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6

You know that $$2 = \dim \ker A + \dim {\rm col} \ A.$$ If $\dim {\rm col} \ A = 1$, our hypothesis, then $\dim \ker A = 1$, so we have non zero vectors in $\ker A = \ker(A - 0 \ {\rm Id})$. So yes, $0$ is an eigenvalue.


4

Remember that $\;\{v_1,...,v_r\}\;$ are lin. dependent iff there is $\;1\le i\le r\;$ s.t. $\;v_i\;$ lind. dep. on $\;v_1,...,v_{i-1}\;$, so let $\;i\;$ be the first such index for which this happens: $$(1)\;\;v_i=\sum_{k=1}^{i-1}a_kv_k\implies\lambda_iv_i=\sum_{k=1}^{i-1}a_k\lambda_iv_k$$ ...


4

Notice that if $D$ is a Hermitian matrix and $v$ is a vector then $v^\dagger D v\in\mathbb{R}$. Thus, if $Av=0$ then $0=v^\dagger Av=v^\dagger Bv+i(v^\dagger CC^\dagger v)$. Since $v^\dagger Bv \in \mathbb{R}$ and $v^\dagger CC^\dagger v\in \mathbb{R}$ then $v^\dagger Bv=v^\dagger CC^\dagger v=0$. But this implies that $C^\dagger v=0$, since ...


3

I will map it out and you can fill in the details. Approach 1: We are given: $$A = \begin{pmatrix} 0 & 1 & 2 \\ -5 &-3 & -7 \\ 1 & 0 & 0 \end{pmatrix} $$ Setting up and solving $$|A - \lambda I| = 0 \implies -(\lambda + 1)^3 = 0 \implies \lambda_{1,2,3} = -1$$ To find the eigenvectors, we would take $[A - \lambda_i I] = 0$, ...


3

Hint If $\lambda$ is an eigenvalue of $A$ associated to the eigenvector $v$, then $$A^mv = A^{m-1}Av = A^{m-1}\lambda v = \ldots = \lambda^m v,$$ i.e. $\lambda^m$ is an eigenvalue of $A^m$. Now, can you find the eigenvalues of $A$?


3

No problem. Just go back to the definition. $Ax = \lambda x$ Find a $\lambda$ and a nonzero x s.t. the above holds. There are in fact infinitely many since the matrix has a determinant of zero/is singular.


2

Condition B is impossible to fulfill. To see this, take the trace of the equation. As $tr(AB)=tr(BA)$, if follows that in order to fulfill B, the trace of $A$ has to be zero. Now, the trace is the sum of the eigenvalues. By the assumptions, all eigenvalues are non-negative, with at least one of them is positive. Hence, the trace of $A$ is positive, and a ...


2

Here is a slightly different proof than usual for fields $\mathbb{R}$ or $\mathbb{C}$: Suppose we order the eigenvalues so that $|\lambda_1| > |\lambda_2| > \cdots > |\lambda_r|$. Suppose $v=\sum_k \alpha_k v_k = 0$, where the $v_k$ correspond to the $\lambda_k$. Then $({1 \over \lambda_1} A)^m v = \sum_k \alpha_k ({\lambda_k \over \lambda_1})^m ...


2

We are given: $$X(t) = \begin{bmatrix} -3 & 0 & 2 \\ 1 & -1 & 0\\ -2 & -1 & 0\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 4 \\ -1\\ 2\end{bmatrix}$$ We can write this as: $$\tag 1 \begin{align} x' &= -3x + 2z \\ y' &= x-y \\ z' &= -2x - y \end{align}$$ Taking the Laplace ...


1

You have a system $\mathbf X'=\mathbf{AX}$. Call $\mathbf x=\mathscr L\{\mathbf X\}$. Then if you apply Laplace transform to the initial differential equations you will obtain $s\mathbf x-\mathbf x(0)=\mathbf{Ax}$. This implies that $$(s\mathbf I-\mathbf A)\mathbf x=\mathbf x(0)\Longrightarrow \mathbf x = (s\mathbf I-\mathbf A)^{-1}\mathbf ...


1

Here, I use $*$ to denote the conjugate transpose. Thoughts so-far: Lemma: for a matrix $M$ and vector $x$, $Mx = 0 \iff M^*Mx = 0$ Note that $$ \psi^*(B + iCC^*)\psi = (\psi^*B\psi) + i(\psi^*CC^*\psi) $$ Noting that $(\psi^*B\psi),(\psi^*CC^*\psi) \in \Bbb R$, we conclude that $$ \psi^* A \psi = 0 \iff (\psi^*B\psi) = (\psi^*CC^*\psi) = 0\\ ...


1

If $\lambda\ne0$ then the polynomial with simple roots $x^2-\lambda x=x(x-\lambda)$ annihilates $A$ and clearly $A\ne \lambda I_n$ and $A\ne0$ so $0$ and $\lambda$ are eigenvalues of $A$ and the multiplicity of $\lambda$ is $k$. If $\lambda=0$ then $A$ is nilpotent and $A$ in it's Jordan canonical form has a Jordan block with size $k$. Can you know now the ...


