumber\] \[\left[\begin{array}{c}\sqrt{3}/3&-\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&\sqrt{2}/2&-\sqrt{6}/6 \\ \sqrt{3}/3&0&\frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\nonumber\] \[=\left[\begin{array}{rrr}1&0&0\\0&-2&0\\0&0&-2\end{array}\right]\nonumber\]

Exercise \(\PageIndex{4}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{rrr}17&-7&-4 \\ -7&17&-4 \\ -4&-4&14\end{array}\right]\nonumber\] Hint: Two eigenvalues are \(18\) and \(24\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 6,\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ -\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 18,\left\{\left[\begin{array}{c}-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2} \\ 0\end{array}\right]\right\}\leftrightarrow 24\nonumber\] The matrix \(U\) has these as its columns.

Exercise \(\PageIndex{5}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{rrr}13&1&4\\1&13&4\\4&4&10\end{array}\right]\nonumber\] Hint: Two eigenvalues are \(12\) and \(18\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ -\frac{1}{6}\sqrt{6} \\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 6,\left\{\left[\begin{array}{c}-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2} \\ 0\end{array}\right]\right\}\leftrightarrow 12, \left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 18.\nonumber\] The matrix \(U\) has these as its columns.

Exercise \(\PageIndex{6}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}-\frac{5}{3}&\frac{1}{15}\sqrt{6}\sqrt{5}&\frac{8}{15}\sqrt{5} \\ \frac{1}{15}\sqrt{6}\sqrt{5}&-\frac{14}{5}&-\frac{1}{15}\sqrt{6} \\ \frac{8}{15}\sqrt{5}&-\frac{1}{15}\sqrt{6}&\frac{7}{15} \end{array}\right]\nonumber\] Hint: The eigenvalues are \(-3,-2,1\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ 0\\ \frac{1}{6}\sqrt{5}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}-\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{5}\sqrt{5} \\ \frac{1}{15}\sqrt{2}\sqrt{15}\end{array}\right]\right\} -2,\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ \frac{2}{5}\sqrt{5} \\ \frac{1}{30}\sqrt{30}\end{array}\right]\right\}\leftrightarrow -3\nonumber\] These vectors are the columns of \(U\).

Exercise \(\PageIndex{7}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}3&0&0 \\ 0&\frac{3}{2}&\frac{1}{2} \\ 0&\frac{1}{2}&\frac{3}{2}\end{array}\right]\nonumber\]

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}0\\-\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}0\\ \frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 2,\left\{\left[\begin{array}{c}1\\0\\0\end{array}\right]\right\}\leftrightarrow 3.\nonumber\] These vectors are the columns of the matrix \(U\).

Exercise \(\PageIndex{8}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}2&0&0\\0&5&1\\0&1&5\end{array}\right]\nonumber\]

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}1\\0\\0\end{array}\right]\right\}\leftrightarrow 2,\left\{\left[\begin{array}{c}0\\ -\frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 4, \left\{\left[\begin{array}{c}0\\ \frac{1}{2}\sqrt{2} \\ \frac{1}{2}\sqrt{2}\end{array}\right]\right\}\leftrightarrow 6.\nonumber\] These vectors are the columns of \(U\).

Exercise \(\PageIndex{9}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}\frac{4}{3}&\frac{1}{3}\sqrt{3}\sqrt{2}&\frac{1}{3}\sqrt{2} \\ \frac{1}{3}\sqrt{3}\sqrt{2}&1&-\frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{2}&-\frac{1}{3}\sqrt{3}&\frac{5}{3}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(0,2,2\) where \(2\) is listed twice because it is a root of multiplicity \(2\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{5}\sqrt{2}\sqrt{5} \\ \frac{1}{5}\sqrt{3}\sqrt{5} \\ \frac{1}{5}\sqrt{5}\end{array}\right]\right\}\leftrightarrow 0,\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ 0\\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right],\left[\begin{array}{c}\frac{1}{5}\sqrt{2}\sqrt{5} \\ \frac{1}{5}\sqrt{3}\sqrt{5} \\ -\frac{1}{5}\sqrt{5}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] The columns are these vectors.

Exercise \(\PageIndex{10}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for \(A\). Diagonalize \(A\) by finding an orthogonal matrix \(U\) and a diagonal matrix \(D\) such that \(U^TAU = D\). \[A=\left[\begin{array}{ccc}1&\frac{1}{6}\sqrt{3}\sqrt{2}&\frac{1}{6}\sqrt{3}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{2}&\frac{3}{2}&\frac{1}{12}\sqrt{2}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{6}&\frac{1}{12}\sqrt{2}\sqrt{6}&\frac{1}{2}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(2,1,0\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{3}\sqrt{3} \\ 0\\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow 0,\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ -\frac{1}{2}\sqrt{2} \\ \frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 1,\left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{2}\sqrt{2} \\ \frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] The columns are these vectors.

Exercise \(\PageIndex{11}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix \[A=\left[\begin{array}{ccc}\frac{1}{3}&\frac{1}{6}\sqrt{3}\sqrt{2}&-\frac{7}{18}\sqrt{3}\sqrt{6} \\ \frac{1}{6}\sqrt{3}\sqrt{2}&\frac{3}{2}&-\frac{1}{12}\sqrt{2}\sqrt{6} \\ -\frac{7}{18}\sqrt{3}\sqrt{6}&-\frac{1}{12}\sqrt{2}\sqrt{6}&-\frac{5}{6}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(1,2,-2\).

