Solve with Instructors
Week - 5
Arun Prakash A
The Length of the complex vector \( u = \begin{bmatrix} 1+i \\ 1-i \\1 \end{bmatrix}\)
||x||^2 = x^*x= |x_1|^2+|x_2|^2+\cdots+|x_n|^2
||u|| =\sqrt{ u^*u}=\sqrt{ |u_1|^2+|u_2|^2+\cdots+|u_n|^2}
||u|| =\sqrt{ u^*u}=\sqrt{ 2+2+1}=2.23
The Length of a complex vector and its conjugate remains same. True or False?
The Length of the complex vector \( u = \begin{bmatrix} 1+i \\ 1-i \\1 \end{bmatrix}\) scaled by the scalar \( \eta = 4+3i \) is?
||\eta u|| =\sqrt{ (\eta u)^*(\eta u)}=|\eta| \ \ ||u||
=5*2.23
=11.15
Check whether the two vectors \(x = \begin{bmatrix} i \\ 3 \\1 \end{bmatrix} \)
and \(y = \begin{bmatrix} i \\ 0 \\1 \end{bmatrix} \) are orthogonal?
Find the unit vector that points in the opposite direction of the vector \(x = \begin{bmatrix} 1-i \\ 3+i \\2 \\1+i \end{bmatrix} \)
Identify non-Hermitian matrix from the following list of matrices and justify why it is not Hermitian?
- \(A = \begin{bmatrix} 3 & i & 2+i\\ -i & -2 & -7\\2-i & 7*\cos(\pi)&1 \end{bmatrix} \)
2.\(B = \begin{bmatrix} 1 & i \\ -i & -i*i \end{bmatrix} \)
Every diagonal matrix is Hermitian.
True or False
Every real diagonal matrix is Hermitian.
True or False
3. \(C = \begin{bmatrix} -2 & 1-2i \\ 0 & 3 \end{bmatrix} \)
4. \(CC^*\) and \(C^*C\) are Hermitian?
The statement that the singular values are always greater than or equal to zero is?True or False
Find the SVD for the matrix \( A = \begin{bmatrix} 1& 2 \\ 3&6 \end{bmatrix}\).
1. Find \( A^TA\)
\begin{bmatrix} 10 & 20 \\ 20 & 40 \end{bmatrix}
2. Eigenvalues of \( A^TA\)
\( \lambda^2-50\lambda=0\)
\( \lambda_1 =50, \lambda_2 = 0\)
\therefore, \Sigma = \begin{bmatrix} \sqrt{50} & 0 \\ 0 & 0 \end{bmatrix}
3. Eigenvectors of \( A^TA\)
\(\lambda_1 = 50\)
N(A^TA-\lambda_1I)=0
\(x_1 = \begin{bmatrix}0.5\\1 \end{bmatrix}\)
\(x_2 = \begin{bmatrix}-2\\1 \end{bmatrix}\)
4. Normalize the Eigenvectors of \( A^TA\) and construct \(Q_2^T\)
Q_2=\begin{bmatrix} 0.447 & -0.894 \\ 0.894 & 0.447 \end{bmatrix}
5. To find \(Q_1\), either find Eigenvectors of \( AA^T\) or use the relation \(y_1 = \frac{1}{\sigma_1}A*x_1\)
\(y_1 = \frac{1}{\sigma_1}A*x_1\)
y_1 = \frac{1}{\sqrt{50} }\begin{bmatrix} 1& 2 \\ 3&6 \end{bmatrix}*\begin{bmatrix}0.447\\0.894 \end{bmatrix}
y_1 = \begin{bmatrix}0.3162 \\ 0.9487\end{bmatrix}
\(y_2 = \frac{1}{\sigma_2}A*x_2\)
However, \(\sigma_2 = 0\)?
y_2 \perp y1
\therefore y_2 =\begin{bmatrix}-0.9487 \\ 0.3162\end{bmatrix}
\because [x,y]^T*[-y*x]=0
What is the geometrical meaning of \(\sigma_2 = 0\)?
Q_1=\begin{bmatrix} 0.3162 & -0.9487 \\ 0.9487 & 0.3162 \end{bmatrix}
\begin{bmatrix} 1& 2 \\ 3&6 \end{bmatrix} = \begin{bmatrix} 0.3162 & -0.9487 \\ 0.9487 & 0.3162 \end{bmatrix} \begin{bmatrix} \sqrt{50} & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0.447 & 0.894 \\ -0.894 & 0.447 \end{bmatrix}
\begin{bmatrix}
0.9994 & 1.9989 \\
2.9986 & 5.9972
\end{bmatrix}
Q_2=\begin{bmatrix} 0.447 & -0.894 \\ 0.894 & 0.447 \end{bmatrix}
MLF_SWI_5
By Arun Prakash
