Question

Let \(A=\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ -1 & 1 \end{pmatrix}\)and let A^T denote the transpose of A. Let \(u=\begin{pmatrix} u_1 \\ u_2 \end{pmatrix}\) and \(v=\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}\)be column vectors with entries in \(\R\) such that \(u^2_1+u_2^2=1\) and \(v_1^2+v_2^2+v^2_3=1\). Suppose
\(Au=\sqrt2v\) and \(A^Tv=\sqrt2u.\)
Then |\(𝑢1 + 2 \sqrt2 𝑣_1\)| is equal to _________. (Rounded off to two decimal places)

Answer 1

Correct Answer: 3

We are given

\[A=\begin{pmatrix} 1 & 1 \\[6pt] 0 & 1 \\[6pt] -1 & 1 \end{pmatrix},\qquad u=\begin{pmatrix} u_1 \\ u_2 \end{pmatrix},\qquad v=\begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix},\]

with the conditions

\[u_1^2+u_2^2 = 1,\qquad v_1^2 + v_2^2 + v_3^2 = 1,\]

\[Au = \sqrt{2}\, v,\qquad A^T v = \sqrt{2}\, u.\]

Step 1: Compute \(A^T A\)

\[A=\begin{pmatrix} 1 & 1 \\ 0 & 1 \\ -1 & 1 \end{pmatrix}\]

\[A^T A = \begin{pmatrix} 1^2 + 0^2 + (-1)^2 & 1\cdot 1 + 0\cdot 1 + (-1)\cdot 1 \\[4pt] 1\cdot 1 + 1\cdot 0 + 1\cdot (-1) & 1^2 + 1^2 + 1^2 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}.\]

Thus the eigenvalues are

\[\lambda_1 = 2,\qquad \lambda_2 = 3,\]

so the singular values are

\[\sigma_1 = \sqrt{2},\qquad \sigma_2 = \sqrt{3}.\]

Since the problem specifies \(\sigma=\sqrt{2}\), the associated eigenvector of \(A^T A\) is

\[u = (\pm 1,\, 0)^T.\]

Step 2: Compute the corresponding \(v\)

\[Au = \pm \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}.\]

Since \(Au = \sqrt{2}\, v\),

\[v = \pm \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}.\]

Hence,

\[v_1 = \pm \frac{1}{\sqrt{2}},\]

with the same sign as \(u_1\).

Step 3: Compute the required expression

\[\left| u_1 + 2\sqrt{2}\, v_1 \right|.\]

Substitute \(u_1 = \pm 1\), \(v_1 = \pm \frac{1}{\sqrt{2}}\) with matching signs:

\[u_1 + 2\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = u_1 + 2.\]

If \(u_1 = 1\): \(1 + 2 = 3\)

If \(u_1 = -1\): \(-1 - 2 = -3\), whose absolute value is \(3\)

Thus,

\[\boxed{3.00}\]