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{3}\sqrt{3} \\ \frac{1}{2}\sqrt{2} \\ \frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 1, \left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ 0\\ \frac{1}{3}\sqrt{2}\sqrt{3}\end{array}\right]\right\}\leftrightarrow -2, \left\{\left[\begin{array}{c}\frac{1}{3}\sqrt{3} \\ \frac{1}{2}\sqrt{2} \\ -\frac{1}{6}\sqrt{6}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] Then the columns of \(U\) are these vectors.

Exercise \(\PageIndex{12}\)

Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix \[A=\left[\begin{array}{ccc}-\frac{1}{2}&-\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{1}{10}\sqrt{5} \\ -\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{7}{5}&-\frac{1}{5}\sqrt{6} \\ \frac{1}{10}&\sqrt{5}&-\frac{1}{5}\sqrt{6}&-\frac{9}{10}\end{array}\right]\nonumber\] Hint: The eigenvalues are \(-1,2,-1\) where \(-1\) is listed twice because it has multiplicity \(2\) as a zero of the characteristic equation.

Answer

The eigenvectors and eigenvalues are: \[\left\{\left[\begin{array}{c}-\frac{1}{6}\sqrt{6} \\ 0 \\ \frac{1}{6}\sqrt{5}\sqrt{6}\end{array}\right],\left[\begin{array}{c}\frac{1}{3}\sqrt{2}\sqrt{3} \\ \frac{1}{5}\sqrt{5} \\ \frac{1}{15}\sqrt{2}\sqrt{15}\end{array}\right]\right\}\leftrightarrow -1,\left\{\left[\begin{array}{c}\frac{1}{6}\sqrt{6} \\ -\frac{2}{5}\sqrt{5} \\ \frac{1}{30}\sqrt{30}\end{array}\right]\right\}\leftrightarrow 2.\nonumber\] The columns of \(U\) are these vectors. \[\left[\begin{array}{ccc}-\frac{1}{6}\sqrt{6}&\frac{1}{3}\sqrt{2}\sqrt{3}&\frac{1}{6}\sqrt{6} \\ 0&\frac{1}{5}\sqrt{5}&-\frac{2}{5}\sqrt{5} \\ \frac{1}{6}\sqrt{5}\sqrt{6} &\frac{1}{15}\sqrt{2}\sqrt{15}&\frac{1}{30}\sqrt{30}\end{array}\right]^T\left[\begin{array}{ccc}-\frac{1}{2}&-\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{1}{10}\sqrt{5} \\ -\frac{1}{5}\sqrt{6}\sqrt{5}&\frac{7}{5}&-\frac{1}{5}\sqrt{6} \\ \frac{1}{10}\sqrt{5}&-\frac{1}{5}\sqrt{6}&-\frac{9}{10}\end{array}\right].\nonumber\] \[\left[\begin{array}{ccc}-\frac{1}{6}\sqrt{6}&\frac{1}{3}\sqrt{2}\sqrt{3}&\frac{1}{6}\sqrt{6} \\ 0&\frac{1}{5}\sqrt{5}&-\frac{2}{5}\sqrt{5} \\ \frac{1}{6}\sqrt{5}\sqrt{6}&\frac{1}{15}\sqrt{2}\sqrt{15}&\frac{1}{30}\sqrt{30}\end{array}\right]=\left[\begin{array}{rrr}-1&0&0\\0&-1&0\\0&0&2\end{array}\right]\nonumber\]

Exercise \(\PageIndex{13}\)

Explain why a matrix \(A\) is symmetric if and only if there exists an orthogonal matrix \(U\) such that \(A = U^TDU\) for \(D\) a diagonal matrix.

Answer

If \(A\) is given by the formula, then \[A^T=U^TD^TU=U^TDU=A\nonumber\] Next suppose \(A = A^T\). Then by the theorems on symmetric matrices, there exists an orthogonal matrix \(U\) such that \[UAU^T=D\nonumber\] for \(D\) diagonal. Hence \[A=U^TDU\nonumber\]

Exercise \(\PageIndex{14}\)

Show that if \(A\) is a real symmetric matrix and \(λ\) and \(µ\) are two different eigenvalues, then if \(\vec{x}\) is an eigenvector for \(λ\) and \(\vec{y}\) is an eigenvector for \(µ\), then \(\vec{x} •\vec{y} = 0\). Also all eigenvalues are real. Supply reasons for each step in the following argument. First \[\lambda \vec{x}^T\overline{\vec{x}}=(A\vec{x})^T\overline{\vec{x}}=\vec{x}^TA\overline{\vec{x}}=\vec{x}^T\overline{A\vec{x}}=\vec{x}^T\overline{\lambda \vec{x}}=\overline{\lambda}\vec{x}^T\overline{\vec{x}}\nonumber\] and so \(\lambda=\overline{\lambda}\). This shows that all eigenvalues are real. It follows all the eigenvectors are real. Why? Now let \(\vec{x},\: \vec{y},\:µ\) and \(λ\) be given as above. \[\lambda (\vec{x}\bullet \vec{y})=\lambda \vec{x}\bullet \vec{y}=A\vec{x}\bullet \vec{y}=\vec{x}\bullet A\vec{y}=\vec{x}\bullet\mu \vec{y}=\mu (\vec{x}\bullet \vec{y})=\mu (\vec{x}\bullet \vec{y})\nonumber\] and so \[(\lambda -\mu )\vec{x}\bullet \vec{y}=0\nonumber\] Why does it follow that \(\vec{x}\bullet \vec{y}=0\)?

Answer

Since \(\lambda\neq\mu\), it follows \(\vec{x}\bullet \vec{y}=0\).

Exercise \(\PageIndex{15}\)

Prove that an \(n\times n\) matrix \(U\) is orthogonal if and only if its columns are an orthonormal set of vectors.