Question

(c) Show directly that iC-Ax, then C(Dx) Dx) 7. Consider any A and a "Givens rotation" M in the 1-2 plane Cl cos θ -sinθ 0 A de fl , M |sin θ cos θ 0 Choose the rotation angle to produce zero in the (3, 1) entry of M- AM 336 Chapter 5 Eigenvalues and Eigenvectors Note. This "zeroing" is not so easy to continue, because the rotations that produce zero in place of d and h will spoil the new zero in the corner. We have to leave one diagonal below the main one, and finish the eigenvalue calculation in a different way Otherwise, if we could make A diagonal and see its eigenvalues, we would be finding the roots of the polynomial det(A -XI) by using only the square roots that determine cos θ-and that is impossible 8. What matrix M changes the basis V-(1,1), V2 (1,4) to the basis (2,5), v2 = (1,4)? The columns of M come from expressing Vi and V2 as combinations 2mijvi of the v's.

Answer 1

1) Given matrix is : A = and M = $\begin{bmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{bmatrix}$ .

Now, M^-1 = M^T   [Since here M is an orthogonal matrix]

i.e., M^-1 = $\begin{bmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{bmatrix}$ .

Then, M^-1AM = $\begin{bmatrix} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{bmatrix}$ $\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}$ $\begin{bmatrix} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{bmatrix}$

We need the (3,1) entry of M^-1AM. So we are not going to calculate three matrices. For (3,1) entry, we just calculate $\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$ $\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}$ $\begin{bmatrix} \cos\theta\\ \sin\theta\\ 0\end{bmatrix}$ .

Now, $\begin{bmatrix} 0 & 0 & 1 \end{bmatrix}$ $\begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i \end{bmatrix}$ $\begin{bmatrix} \cos\theta\\ \sin\theta\\ 0\end{bmatrix}$ = $\begin{bmatrix} g & h & i \end{bmatrix}$ $\begin{bmatrix} \cos\theta\\ \sin\theta\\ 0\end{bmatrix}$ = $g\cos\theta+h\sin\theta$

If both of g and h are non-zero, then (3,1) entry cannot be zero.

If h = 0 and g $\neq$ 0, then the entry becomes $g\cos\theta$ and it will be 0 if $\theta=\frac{n\pi}{2}$ , where n is odd.

If g = 0 and h $\neq$ =0, then the entry becomes $h\sin\theta$ and it will be 0 if $\theta=\frac{n\pi}{2}$ , where n is even.

If h = 0 and g = 0, then the entry becomes 0 and it does not depend on $\theta$ .

2) Given, V₁ = (1,1), V₂ = (1,4) and v₁ = (2,5), v₂ = (1,4).

Now, V₁ = (1,1) = 1*(2,5) + (-1)*(1,4) = 1*v₁+(-1)*v₂

And, V₂ = (1,4) = 0*(2,5) + 1*(1,4) = 0*v₁+1*v₂

Therefore, the required matrix is : M = $\begin{bmatrix} 1 & 0\\ -1 & 1 \end{bmatrix}$ .