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Determine which of the matrices in Exercises 1–6 are symmetric.

2. \(\left( {\begin{aligned}{{}}3&{\,\, - 5}\\{ - 5}&{ - 3}\end{aligned}} \right)\)

Short Answer

Expert verified

The given matrix is symmetric.

Step by step solution

01

Find the transpose

A matrix\(A\) with, \(n \times n\) dimension, is symmetric if it satisfies the equation\({A^T} = A\).

It is given that\(A = \left( {\begin{aligned}{{}}3&{\,\, - 5}\\{ - 5}&{ - 3}\end{aligned}} \right)\). Find the transpose of\(A\), as shown below:

\(\begin{aligned}{}{A^T} &= \left( {\begin{aligned}{{}}3&{\,\, - 5}\\{ - 5}&{ - 3}\end{aligned}} \right)\\ &= A\end{aligned}\)

02

Draw the conclusion

As \({A^T} = A\), so it can be concluded that \(A\) is a symmetric matrix.

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Most popular questions from this chapter

(M) Orhtogonally diagonalize the matrices in Exercises 37-40. To practice the methods of this section, do not use an eigenvector routine from your matrix program. Instead, use the program to find the eigenvalues, and for each eigenvalue \(\lambda \), find an orthogonal basis for \({\bf{Nul}}\left( {A - \lambda I} \right)\), as in Examples 2 and 3.

38. \(\left( {\begin{aligned}{{}}{.{\bf{63}}}&{ - .{\bf{18}}}&{ - .{\bf{06}}}&{ - .{\bf{04}}}\\{ - .{\bf{18}}}&{.{\bf{84}}}&{ - .{\bf{04}}}&{.{\bf{12}}}\\{ - .{\bf{06}}}&{ - .{\bf{04}}}&{.{\bf{72}}}&{ - .{\bf{12}}}\\{ - .{\bf{04}}}&{.{\bf{12}}}&{ - .{\bf{12}}}&{.{\bf{66}}}\end{aligned}} \right)\)

Question: If A is \(m \times n\), then the matrix \(G = {A^T}A\) is called the Gram matrix of A. In this case, the entries of G are the inner products of the columns of A. (See Exercises 9 and 10).

9. Show that the Gram matrix of any matrix A is positive semidefinite, with the same rank as A. (See the Exercises in Section 6.5.)

Determine which of the matrices in Exercises 7–12 are orthogonal. If orthogonal, find the inverse.

12. \(P = \left( {\begin{aligned}{{}}{.5}&{.5}&{ - .5}&{ - .5}\\{.5}&{.5}&{.5}&{.5}\\{.5}&{ - .5}&{ - .5}&{.5}\\{.5}&{ - .5}&{.5}&{ - .5}\end{aligned}} \right)\)

Let \(A = \left( {\begin{aligned}{{}}{\,\,\,2}&{ - 1}&{ - 1}\\{ - 1}&{\,\,\,2}&{ - 1}\\{ - 1}&{ - 1} &{\,\,\,2}\end{aligned}} \right)\),\({{\rm{v}}_1} = \left( {\begin{aligned}{{}}{ - 1}\\{\,\,\,0}\\{\,\,1}\end{aligned}} \right)\) and and\({{\rm{v}}_2} = \left( {\begin{aligned}{{}}{\,\,\,1}\\{\, - 1}\\{\,\,\,\,1}\end{aligned}} \right)\). Verify that\({{\rm{v}}_1}\), \({{\rm{v}}_2}\) an eigenvector of \(A\). Then orthogonally diagonalize \(A\).

Compute the quadratic form \({x^T}Ax\), when \(A = \left( {\begin{aligned}{{}}3&2&0\\2&2&1\\0&1&0\end{aligned}} \right)\) and

a. \(x = \left( {\begin{aligned}{{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right)\)

b. \(x = \left( {\begin{aligned}{{}}{ - 2}\\{ - 1}\\5\end{aligned}} \right)\)

c. \(x = \left( {\begin{aligned}{{}}{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\\{\frac{1}{{\sqrt 2 }}}\end{aligned}} \right)\)

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