"Consider the matrix $H(v) = \\hat{1} - 2vv^T$, where $v$ is a unit column vector. What is the rank of the matrix $H(v)$? Prove that it is orthogonal.\n"
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"### **Solution:**\n",
"\n",
"Let we have a vector $v = (v_{1} v_{2} ... v_{n})^T$ that, by condition, is unit, i.e. $v^Tv=1$.Then the matrix $\\underset{n\\times{}n}H$ has the following form:\n",
"It can be seen that the matrix $H$ is symmetrical. Let us prove that it is orthogonal, which also means that the equality $HH^T = \\hat{1}$. On the other hand, due to symmetry:\n",
"It follows that the matrix is orthoganal. It also follows that the $rk(H) = n$.\n",
"\n"
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"### **Problem 2(10)** \n",
"Prove the following inequalities and provide examples of x and A when they turn into equalities:\n",
"\n",
"* $∥x∥_{2} ≤ \\sqrt{m}∥x∥_{∞}$\n",
"* $∥A∥_{∞} ≤ \\sqrt{n}∥A∥_{2}$\n",
"\n",
"where $x$ is a vector of $m$ components and $A$ is $m × n$ matrix.\n",
"\n"
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"**1)** By definition we have: \n",
"$\\parallel x∥_{2} = \\sqrt{\\sum_{n=1}^{m} x_i^2}$ and $\\parallel x∥_{∞} = \\underset{1 ≤ i ≤ m}\\max{|x_i|}$ \n",
"Then the original inequality can be rewritten in the form: \n",
"$\\sqrt{\\sum_{n=1}^{m} x_i^2} ≤ \\sqrt{m}\\underset{1 ≤ i ≤ m}\\max{|x_i|}$ \n",
"Let's square both sides: \n",
"$\\sum_{n=1}^{m} x_i^2 ≤ m (\\underset{1 ≤ i ≤ m}\\max{|x_i|})^2 ⇒ \\sum_{n=1}^{m} x_i^2 ≤ m \\underset{1 ≤ i ≤ m}\\max{x_i^2} ⇔ \\sum_{n=1}^{m} x_i^2 ≤ \\sum_{n=1}^{m} \\underset{1 ≤ i ≤ m}\\max{x_i^2}$ \n",
"Hence its validity is obvious and it is clear that equality is achieved for arbitrary $m ∈ ℕ$ if $x_{1} = x_{2} = ... = x_{m}$, otherwise if $m = 1$."
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"### **Problem 3(5)**:\n",
"\n",
"Assuming $u$ and $v$ are $m$-vectors, consider the matrix $A = 1 + uv^T$ which is a rank-one perturbation of identity. Can it be singular? Assuming it is not, compute its inverse. You may look for it in a form of $A^{-1} = 1 + \\alpha{}uv^T$ for some scalar $\\alpha$ and evaluate $\\alpha$.\n"
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"### **Solution:** \n",
"**1)** Matrix $A$ is singular if there is a $m$-vector $x \\ne 0$ such that: \n",
"Since $v^Tu$ is a scalar, we denote it as $γ \\ne -1$ (otherwise the matrix will be singular). Then the resulting equality will be written in the form: \n",