1

Obviously, the norm is at least what you anticipate by setting $f = e_n$ where $n$ almost realizes the supremum of $|\lambda_n - \lambda|^{-1}$. Conversely, write $f$ as $f = \sum_n c_n e_n$. Then $$\|(K - \lambda I)^{-1} f\|^2 = \sum_n |c_n|^2 |\lambda_n - \lambda|^{-2} \leq (\sup_n |\lambda_n - \lambda|^{-2}) (\sum_n |c_n|^2).$$ Taking the square root ...


1

The proof is equivalent to giving a counterexample of the statement. Note that this satisfies the logical properties of a proof. Because indeed it proves that the negation of the statement is true. To this end, see if you can form the logical negation of the statement (have you tried writing this statement mathematically, even?): The following is NOT true ...


1

Use the usual method $$\det (A - \lambda I) = 0$$ that is $$\begin{vmatrix}2-\lambda & 0 & 0\\ 0 &2- \lambda & 4 \\ 0 &-1 &2 - \lambda \end{vmatrix} = (2 - \lambda)\begin{vmatrix}2- \lambda & 4 \\ -1 &2 - \lambda \end{vmatrix} = (2 - \lambda)\Bigg((2 - \lambda)^2 + 4\Bigg) = 0$$ You can take it from here.


1

Let's say $Au=\lambda u$ and $Bu= \mu u$. Then $ABu=A(\mu u)=\lambda \mu u$. So, $u$ is an eigenvector of $AB$. Similar argument applies for $BA$.


1

For n = 7 consider jordan blocks $(3,2,2)$ and $(3,3,1)$ both these have algebraic and geometric multiplicity $3$, but are not similar. That is, consider matrices $A, B$ such that, $\chi_A = (x- \lambda)^3(x- \lambda)^3(x- \lambda)^1$ and $\chi_B = (x- \lambda)^3(x- \lambda)^2(x- \lambda)^2$ For $n > 7$ append rest to these jordan blocks as i.e. blocks ...


1

You'll have $|\langle v_1,v_2 \rangle | = 1$ if and only if they are parallel (i.e. $v_1 = a v_2$, $a \in \Bbb C$, $|a| = 1$). One direction is clear. On the other hand, suppose $v_1 \not\parallel v_2$. Then $v_1 = c_1 v_2 + c_2 v_2^\perp$ for some unit length vector $v_2^\perp$ that is orthogonal to $v_2$, and such that $|c_1|^2 + |c_2|^2 = 1$. Thus ...


1

There is a general method for finding sequences which satisfy linear recurrent relations with constant coefficients. In this case, we have the relation $$ x_{n+1}+x_{n-1} = \mu x_n \text{ for } n\geq 1. \text{ (*)} $$ Here is how we solve it: consider all the sequences (not nesessarily from $l^2$) which satisfy (*). They form a linear space (it's easy to ...


1

Yes, continuity should be guaranteed, as noted by Ilmari Karonen. But if you care about smoothness (which is what people usually want), this is a more complicated question. If the eigenvalues are distinct for all the $x$ you are interested in, then the eigenvalues are all smooth functions of $x$. However, if your eigenvalues can become degenerate for certain ...


1

The SVD of $A=U\Sigma V^{\star}$. Your matrix is $\Sigma^{\prime}=i\Sigma$. Then you can just write $$ A=-iU\Sigma^{\prime}V^{\star}. $$


1

Hints: The elements of the $W_i$ and $W_j$, $j \neq i$, span $v$ so you only have to show that those maps are equal in two cases: 1) when $v \in W_i$ and 2) when $v \in W_j$ for $j \neq i$. In both those cases you know exactly what $T$ and $E_i$ does to $v$ so you should be able to just plug into that formula and observe that both sides are equal.


1

It's enough to show the equality for each $\;w_i\in W_i\;,\;\;i=1,...,r\;$ : $$\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}\left(T-a_j I\right)(w_i)=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}\left(Tw_i-a_jw_i\right)=$$ $$=\frac1{\prod\limits_{j\neq i}(a_i-a_j)}\prod_{j\neq i}(a_iw_i-a_jw_i)=\frac1{\prod\limits_{j\neq ...


1

$Ax=\lambda x\Rightarrow (pId+qA)x=px+qAx=px+q\lambda x=(p+q\lambda)x$


1

The options you've chosen to be incorrect are the right ones, but for the wrong reason in 5. You have to consider the eigenvalues of $T$, the matrix transformation, not the eigenvalues of the matrices it acts on. To prove 6 wrong, you therefore have to prove that $T(A) = 2A$ has no non-zero solutions. Edit: The reasoning for 1 happens to be wrong also, for ...


1

This is a kind of question which makes us confused easily. First, what is $T?$ It is a transformation from a vector space to a vector space. Since the space is of finite dimension, $T$ can be written as a matrix and we can talk about its determinant and everything. Now we have two kinds of matrices here. One is 2x2 matrices on which $T$ maps, the other is ...


1

A basis of the vector space of symmetric $2\times2$ matrices is $$ \mathscr{B}= \left\{ A_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}; A_2=\begin{bmatrix}0&1\\1&0\end{bmatrix}; A_3=\begin{bmatrix}0&0\\0&1\end{bmatrix} \right\} $$ If $T$ is linear, then its matrix with respect to $\mathscr{B}$ is given by $$ X= \begin{bmatrix} ...



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