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Problem 1(5):¶
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.
Solution:¶
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:
$H(v) = \hat{1} - 2vv^T = (h_{ij}), 1\leq{i,j}\leq{n}$, where $h_{ij} = \begin{equation} \begin{cases} 1-2v_{i}^2, i = j\\ -2v_{i}v_{j}, i \ne j \end{cases}. \end{equation}$
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:
$HH^T = H^2 = (\hat{1} - 2vv^T)^2 = \hat{1} - 4vv^T + 4(vv^T)(vv^T) \overset{v^Tv=1} = \hat{1} - 4vv^T + 4vv^T = \hat{1}$.
It follows that the matrix is orthoganal. It also follows that the $rk(H) = n$.
Problem 2(10)¶
Prove the following inequalities and provide examples of x and A when they turn into equalities:
- $∥x∥_{2} ≤ \sqrt{m}∥x∥_{∞}$
- $∥A∥_{∞} ≤ \sqrt{n}∥A∥_{2}$
where $x$ is a vector of $m$ components and $A$ is $m × n$ matrix.
1) By definition we have:
$\parallel x∥_{2} = \sqrt{\sum_{n=1}^{m} x_i^2}$ and $\parallel x∥_{∞} = \underset{1 ≤ i ≤ m}\max{|x_i|}$
Then the original inequality can be rewritten in the form:
$\sqrt{\sum_{n=1}^{m} x_i^2} ≤ \sqrt{m}\underset{1 ≤ i ≤ m}\max{|x_i|}$
Let's square both sides:
$\sum_{n=1}^{m} x_i^2 ≤ m (\underset{1 ≤ i ≤ m}\max{|x_i|})^2 ⇒ \sum_{n=1}^{m} x_i^2 ≤ m \underset{1 ≤ i ≤ m}\max{x_i^2} ⇔ \sum_{n=1}^{m} x_i^2 ≤ \sum_{n=1}^{m} \underset{1 ≤ i ≤ m}\max{x_i^2}$
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 $m = 1$.
Problem 3(5):¶
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 $\alpha$.
Solution:¶
1) Matrix $A$ is singular if there is a $m$-vector $x \ne 0$ such that:
$(1 + uv^T)x = 0 ⇔ x + uv^Tx = 0 ⇔ uv^Tx = -x.$
Note that $v^Tx$ is a scalar, so the matrix $A$ is singular if $x = \beta{}u$, $\beta{} \ne 0$.
From here we can see:
$uv^Tx = -x ⇔ \beta{}uv^Tu = -\beta{}u ⇔ v^Tu = -1$.
Thus, matrix $A$ is singular if $v^Tu = -1$
2) Assume that matrix $A$ has an inverse of the form $A^{-1} = 1 + αuv^T$.
$AA^{-1} = 1 ⇔ (1 + uv^T)(1 + αuv^T) = 1 ⇔ αuv^T + uv^T + αuv^Tuv^T = O$.
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:
$αuv^T + uv^T + αuv^Tuv^T = O ⇔ αuv^T + uv^T + αγuv^T = O ⇔ (α + 1 + αγ)uv^T = O ⇔ α + 1 + αγ = 0 ⇔ α = -\frac{1}{1 + γ} = -\frac{1}{1 + v^Tu}$.
Thus, the matrix $A$ has the inverse of the form $A^{-1} = 1 + αuv^T$ and $α = -\frac{1}{1 + v^Tu}, v^Tu \ne -1$
Problem 4(5):¶
Prove that for any unitary matrix $U$ one has $\parallel UA \parallel_{F} = \parallel AU\parallel_{F} = \parallel A \parallel_{F}$.
Solution:¶
Let's write down the expression by definition:
$\parallel UA \parallel_{F} = \sqrt{tr[(UA)^*(UA)]} = \sqrt{tr[A^*U^*UA]}$.
Let us take advantage of the fact that $U$ is unitary, i.e. $U^*U = UU^*= I$. That:
$\parallel UA \parallel_{F} = \sqrt{tr[A^*U^*UA]} = \sqrt{tr[A^*A]} = \parallel A \parallel_{F}$
Similarly for $AU$ we get:
$\parallel AU \parallel_{F} = \sqrt{tr[(AU)^*(AU)]} = \sqrt{tr[U^*A^*AU]}$.
Let's use the trace property $tr(AB) = tr(BA)$. So: $\parallel AU \parallel_{F} = \sqrt{tr[U^*A^*AU]} = \sqrt{tr[AUU^*A^*]} = \sqrt{tr[AA^*]} = \parallel A \parallel_{F}$.
Thus $\parallel UA \parallel_{F} = \parallel AU\parallel_{F} = \parallel A \parallel_{F}$.