Worksheet 1-2 (with Solutions)

Worksheet 1-2 (with Solutions)#

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Worksheet 1-2: Q1#

For the matrix \(A=\begin{bmatrix} 1 & -2 \\ 3 & 4 \end{bmatrix}\), find

  1. The induced \(1\)-norm, \(\|A\|_1\).

    Solution

    The induced 1-norm is the maximum absolute column sum:

    \[\|A\|_1 = \max_j \sum_i |A_{ij}| = \max(|1|+|3|,\; |-2|+|4|) = \max(4, 6) = 6\]

  2. The induced \(\infty\)-norm, \(\|A\|_\infty\).

    Solution

    The induced \(\infty\)-norm is the maximum absolute row sum:

    \[\|A\|_\infty = \max_i \sum_j |A_{ij}| = \max(|1|+|-2|,\; |3|+|4|) = \max(3, 7) = 7\]

  3. The Frobenius norm, \(\|A\|_F\).

    Solution

    The Frobenius norm is the entrywise 2-norm:

    \[\|A\|_F = \sqrt{\sum_{ij} A_{ij}^2} = \sqrt{1^2 + 2^2 + 3^2 + 4^2} = \sqrt{30}\]

Worksheet 1-2: Q2#

We are going to determine the induced 2-norm for the matrix \(A=\begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix}\).

  1. Compute \(A^TA\).

    Solution
    \[\begin{split}A^T A = \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 1+4 & -2-2 \\ -2-2 & 4+1 \end{bmatrix} = \begin{bmatrix} 5 & -4 \\ -4 & 5 \end{bmatrix}\end{split}\]

  2. Find the eigenvalues of \(A^TA\). We generally sort these in decreasing order so that \(\lambda_1 \geq \lambda_2\).

    Solution
    \[\begin{split}\det(A^T A - \lambda I) = \det\begin{bmatrix} 5-\lambda & -4 \\ -4 & 5-\lambda \end{bmatrix} = (5-\lambda)^2 - 16 = \lambda^2 - 10\lambda + 9 = (\lambda - 9)(\lambda - 1)\end{split}\]

    So the eigenvalues are \(\lambda_1 = 9\) and \(\lambda_2 = 1\).

  3. Find the singular values of \(A\).

    Solution

    The singular values are the square roots of the eigenvalues of \(A^TA\):

    \[\sigma_1 = \sqrt{\lambda_1} = 3, \qquad \sigma_2 = \sqrt{\lambda_2} = 1\]

  4. Compute the induced 2-norm of \(A\).

    Solution

    The induced 2-norm is the largest singular value of \(A\):

    \[\|A\|_2 = \sigma_1 = 3\]

Worksheet 1-2: Q3#

  1. Show that for \(A \in \mathbb{R}^{m \times n}\), if \(\lambda\) is a non-zero eigenvalue of \(A^TA\) with eigenvector \(v\), then \(\lambda\) is a non-zero eigenvalue of \(AA^T\) with eigenvector \(Av\).

    Solution

    If \(\lambda \neq 0\) with eigenvector \(v\), then \(A^T A v = \lambda v\). Multiply both sides by \(A\):

    \[AA^T (A v) = \lambda(Av)\]

    This is exactly what is needed to show that \(\lambda\) is an eigenvalue of \(AA^T\) with eigenvector \(Av\).

  2. Use the previous statement to show that for \(A \in \mathbb{R}^{m \times n}\), \(\|A\| = \|A^T\|\) for the spectral norm (AKA induced 2-norm).

    Solution

    By the previous question, the non-zero eigenvalues of \(A^TA\) and \(AA^T\) are the same. Thus, the singular values of \(A\) and \(A^T\) are the same (they are the square roots of those eigenvalues). Therefore, the induced 2-norms are equal: \(\|A\|_2 = \|A^T\|_2\), since they are each the largest singular value.

  3. Show that \(\|A\|_F^2 = \sum_{i=1}^n \lambda_i(A^TA)\). (Hint: Consider the trace of \(A^TA\).)

    Solution

    Entry \((j,j)\) of \(A^TA\) is \(\sum_{i=1}^m a_{ij}^2\), so

    \[\|A\|_F^2 = \sum_{i=1}^m \sum_{j=1}^n a_{ij}^2 = \sum_{j=1}^n (A^TA)_{jj} = \operatorname{trace}(A^TA)\]

    Since the trace of a matrix equals the sum of its eigenvalues, \(\|A\|_F^2 = \sum_{i=1}^n \lambda_i(A^TA)\).

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