8 Linear Algebra V

In this class,

• Questions must be answered in order asked.
• Solutions must be neat.
• Solutions must be exact and fully simplified.
• You must use the given symbols.
• You must show all reasoning.
• Copying is never allowed, even when working together.

1. UNGRADED: An anti-symmetric matrix is a matrix for which $A^{\rm T}=-A$. Are the eigenvalues of an antisymmetric real matrix real too? To check, write down the simplest nontrivial anti-symmetric $2\times2$ matrix you can think of (which may not be symmetric) and see. In fact, the eigenvalues of an antisymmetric matrix are always purely imaginary, i.e. proportional to $i=\sqrt{-1}$. The (complex) eigenvectors are orthogonal, as long as you remember that in the first vector of a dot product, you must take complex conjugate, i.e. replace every $i$ by $-i$. Verify this for your antisymmetric matrix.

2. Analyze and accurately draw the quadratic curve
   
   $$- 2 x^2 + xy + 3 y^2 = 5$$
   
   
   using matrix diagonalization. Show the exact, as well as the approximate values for all angles. Repeat for the curve,
   
   $$1 x^2 + xy + 6 y^2 = 5$$
   
   
   Note: if you add say 3 times the unit matrix to a matrix $A$, then the eigenvectors of $A$ do not change. It only causes the eigenvalues to increase by 3, as you can readily verify from the definition of eigenvectors and eigenvalues. Use this to your advantage.

3. Given
   
   $$A = \left( \begin{array}{rrrr} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{array} \right)$$
   
   
   Without doing any mathematics, what can you say immediately about the eigenvalues and eigenvectors of this matrix? Now find the equation for the eigenvalues. It is a cubic one. However, one eigenvalue is immediately obvious from looking at $A$. What eigenvalue $\lambda_i$, that makes $A-\lambda_iI$ singular, is immediately obvious from looking at A? Explain. Factor out the corresponding factor $(\lambda-\lambda_i)$ from the cubic, then find the roots of the remaining quadratic. Number the single eigenvalue $\lambda_1$, and the double one $\lambda_2$ and $\lambda_3$. The found two basis vectors of the null space of $A-\lambda_2I$, call them $\vec e_2^{ *}$ and $\vec e_3^{ *}$, will not be orthogonal to each other. To make them orthogonal, you must eliminate the component that $\vec e_3^{ *}$ has in the direction of $\vec e_2^{ *}$. In particular, if $\vec e_2$ is the unit vector in the direction of $\vec e_2^{ *}$, then $\vec e_3^{ *}\cdot \vec e_2$ is the scalar component of $\vec e_3^{ *}$ in the direction of $\vec e_2^{ *}$. Multiply by the unit vector $\vec e_2$ to get the vector component, and substract it from $\vec e_3^{ *}$:
   
   $$\vec e_3^{ **} = \vec e_3^{ *} - (\vec e_3^{ *} \cdot \vec e_2)\vec e_2$$
   
   
   (This trick of making vectors orthogonal by substracting away the components in the wrong directions is called Gram-Schmidt orthogonalization.) Now make this vector of length 1. Then describe the transformation of basis that turns matrix $A$ into a diagonal one. What is the transformation matrix $P$ from old to new and what is its inverse? What is the diagonal matrix $A'$? Do your eigenvectors form a right or left-handed coordinate system?